IV 1 2 3 4 5 6 7 8
1 X X X X X X X X
2 X X X X X X X X
3 X X X X X X X X
4 X X X X X X X X
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 X X X X X X X X X X X X X X
X X X X X X X X X
1
X
X
I.1.1 Identify and sketch the set of (a) jz 1 ij = 1 (f) (b) 1 < j2z 6j < 2 (g) (c) jz 1j2 + jz + 1j2 < 8 (h) (d) jz 1j + jz + 1j 2 (i) (e) jz 1j < jzj (j)
points satisfying. 0 < Im z < < Re z < jRe zj < jzj Re (iz + 2) > 0 jz ij2 + jz + ij2 < 2
Solution Let z = x + iy, where x; y 2 R. (a) Circle, centre 1 + i, radius 1. jz
1
ij = 1 , j(x
1)j = 1 , (x
1) + i (y
1)2 + (y
1)2 = 12
(b) Annulus with centre 3, inner radius 1=2, outer radius 1.
1 < j2z
6j < 2 , 1 < 2 jz
3j < 2 ,
jx + iy
1j2 + jx + iy + 1j2 < 8 ,
3j < 1 , (1=2)2 < (x
, 1=2 < jz p (c) Disk, centre 0, radius 3.
, (x
3)2 + y 2 < 12
1)2 + y 2 + (x + 1)2 + y 2 < 8 , x2 + y 2 <
p
2
3
(d) Interval [ 1; 1].
jz
q 2 , (x
1j + jz + 1j q , (x
q , (x + 1)2 + y 2
2
1) +
2
1) + 2
y2
x+1,
y2 2
q
q
2 (x + 1)2 + y 2 , q 2 2 2 (x + 1) + y , 2
2
(x + 1) +
y2
(x + 1)2 , y = 0
Now, take y = 0 in the inequality, and compute the three intervals 2
x < 1; 1 x x > 1;
then jx 1 then jx then jx
1j + jx + 1j = (x 1) (x + 1) = 2x 2; 1j + jx + 1j = (x 1) + (x + 1) = 2 2 1j + jx + 1j = (x 1) + (x + 1) = 2x 2:
(e) Half–plane x > 1=2.
jz
1j < jzj , jz
1j2 < jzj2 , jx + iy , (x
1j2 < jx + iyj2 ,
1)2 + y 2 < x2 + y 2 , x > 1=2
(f) Horizontal strip, 0 < y < . (g) Vertical strip,
0 (i) Half plane y < 2. Re (iz + 2) > 0 , Re (i (x + iy) + 2) > 0 ,
y+2>0,y <2
(j) Empty set.
jz
ij2 + jz + ij2 < 2 , jx + iy , x2 + (y
ij2 + jx + iy + ij2 < 2 ,
1)2 + x2 + (y + 1)2 < 2 , x2 + y 2 < 0
3
I.1.2 Verify from the de…nitions each of the identities (a) z + w = z + w (b) zw = z w (c) jzj = jzj Draw sketches to illustrate (a) and (c).
(d) jzj2 = z z
Solution Substitute z = x + iy and w = u + iv, and use the de…nitions. (a)
z + w = (x + iy) + (u + iv) = (x + u) + (y + v) i = = (x + u) (y + v) i = (x iy) + (u
iv) = z + w:
(b)
zw = (x + iy) (u + iv) = (xu yv) + (xv + yu) i = = (xu yv) (xv + yu) i = (x
iy) (u
iv) = z w:
(c)
jzj = x + iy = jx (d)
jzj2 = jx + iyj2 =
iyj =
q
p x2 + y 2
x2 + ( y)2 =
2
p
x2 + y 2 = jx + iyj = jzj :
= x2 + y 2 = = x2
4
i2 y 2 = (x + iy) (x
iy) = z z:
I.1.3 Show that the equation jzj2 centered at a with radius . Solution Let z = x + iy and a = jzj2
= x2 + y 2
=x +y
2
2
represents a circle
+ i , we have
2 Re (az) + jaj2 = 2
2 Re (az) + jaj2 =
2 Re ((
i ) (x + iy)) +
2 Re (( x + y) + i ( y 2
=x +y
2
2 ( x + y) +
2
x)) + 2
+
2
2 Re (az) + jaj2 = (x
)2 + (y
2
5
=
+
2
=
)2 + (y
)2 :
, becomes
)2 =
which is the equation for a circle of radius complex notation is the point a = + i .
2
=
= (x Thus the equation jzj2
2
+
2
;
centered at ( ; ), which in
I.1.4 Show that jzj jRe zj + jIm zj, and sketch the set of points for which equality holds.
Solution Apply triangle inequality to z = Re z + i Im z, to obtain jzj 6 jRe zj + jIm zj. Now set z = x + iy, and see then equality holds p p p p p p 2 x2 + y 2 = x2 + y 2 jzj = jRe zj+jIm zj , x2 + y 2 = x2 + y 2 , p p p p 2 2 x2 + x2 +2 x2 y 2 , x2 y 2 = 0 , x = 0 or y = 0: , x2 +y 2 =
Equality holds only when z is real or pure imaginary, which are all the points on the real and imaginary axis.
6
2
,
I.1.5 Show that jRe zj
jzj ;
jIm zj
jzj :
Show that jz + wj2 = jzj2 + jwj2 + 2 Re (z w) : Use this to prove the triangle inequality jz + wj
jzj + jwj.
Solution Let z = x + iy. Then since the square root function is monotone, we have p p jRe zj = jxj = x2 6 x2 + y 2 = jzj ; p p jIm zj = jyj = y 2 6 x2 + y 2 = jzj :
Now for z; w we have jz + wj2 =
= (z + w) (z + w) = (z + w) (z + w) = z z + z w + wz + ww = = jzj2 + 2 Re (z w) + jwj2 ;
where we have used the fact that 2 Re (z w) = z w +z w = z w + wz = z w +wz. We use both of the above facts and the trivial identities jzj = jzj and jzwj = jzj jwj to prove the triangle inequality for z; w. We have jz + wj2 = jzj2 + 2 Re (z w) + jwj2
jzj2 + 2 jRe (z w)j + jwj2
jzj2 + 2 jz wj + jwj2 = jzj2 + 2 jzj jwj + jwj2 = (jzj + jwj)2 :
The desired inequality now follows by taking square root of both sides.
7
I.1.6 For …xed a 2 C, show that jz
aj = j1
azj = 1 if jzj = 1 and 1 az 6= 0.
Solution If jzj = 1, then jzj = 1 and z z = 1. Use this and get jz
aj = jz
aj jzj = jz z
azj = j1
We have
as was to be shown.
jz j1
aj = 1; azj
8
azj = j1
azj :
I.1.7 Fix > 0, = 6 1, and …x z0 ; z1 2 C. Show that the set of z satisfying jz z0 j = jz z1 j is a circle. Sketch it for = 12 and = 2, with z0 = 0 and z1 = 1. What happens when = 1? Solution Recall that a circle in R2 centered at (a; b) with radius r is given by the equation a)2 + (y
(x
b)2 = r2 :
We manipulate the equation jz
z0 j = jz
z1 j :
The solutions set of the equation above remains the same if we square both sides, z0 j2 =
jz
2
z1 j2 :
jz
Let z = x + iy, z0 = x0 + iy0 and z1 = x1 + iy1 . Thus our equation becomes x0 )2 + (y
(x
y0 )2 =
2
x1 )2 + (y
(x
y1 )2 :
Expanding the squares, and grouping terms, we have 1
2
x2 2 x0
2
x1 x+ x20 2
Dividing both sides by (1
x2
2
(x0 1
2
x1 )
2
x+
(x20 1
2 2 x1
+ 1
2
y 2 2 y0
2
y1 y+ y02
), we have 2 2 x1 ) 2
+ y2
2
(y0 1
2
y1 )
2
y+
(y02 1
2 2 y1 ) 2
Now complete the squares for both the x and y terms. Recall that x2
2ax + b = (x
So we have
9
a)2
a2 + b:
=0
2 2 y1
=0
x
2
x0 1
2
x1
+
2
y0 1
+ y
2
(1
) (x20 (1
2
2
y1
+
2
2 2 x1 ) 2 )2 2
(1
2
(x0
) (y02 (1
2
x1 )
2 2 y1 ) 2 )2
+ 2
(y0
2
y1 )
= 0:
This becomes
x =
x0 1
2
(x0
2
2
x1
+ y
2
2
y0 1
2
2
x1 ) + (y0
2
y1
=
2 2
y1 )
2
(1
) (x20 + y02
2
(x21 + y12 ))
2 )2
(1
2
=
x0 )2 + (y1
(x1
(1
2 )2
=
y0 )2
;
which is the equation for a circle. If z0 = 0, and z1 = 1, we have 2
2
x
2 2
2
1
+y =
1
2
:
1 ; 0 , and when = 2, When = 21 we have a circle of radius 23 centered at 3 4 2 we have a circle of radius 3 centered at 3 ; 0 . When = 1, we have the equation
jz
z0 j = jz
z1 j ;
which is the line bisecting the two points. When z0 = 0; z1 = 1, this is the line Re z = 21 .
10
I.1.7
I.1.7
Im z
2
I.1.7
Im z
1
-1
2
-1
Re z
1 2
2
-1
Re z
-2
=2
11
1 -1
-2
=1
2 1
1 -1
-2
=
Im z
1
1 -1
2
2
Re z
I.1.8 Let p (z) be a polynomial of degree n 1 and let z0 2 C. Show that there is a polynomial h (z) of degree n 1 such that p (z) = (z z0 ) h (z)+p (z0 ). In particular, if p (z0 ) = 0, then p (z) = (z z0 ) h (z). Solution Set p (z) = an z n + an 1 z n
1
+ an 2 z n
2
+
+ bn 3 z n
3
+ a2 z 2 + a1 z + a0 :
and h (z) = bn 1 z n
1
+ bn 2 z n
2
+ : : : + b2 z 2 + b1 z + b0 :
We equate coe¢ cients in the polynomial identity p (z) = (z p (z0 ), and get an z n + an 1 z n
1
+ an 2 z n
2
z0 ) h (z) +
+ a2 z 2 + a1 z + a0 =
+
= (z z0 ) bn 1 z n 1 + bn 2 z n 2 + bn 3 z n 3 + : : : + b2 z 2 + b1 z + b0 +p (z0 ) = = bn 1 z n + bn 2 z n 1 + bn 3 z n 2 + : : : + b2 z 3 + b1 z 2 + b0 z + p (z0 ) bn 1 z0 z n 1 bn 2 z0 z n 2 bn 3 z0 z n 3 : : : b2 z0 z 2 b1 z0 z b0 z0 = bn 1 z n +(bn 2 bn 1 z0 ) z n 1 +(bn 3 bn 2 z0 ) z n 2 +(bn 4 bn 3 z0 ) z n 3 + + (b2 b3 z0 ) z 3 + (b1 b2 z0 ) z 2 + (b0 b1 z0 ) z + p (z0 ) b0 z0 : Equate and solve for the bj ´s in terms of aj ´s. 8 < an = b n 1 ak = bk 1 bk z0 ; 0 : a0 = p (z0 ) b0 z0
k
n
8 b n 1 = an > > > nP k 1 > < bk = ak+1+i z0i ; 0 1 ) i=0 > n > P > > ai z0i : p (z0 ) = i=0
Proof by induction on degree n of p (z), set p (z) = an z n + an 1 z n where an 6= 0. 12
1
: : : + a0 ;
k
n
2
Fix z0 and write z0 ) z n
p (z) = an (z
1
+ r (z) ;
where deg r (z) n 1. By using the induction hypothesis, we can assume that r (z) = q (z) (z where deg q (z)
z0 ) + c;
deg r (z). Then
p (z) = an z n
1
+ q (z) (z
Since deg q (z) n 2, deg r (z) Plug in z0 , get p (z0 ) = c.
z0 ) + c = h (z) (z
n
13
1.
z0 ) + c
I.1.9 Find the polynomial h (z) in the preceding exercise for the following choices of p (z) and z0 (a) p (z) = z 2 and z0 = i (b) p (z) = z 3 + z 2 + z and z0 = 1 (c) p (z) = 1 + z + z 2 + + z m and z0 = 1 Solution From the preceding exercise we have p (z) = (z
z0 ) h (z) + p (z0 ) :
We solve for h (z) then h (z) =
p (z) z
p (z0 ) : z0
(a) We have that p (z) = z 2 and z0 = i, thus p (z0 ) = p (i) = Thus p (z) h (z) = z
1.
p (z0 ) z2 + 1 = = z + i; z0 z i
and z 2 = (z
i) (z + i)
(b) We have that p (z) = z 3 + z 2 + z and z0 = Thus
h (z) =
p (z) z
1:
1, thus p (z0 ) = p ( 1) =
z3 + z2 + z + 1 p (z0 ) (z + 1) (z 2 + 1) = = = z 2 + 1; z0 z+1 z+1
and z 3 + z 2 + z = (z + 1) z 2 + 1 (c) We have that p (z) = 1 + z + z 2 +
+ z m and z0 = 14
1: 1, thus
1.
0; m odd 1; m even
p (z0 ) = p ( 1) = Thus
h (z) = 8 < (zm = : (zm
p (z) z 1 +z m 3
p (z0 ) = z0 +z 2 +1)(z+1)
= zm
z+1 +z 3 +z )(z+1)+1 z+1
1 +z m 3
1
= zm
+ zm 1
3
+
+ zm
3
+ z2 + 1
if m is odd
+ : : : + z 3 + z if m is even
and
zm + zm
1
+ =
z2 + z + 1 = (z + 1) (z m (z + 1) (z m
+ zm 1 + zm 1
3
+ 3 +
15
+ z 2 + 1) if m is odd 3 + z + z) + 1 if m is even
I.1.10 Let q (z) be a polynomial of degree m p (z) can be expressed in the form
1. Show that any polynomial
p (z) = h (z) q (z) + r (z) ; where h (z) and r (z) are polynomials and the degree of the remainder r (z) is strictly less than m. Hint. Proceed by induction on the degree of p (z). The resulting method is called the division algorithm. Solution First suppose p (z) is the zero polynomial. (So, the degree of p (z) is 1.) The degree of r (z) must be less than the degree of q (z). If h (z) 6= 0, it follows that the degree of h (z) q (z) is greater than the degree of r (z). This then implies that h (z) q (z)+r (z) 6= 0. So, h (z) q (z)+r (z) = 0 implies that h (z) = 0, and thus r (z) = 0. So, the polynomials are h (z) = 0 and r (z) = 0, and these polynomials are the only ones that satisfy both conditions. Now assume that the division algorithm is true for all polynomials p (z) of degree less than n. (Where n 0.) If the degree of q (z) is greater than the degree of p (z), and h (z) is nonzero, then h (z) q (z) + r (z) has degree greater than p (z). So, if the degree of q (z) is greater than the degree of p (z), then h (z) = 0 and thus r (z) = p (z). This proves both existence and uniqueness of h (z) and r (z), in this case. Now, suppose that the degree of q (z) is less than or equal to the degree of p (z). Set p (z) = an z n + an 1 z n
1
+
+ a1 z + a0
+
+ b1 z + b0 ;
and q (z) = bm z m + bm 1 z m where an 6= 0, bm 6= 0 and m Let p1 (z) =
1
n. an n z bm
m
16
q (z)
p (z) ;
then p1 (z) =
an n z bm
m
bm z m + bm 1 z m
1
+
+ b1 z + b0
an z n + an 1 z n
1
+
The monomials of degree n cancel, and therefore p1 (z) is a polynomial of degree at most n 1. It follows, by assumption, that p1 (z) = h1 (z) q (z)+r1 (z), where h1 (z) and r1 (z) are the unique polynomials satisfying the conditions above. Let h (z) =
an n z bm
m
h1 (z) ;
and let r (z) =
r1 (z) :
Then
h (z) q (z) + r (z) = = =
an n z bm
an n z bm
m
=
m
q (z)
an n z bm
h1 (z) q (z)
r1 (z) =
(h1 (z) q (z) + r1 (z)) = m
q (z)
p1 (z) = = p (z) :
Thus given p (z) and q (z), there exist polynomials h (z) and r (z) satisfying the above two conditions.
17
+ a1 z + a0 :
I.1.11 Find the polynomials h (z) and r (z) in the preceding exercise for p (z) = z n and q (z) = z 2 1.
Solution Require z n = h (z) z 2
1 + r (z) ;
deg r (z)
1:
If n is even zn = zn
2
+ zn
4
+ zn
6
+
+ z2 + 1
z2
1 + 1:
zn = zn
2
+ zn
4
+ zn
6
+
+ z3 + z
z2
1 + z:
If n is odd
Thus h (z) =
zn zn
2 2
+ zn + zn
4 4
+ zn + zn
6 6
+ +
+ z 2 + 1; if n is even, ; + z 3 + z; if n is odd.
and r (z) =
1; if n is even, : z; if n is odd.
18
I.2.1 Express all values of the following expressions in both polar and Cartesian and plot them. p coordinates, p 4 (a) i (c) 1 (e) ( 8)1=3 (g) (1 + i)8 25 p p 1+i p (b) i 1 (d) 4 i (f) (3 4i)1=8 (h) 2 Solution (a) p
i=
(b) p
i
n
i( =2+2k ) 1=2
i( =4+k )
=e
e
p
1 =
2ei(3
21=4 ei3
=
=4+2k )
=8
o ; k = 0; 1 = ei
1=2
; 21=4 ei11
= 21=4 ei(3
=8
=8+k )
=4
i5 =4
;e
; k = 0; 1
=
n
p o (1 + i) = 2 :
=
21=4 (cos (3 =8) + i sin (3 =8)) :
=
(c) p 4
1 = =
n
ei(
+2k ) 1=4
i =4
i3 i=4
e
;e
=4+k =2)
i5 =4
i7 =4
;e
;e
(d) p 4
i=
n
ei(
o ; k = 0; 1; 2; 3 = n p = (1 i)= 2; ( 1
= ei(
=2+2k ) 1=4 i( =8+k =2)
e
; k = 0; 1; 2; 3 = ei
= f (cos ( =8) + i sin ( =8)) ;
=8
; ei5
p o i) = 2 : =8
; ei9
=8
; ei13
(cos (5 =8) + i sin (5 =8))g :
(e) ( 8)1=3 =
n
23 ei(
+2k ) 1=3
= 2ei
= 2ei( =3
o ; k = 0; 1; 2 = n p =3 = 1 + i 3; 2; 1
=3+2k =3)
; 2ei ; 2ei5
(f) 19
p o i 3 :
=8
o
=
Draw …gure and get tan (3
4i)1=8 =
=
51=8 ei(
1=8
5
n
5ei(
0 =8)
0
=
0 +2k
4=3 )
) 1=8
; 51=8 ei(
0 =8+
0
= 51=8 ei( =4)
= tan
1
0 =8+k
=4)
; 51=8 ei( 1=8
(cos ( 0 =8) + i sin ( 0 =8)) ; 5
0 =8+
4 3
.
o ; k = 0; 1; : : : ; 7 = =2)
; 51=8 ei(
0 =8+3
=4)
=
(cos ( 0 =8 + =4) + i sin ( 0 =8 + =4)) ;
51=8 (cos ( 0 =8 + =2) + i sin ( 0 =8 + =2)) ; 51=8 (cos ( 0 =8 + 3 =4) + i sin ( 0 =8 + 3 =4)) : (g) p
(1 + i)8 =
2ei(
=4+2k )
8
= 16ei(2
+16k )
= 16ei2 = 16:
(h) 25
1+i p 2
= ei(
=4+2k ) 25
I.2.1a
= ei(25
=4+50k )
= ei(
=4+6 +50k )
I.2.1b 1
1
-1
-1
1
-1
1 -1
I.2.1e
I.2.1f 2
1
1
1+i = p : 2
1
-1
I.2.1d
-1
=4
I.2.1c
1
-1
= ei
-2
1
2
-1
-2
20
-1
1 -1
I.2.1g
I.2.1h 20 1
-20
20
-1
1 -1
-20
21
I.2.2 Sketch the following sets (a) jarg zj < =4 (c) jzj = arg z (b) 0 < arg (z 1 i) < =3 (d) log jzj = 2 arg z Solution a) Sector. b) Sector. c) Is a spiral curve starting at 0, spiraling to 1. (d) Is a spiral curve, spiraling to 0, and to 1.
I.1.2b
I.1.2a
I.2.2c 5
5
π/3
(π/2,0)
π/4 π/4
-5
5
-5
5
(−π,0)
(2π,0) (−3π/2,0)
-5
-5
I.2.2d (-3 < x < 3)
I.2.2d (-30 < x < 30)
(1,0)
22
I.2.2d (-800 < x < 800)
I.2.3 For a …xed complex number b, sketch the curve ei + be i : 0 2 Di¤erentiate between the cases jbj < 1; jbj = 1 and jbj < 1. Hint. First consider the case b > 0, and then reduce the general case to this case by a rotation.
.
Solution For 0 < b < 1, an ellipse x2 = (1 + b)2 + y 2 = (1 b)2 = 1, traversed in positive direction with increasing . For b = 1, an interval [ 2; 2]. For 1 < b < +1, an ellipse traversed in negative direction. For b = ei' , express equation as ei'=2 ei( '=2) + e i( '=2) to see that curve is rotate of ellipse or interval by '=2.
I.2.7 (b = 0.5) y
I.2.7 (b = 1)
3
y
2 1
-3 -2 -1 -1
I.2.7 (b = 1.5)
3
y
2 1
1
2
3
x
-3 -2 -1 -1
3 2 1
1
2
3
x
-3 -2 -1 -1
-2
-2
-2
-3
-3
-3
23
1
2
3
x
I.2.4 For which n is i an n-th root of unity?
Solution. i is an nth root of unity for in = 1, n = 4k, k = 1; 2; 3; : : :
24
I.2.5 For n 1, show that (a) 1 + z + z 2 + + z n = (1 (b) 1 + cos + cos 2 +
z n+1 ) = (1
+ cos n =
1 2
+
z), 1 sin n+ 2 2 sin =2
z 6= 1 .
Solution (a) Set + zn;
sn = 1 + z + z 2 + and multiply sn with z and have
+ z n+1 :
zsn = z + z 2 + z 3 + Now subtract zsn from the sum sn , and have sn (1
If z 6= 1 we have after division by 1 sn = (b) Apply (a) to z = ei and to z = e
1 + ei + ei2 + 1+e
i
+e
i2
+
z n+1 :
z) = 1
i
z; 1
z n+1 : 1 z
; ei(n+1) ; 1 ei 1 e i(n+1) = : 1 e i
+ ein = +e
in
1
Add the identities, and use the de…nitions of sine and cosine.
25
2 (1 + cos + cos 2 +
+ cos n ) =
ei(n+1) 1 e i(n+1) + = 1 ei 1 e i 1 e i(n+1) ei =2 1 ei(n+1) e i =2 + = = e i =2 ei =2 ei =2 e i =2 1 ei =2 1 e i(n+1) 1 e i =2 1 ei(n+1) + = = 2i sin ( =2) 2i sin ( =2) " # 1 1 sin ( =2) + sin n + 1 ei =2 e i =2 + ei(n+ 2 ) e i(n+ 2 ) = = 2i sin ( =2) sin ( =2) =
1
1 2
=
sin n + 12 =1+ : sin ( =2) Divide both sides with 2 and get 1 + cos + cos 2 +
+ cos n =
26
1 sin n + 21 + : 2 2 sin =2
I.2.6 Fix n 1. Show that the n th roots of unity w0 ; : : : ; wn (a) (z w0 ) (z w1 ) : : : (z wn 1 ) = z n 1: (b) w0 + + wn 1 = 0 if n 2: (c) w0 wn 1 = ( 1)n 1 : nP1 0; 1 k n 1; wjk = (d) n; k = n: j=0 Solution (a) Let w0 ; : : : ; wn 1 be the n a roots of z n 1 since wjn
th roots of unity then wj = e2
1 = e2
ji=n n
ji
1 = e2
ji=n
1
satisfy
, and wj are
1 = 0:
By the fundamental theorem of algebra, and since the root are simple and the coe¢ cient for the z n term is 1 it follows that zn
1 = (z
w0 ) (z
w1 )
(z
wn 1 ) :
(b) Using exercise (a) and multiply the factors in the product we get zn
1 = zn
wn 1 ) z n
(w0 + w1 +
1
+c2 z n
2
+
+( 1)n w0 w1
wn 1 :
By identi…cation of the coe¢ cients we see that w0 + w1 +
+ wn
1
if
=0
n
2:
(c) From (b) follows that 1 = ( 1)n w0 w1 (d) If 1 n 1 X j=0
k
wjk
n
=
n 1 X j=0
wn
1
, w0 w1
wn
1
= ( 1)n
1
:
1
2 ji=n k
e
=
n 1 X
2 ki=n j
e
j=0
27
=
1
e2 ki=n 1 e2 ki=n
n
=
1 e2 ki = 0: 1 e2 ki=n
If k = n n 1 X j=0
wjk
=
n 1 X
2 ji=n n
e
=
j=0
n 1 X j=0
28
2 ji
e
=
n 1 X j=0
1 = n:
I.2.7 Fix R > 1 and n
1, m
0. Show that
zm zn + 1
Rm Rn
1
jzj = R:
;
Sketch the set where equality holds. Hint. See (1.1) p.2. Solution
I.2.7
I.2.7 (n = 1)
I.2.7 (n = 2)
1
a+b
1
b -1
1
-1
1
a -1
I.2.7 (n = 3)
I.2.7 (n = 4)
1
I.2.7 (n = 5)
1
-1
1
1
-1
1
-1
We use that jz
-1
-1
-1
wj
jzj
1 -1
jwj see (1.1) on page 2 in CA, and have that
jz n + 1j > jz n j
1 = Rn
1;
where the last equaltity is due to that jzj = R. Some rearrangement gives jz n
1 + 1j
1 (Rn 29
1)
;
and mulitplication by jz m j = Rm gives zm zn + 1
Rm Rn
1
jzj = R:
;
For equality, we must have jz n + 1j = Rn
1;
because jzjm = Rm . We rearrarange this equality to
j 1j + jz n + 1j = jz n j and we conclude that 1 and z n +1 must lie on the same ray. (We have used the fact that jaj+jbj = ja + bj implies that a; b lie on the same ray, see …rst …gure I.2.7) If z = Rei then z n = Rn ein : Since R > 1, we require that ein =
1 thus
z = wk Re where wk is an n If
i=n
;
i=n
;
th root of unity.
z = wk Re then zn + 1 =
Rn + 1;
and Rm zm = : zn + 1 Rn 1
30
I.2.8 Show that cos 2 = cos2 sin2 and sin 2 = 2 cos sin using de Moivre’s formulae. Find formulae for cos 4 and sin 4 in terms of cos and sin . Solution Let 2 R be given. Then by de Moivre’s formulae (on page 8 in CA) we have for all n 2 Z cos n + i sin n = (cos + i sin )n Hence for n = 2 we get
cos 2 + i sin 2 = = (cos + i sin )2 = cos2 + i2 cos sin = cos2
sin2 =
sin2
+ i (2 cos sin ) ;
then, by setting the real and imaginary parts equal to each other we obtain cos 2 sin 2
= cos2 sin2 = 2 cos sin :
Similarly, applying de Moivre’s formulae for n = 4 we get (using the Binomial Theorem) cos 4 + i sin 4 = (cos + i sin )4 = = cos4 + i4 cos3 sin = cos4
6 cos2 sin2
i4 cos sin3 + sin4 =
6 cos2 sin2 + sin4 + i 4 cos3 sin
4 cos sin3
then, by setting the real and imaginary parts equal to each other we obtain cos 4 sin 4
= cos4 6 cos2 sin2 + sin4 = 4 cos3 sin 4 cos sin3 :
as required.
31
;
I.3.1 Sketch the image under the spherical projection of the following sets on the sphere (a) the lower hemisphere Z 0 (b) the polar cap 43 Z 1 p p (c) lines of lattitude X = 1 Z 2 cos , Y = 1 Z 2 sin , for Z …xed and 0 2 p p (d) lines of longitude X = 1 Z 2 cos , Y = 1 Z 2 sin , for …xed and 1 Z 1 (e) the spherical cap A X 1, with center lying on the equator, for …xed A. Separate into cases, according to various ranges of A. Solution
I.3.1a
I.3.1b y
I.3.1c
2
y
1
-2
4
y
2
-1
1
2
-4
1
-2
2
4
-2
-1
1
-1
-2
-1
-2
-4
-2
I.3.1d
I.3.1e (A = -1/2) y
2
2
y
1
2
I.3.1e (A = 0)
4
y
2
2 1
θ -2
-1
1
2
-6
-4
-2
2
-2
-1
1
-1
-2
-1
-2
-4
-2
32
2
I.3.1e (A = 1/2) y
4 2
-2
2
4
6
-2 -4
(a) Image is the unit disk. (b) p 2 Set Y = 0 and Z = 3=4 in X 2 + Y 2 + Zp = 1, we have X = 7=4. We have the formula x = X= (1 p Z), thus x = 7. Image is the exterior of a disk, with centre 0 and radius 7. p p 1 + z= 1 z. We have that (c) Image is the disk, with centre 0 and radius p p 2 2 Y = p 1 Z sin , we can take the radius to image X = 1 Z cos and p 1 Z2 . for = 0, thus r = 1 Z = p1+Z 1 Z (d) Image is lines of longitude issuing from 0. (e) Case 1: 1 < A < 0 p Image is the exterior of the disk centered at 1=A with radius 1 A2 = jAj. Case 2: A = 0 Image is the left half–plane. Case 3: 0 < A < 1 p Image is a disk centered at 1=A p with radius p 1 A2 = jAj. Set Y = 0 then Z goes from 1 A2 top 1 A2 . We have p the formula 2 x = X= (1 Z) thus x goes from A= 1 + 1 A = 1 1 A2 =A to p p A= 1 1 A2 = 1 + 1 A2 =A.
33
I.3.2 If the point P on the sphere corresponds to z under the stereographic projection, show that the antipodal point P on the sphere corresponds to 1=z. Solution For z = x + iy the corresponding point on the sphere under transformation given on p. 12 in CA is given by (X; Y; Z) where 2x ; jzj2 + 1 2y ; Y = 2 jzj + 1 jzj2 1 Z = : jzj2 + 1
X =
For z 6= 0, we have y x i 2 2 x iy (x jzj jzj which corresponds to the point on the sphere given by (X 0 ; Y 0 ; Z 0 ) where 1 = z
1
x + iy = iy) (x + iy)
=
X
0
=
Y0 = Z
0
=
2x jzj2 1 2 + z 2y jzj2 1 2 + z 1 2 z 1 2 + z
1
1 1 1
x + iy = x2 + y 2
=
2x = jzj + 1
X;
=
2y = jzj + 1
Y;
=
2
2
jzj2 1 = jzj2 + 1
Z:
Hence, the map (X; Y; Z) 7 ! ( X; Y; Z) on the sphere corresponds to z 7 ! 1=z of C.
34
I.3.3 Show that as z travers a small circle in the complex plane in the positive (counterclockwise) direction, the corresponding point P on the sphere traverses a small circle in the negative (clockwise) direction with respect to someone standing at the center of the circle and with body outside the sphere. (Thus the stereographic projection is orientation reversing, as a map from the sphere with orientation determined by the unit outer normal vector to the complex plane with the usual orientation.)
Solution Draw the picture. Or argue as follows. The orientation of the image circle is the same for all circles on the sphere orientated so that N is outside the circle. This can be seen by moving one circle continuously to the other, and seeing that the image circles moves continuously. Thus we need to shrink it only for the equator of the sphere oriented as indicated ( ), if the South Pole is inside it. And the image circle is the unity and positive direction of the unit circle ( ) is the converse. So the orientation of the image is clockwise (negative).
35
I.3.4 Show that a rotation of the sphere of 180 about the X-axis corresponds under stereographic projection to the inversion z 7 ! 1=z of C. Solution For z = x + iy the corresponding point on the sphere under transformation given on p. 12 in CA is given by (X; Y; Z) where 2x ; jzj2 + 1 2y Y = ; 2 jzj + 1 jzj2 1 Z = : jzj2 + 1
X =
For z 6= 0, we have 1 1 x iy x iy x = = = 2 = z x + iy (x + iy) (x iy) x + y2 jzj2
i
y jzj2
which corresponds to the point on the sphere given by (X 0 ; Y 0 ; Z 0 ) where
X
0
=
Y0 = Z0 =
2x jzj2 = 1 2 + 1 z 2y jzj2 = 1 2 + 1 z 1 1 jzj2 = 1 2 + 1 jzj
2x = X; jzj2 + 1 2y = jzj + 1 2
jzj2 1 = jzj2 + 1
Y; Z:
Hence, the map (X; Y; Z) 7 ! (X; Y; Z) on the sphere corresponds to inversion z 7 ! 1=z of C. Moreover, the map (X; Y; Z) 7 ! (X; Y; Z) is given by rotation of the sphere by 180 about the X-axis.
36
I.3.5 Suppose (x; y; 0) is the spherical projection of (X; Y; Z). Show that the product of the distances from the north pole N to (X; Y; Z) and from N to (x; y; 0) is 2. What is the situation when (X; Y; Z) lies on the equator on the sphere?
Solution The distance from from the north pole N = (0; 0; 1) to (X; Y; Z) is q X 2 + Y 2 + (Z 1)2 ;
and the distance from the north pole N = (0; 0; 1) to (x; y; 0) = is s Y2 X2 + + 1: (1 Z)2 (1 Z)2
X ; Y ;0 1 Z 1 Z
The product of distances is
X 2 + Y 2 + (Z
1)2
1=2
X2 Y2 + +1 (1 Z)2 (1 Z)2
1=2
=
X 2 + Y 2 + (Z 1 Z
1)2
When (X; Y; Z) lie on the equator, the product is simply the square of the distance from N to a point on the equator. By the Pythagorean Law, this is 1 + 1 = 2.
37
= 2:
I.3.6 We de…ne the chordal distance d (z; w) between two points z; w 2 C to be the length of the straight line segment joining to the points P and Q on the unit sphere whose stereographic projections are z and w, respectively. (a) Show that the chordal distance is a metric, that is, it is symmetric, d (z; w) = d (w; z); satis…es the triangle inequality d (z; w) d (z; ) + d ( ; w); and d (z; w) = 0 if and only if z = w. (b) Show that the chordal distance from z to w is given by 2 jz wj q ; d (z; w) = q 2 2 1 + jzj 1 + jwj
z; w 2 C:
(c) What is d (z; 1)? Remark. The expression for d (z; w) shows that in…nitesimal arc length corresponding to the chordal metric is given by d (z) =
2ds ; 1 + jzj2
where ds = jdzj is the usual Euclidean in…nitesimal arc length. The in…nitesimal arc length d (z) determines another metric, the spherical metric (z; w), on the extended complex plane. See Section IX.3. Solution (a) Follows from fact that Euclidean distance in R3 is a metric on the sphere. (b) Set z = x1 + iy1 ; w = x2 + iy2 , we have jz
wj2 = = (z = zz zw 2
2
= jzj + jwj
w) (z
w) = (z
w) (z
2
2
zw + ww = jzj + jwj
(x1 + iy1 ) (x2
iy2 )
(x1
2
w) = zw
iy1 ) (x2 + iy2 ) =
= jzj + jwj2 38
zw = 2x1 x2
2y1 y2 : (1)
We take square of distance between P and Q 2
2
d (z; w)2 = (X X 0 ) + (Y Y 0 ) + (Z = X 2 + Y 2 + Z 2 + X 02 + Y 02 + Z 02
=2
2
Z 0) = 2 (XX 0 + Y Y 0 + ZZ 0 ) = ! 4x1 x2 + 4y1 y2 + jzj2 1 jwj2 1 = =2 2 jzj2 + 1 jwj2 + 1 ! jzj2 + 1 jwj2 + 1 4x1 x2 4y1 y2 jzj2 1 jwj2 1 jzj2 + 1
2jzj2 + 2 jwj2 jzj2 + 1
=2
jwj2 + 1 ! 4 jz 4x1 x2 4y1 y2 (1) = 2 2 jwj + 1 jzj + 1
=
wj2 : jwj2 + 1
Taking the positive square root we have 2 jz wj q d (z; w) = q ; 1 + jzj2 1 + jwj2
(c)
wj 2 jz=w 1j 2 q q = lim q =q : w!1 2 2 2 2 2 1 + jzj 1 + jwj 1 + jzj 1= jwj + 1 1 + jzj
d (z; 1) = lim q w!1
2 jz
z; w 2 C:
39
I.3.7 Consider the sphere of radius 21 in (X; Y; Z) space, resting on the (X; Y; 0) plane, with south pole at the origin (0; 0; 0) and north pole at (0; 0; 1). We de…ne a stereographic projection of the sphere onto the complex plane as before, so that corresponding points (X; Y; Z) and z (x; y; 0) lie on the same line through the north pole. Find the equations for z = x + iy in terms of X; Y; Z, and the equations for X; Y; Z in terms of z. What is the corresponding formula for the chordal distance? Note. In this case, the equation of the sphere is 2 X 2 + Y 2 + Z 12 = 41 . Solution The line through P = (X; Y; Z) and N = (0; 0; 1) is on the form N + t (P N ), and it meats the xy plane when (x; y; 0) = (0; 0; 1) + t ((X; Y; Z)
(0; 0; 1)) = (tX; tY; 1 + t (Z
By simultaneous equations we have, and solve for X, Y and Z 8 8 < x = tX < X = xt y = tY Y = yt ) : : Z = 1 1t 1 + t (Z 1) = 0
We solve for the t that is a point on the sphere X 2 + Y 2 + Z have x2 y 2 + 2 + t2 t
1 2 , jzj2
1 t
2
1 = , jzj2 + 4
t 2
2
1
t + 1 = 0 , t = jzj2 + 1
thus we have 8 x X = 1+jzj 2 > < y Y = 1+jzj2 > : Z = jzj2 : 1+jzj2
40
=
t2 , 4
1 2 2
1)) :
=
1 4
we
Set z = x1 + iy1 ; w = x2 + iy2 , we have
jz
wj2 = = (z = zz zw 2
2
= jzj + jwj
w) (z
w) = (z
w) (z
2
2
iy2 )
(x1
zw + ww = jzj + jwj
(x1 + iy1 ) (x2
w) = zw
= jzj2 + jwj2
And the coordical distance
41
zw =
iy1 ) (x2 + iy2 ) = 2 (x1 x2 + y1 y2 ) : (1)
d (z; w)2 = (X1 X 2 )2 + (Y1 Y 2 )2 + (Z1 Z 2 )2 = =
x1 1 + jwj2
=
=
jzj2 1 + jwj2 +
2
2
2
y2 1 + jwj2
+ y1 1 + jwj2 2
1 + jzj2
1 + jwj2 2
+ (x22 + y22 ) 1 + jzj2 1 + jzj2
2
1 + jzj2
2
1 + jzj2
2
1 + jzj2
2
1 + jzj2
jz1 j2 + 1 + jzj2
1 + jwj2
2 jzj2 jwj2 + jwj4
2y1 y2 1 + jwj2
1 + jwj2
2
+ jwj2 1 + jzj2
2
+ jzj4
2 (x1 x2 + y1 y2 ) 1 + jwj2
1 + jwj2
1 + jzj2
jzj2 + jwj2 =
2
1 + jzj2
1 + jzj2
2
2 (x1 x2 + y1 y2 ) 1 + jwj2
1 + jwj2
jzj2 + jwj2 =
1 + jzj2
2
1 + jzj2
2
1 + jzj2
2
1 + jzj2
1 + jwj2
2
1 + jwj2
2
1 + jwj2
2
jzj2 + jwj2
2
1 + jzj
1 + jwj2
2 (x1 x2 + y1 y2 ) 2
1 + jwj
(1)
=
This gives 42
2
=
+
+
2 jzj2 jwj2 + jwj4
=
2 (x1 x2 + y1 y2 ) 1 + jwj2
2 (x1 x2 + y1 y2 )
jwj2
=
2
1 + jzj2
=
=
2 jzj2 jwj2 + jwj4
1 + jwj2
!2
+ jzj2
2
2
2
2
y2 1 + jzj2
+ jzj4
1 + jwj2
jz2 j2 1 + jwj2
jzj2 + 2 jzj2 jwj2 + jzj2 jwj4 + jwj2 + 2 jwj2 jzj2 + jwj2 jzj4 + jzj4 +
=
2
2x1 x2 1 + jwj2
+
=
y1 + 1 + jzj2
x2 1 + jzj2
(x21 + y12 ) 1 + jwj2
=
2
x2 1 + jwj2
x1 1 + jzj2
1 + jzj2
1 + jzj2 =
(1)
=
jz wj2 : jzj2 + 1 jwj2 + 1
=
+
d (z; w) = q
jz
wj q : 2 2 jzj + 1 jwj + 1
43
I.4.1 Sketch each curve and its image under w = z 2 . (a) jz 1j = 1 (c) y = 1 (e) y 2 = x2 1; x > 0 (b) x = 1 (d) y = x + 1 (f) y = 1=x; x 6= 0 Solution
I.4.1a z-plane y
I.4.1b z-plane y
4 2
-4
-2
2
4
-4
x
-2
y
2
2
4
-4
x
y
2
4
x
-4
2
4
x
-4
-2
-4
-4
44
x
-4
I.4.1f z-plane y
4 2
2 -2
4
4
-2
4
-2
2
x
2
2
2
4
I.4.1c w-plane
I.4.1e z-plane y
2 -2 -4
-2
4
-2
-2
-4
I.4.1d z-plane
-2
-4
4
-2
-4
-4
x
2
-2
y
4
I.4.1b w-plane
4
-2
2
-4
I.4.1a w-plane
4 2
-2
-4
-4
y
4 2
-2
y
I.4.1c z-plane
4
x
-4
-2
2 -2 -4
4
x
I.4.1d w-plane y
I.4.1e w-plane y
4 2
-4
-2
y
4 2
2 -2
I.4.1f w-plane
4
x
-4
-2
-4
-4
45
2
2 -2
4
4
x
-4
-2
2 -2 -4
4
x
I.4.2 Sketch the image of eachpcurve in the preceding problem under the principal branch of w = z, and also sketch, on the same grid but in apdi¤erent color, the image of each curve under the other branch of z. Solution w = z 1=2 = jzj ei argjzj
I.4.2a z-plane y
1=2
I.4.2b z-plane y
4 2
-4
-2
2
4
x
-4
y
2
4
x
-4
-2
2
4
x
-4
y
4
-2
-4
-4
46
2
4
x
4 2
2 -2
4
I.4.2c w-plane
2
2
2 -2 -4
I.4.2b w-plane y
4 2
-2
4
-2
I.4.2c z-plane
-4
I.4.2a w-plane
-2
p jzjei Arg z=2
4
-2
-4
-4
=
2
-2
y
n 1=2
= jzj ei Argjzj+i2
4
x
-4
-2 -2 -4
x
I.4.2e z-plane
I.4.2d z-plane y
y
4
-2
2
4
x
-4
-2
2
x
-4
y
2
-2
2 -2
4
x
-4
y
4
-2
-4
-4
47
4 2
2 -2
x
I.4.2f w-plane
2
2
4
-4
I.4.2e w-plane
4
-2
4
-4
I.4.2d w-plane
-2
2
-2
-4
-4
4
2
-2
y
y
4
2
-4
I.4.2f z-plane
4
x
-4
-2
2 -2 -4
4
x
I.4.3 (a) Give a brief description of the function z 7 ! w = z 3 , considered as a mapping from the z plane to the w- plane. (Describe what happens to w as z traverses a ray emanating from the origin, and as z traverses a ray a circle centered at the origin.) (b) Make branch cuts and de…ne explicitly three branches of the inverse mapping. (c) Describe the construction of the Riemann surface of z 1=3 .
Solution (a) The function w = f (z) = z 3 . For z = rei , we have z 3 = r3 ei3 . The radical rays at angle are mapped to rays at angle 3 , that is, arg w = 3 arg z. The magnitude of a complex number is cubed, jwj = jzj3 . Circles, centered at the origin of radius r, are mapped to cocentric circles with radius r3 . (b) We make branch cuts at ( 1; 0], z = w1=3 = jwj ei arg w
1=3
= jwj ei Arg w+i2
we choose g (z) = z 1=3 = r1=3 ei =3 , < < . Sheet 1 : Take f1 (z) = g (z), Sheet 2 : Take f2 (z) = e2 i=3 g (z), Sheet 3 : Take f3 (z) = e4 i=3 g (z).
n 1=3
= ei2
n=3
p jwjei Arg w=3
(c) Top edge of cut on sheet 1 to bottom edge of cut on sheet 2. Top edge of cut on sheet 2 to bottom edge of cut on sheet 3. Top edge of cut on sheet 3 to bottom edge of cut on sheet 1. The endpoints for f (z) is continuous on surface Spara
48
I.2.1b
I.2.1c ±
±±
±
I.4.3a
± ±
±
± ±± ±
±±±±±±±
±
±
1
±
1
±± ±
-1
1
±± ±
-1
Spara
49
-1
±
±
±
±
±
±
1 -1
I.4.4 Describe how to construct the following functions p Riemann surfaces for the 2=5 (a) w = z 1=4 ; (b) w = z i; (c) w = (z 1) : Remark. To describe the Riemann surface of a multivalued function, begin with one sheet for each branch of the function, make branch cuts so that the branches are de…ned continuously on each sheet, and identify each edge of a cut on one sheet to another appropriate edge so that the function values match up continuously. Solution (a) Use four sheets, can make branch cuts along real axis from 1 to 0. (b) Use two sheets, can make branch cuts along horizontal line from 1 + i to i. (c) Use …ve sheets, can make branch cuts along real axis from 1 to 1.
50
I.5.1 Calculate and plot for ez for the following points z. (a) 0 (c) (i 1) =3 (e) i=m; m = 1; 2; 3; : : : (b) i + 1 (d) 37 i (f) m (i 1) m = 1; 2; 3; : : : Solution
I.5.1e
I.5.1f 1
0.2
-1
1
-0.4
-0.2
-1
0.2 -0.2
(a) e0 = 1: (b) e
i+1
= e i e1 =
e:
(c) e
(i 1)3
=e
=3
e
i=3
p ! 3 1 + i = 0:351 + 0:006i: 2 2
=3
=e
(d) e37 i = e36 i e i = (e) We take the limit for the sequence e have e
i=m
i=m
, m = 1; 2; 3; : : : as m ! 1, and
! 1;
51
1:
as m ! 1. Because e i=m = 1 the sequence approaches its limit along a circle with radius 1. (f) We take the limit for the sequence em(i 1) = e m emi , m = 1; 2; 3; : : : as m ! 1, and have e as m ! 1. Because je origo.
m mi
m mi
e j=e
! 0;
e
m
the sequence spiraling to its limit in
52
I.5.2 Sketch each of the following …gures and its image under the exponential map w = ez . Indicate the images of horizontal and vertical lines in your sketch. (a) the vertical strip 0 < Re z < 1, (b) the horizontal strip 5 =3 < Im z < 8 =3, (c) the rectangle 0 < x < 1, 0 < y < =4, (d) the disk jzj =2, (e) the disk jzj , (f) the disk jzj 3 =2. Solution
I.5.2a z-plane
I.5.2b z-plane
4
-2
6 2
1
2 -4
-4
I.5.2a w-plane
-2
0
2
4
I.5.2b w-plane
4
4
2
2
-2
2
4
4
-2
-4
3
8
2
-4
I.5.2c z-plane
2
4
-4
-2 -2
-4
-4
53
2
3
2
3
I.5.2c w-plane 3 2 2
-2
1
4
1
1
I.5.2e z-plane
I..5.2d z-plane
-4
4
4
4
2
2
2
-2
2
4
-4
-2
2
4
-4
-2
2
-2
-2
-2
-4
-4
-4
I.5.2e w-plane
I..5.2d w-plane
2 -2 -4
100
10
2
-2
I.5.2f w-plane
20
4
-4
I.5.2f z-plane
4
-10
10
20
30
100
-10 -20
54
-100
4
I.5.3 Show that ez = ez .
Solution
ez = ex
iy
= ex e
iy
= ex (cos ( y) + i sin ( y)) = ex (cos y
i sin y) =
= ex (cos y + i sin y) = ex (cos y + i sin y) = ex eiy = ex+iy = ez :
55
I.5.4 Show that the only periods of ez are the integral multiples of 2 i, that is, if ez+ = ez for all z, then is an integer times 2 i.
Solution Show that the only periods of ez are the integral muliples of 2 i that is, if ez+ = ez for all z, then is an integer times 2 i. ez = ez+ = ez e ) e = 1 )
56
= 2 mi
I.6.1 Find and plot log z for the following complex numbers z. Specify the principal value. p (a) 2 (b) i (c) 1 + i (d) 1 + i 3 =2 Solution (a) Suppose that n = 0; 1; 2; : : : log 2 = log j2j + i Arg 2 + 2 ni = log 2 + i2 n: (b) Suppose that n = 0; 1; 2; : : : log i = log jij + i Arg i + 2 ni = i =2 + i2 n: (c) Suppose that n = 0; 1; 2; : : :
log (1 + i) = = log j1 + ij + i Arg (1 + i) + 2 ni = log
p
2 + i =4 + i2 n = = log 2=2 + i =4 + i2 n:
(d) Suppose that n = 0; 1; 2; : : :
log
p 1+i 3 2
= = log
p 1+i 3 2
+ i Arg
p 1+i 3 2
+ 2 ni = = i=3 + i2 n:
57
I.6.1a
I.6.1b
I.6.1c
20
20
20
10
10
10
-1
1
2
-1
1
2
-1
1
-10
-10
-10
-20
-20
-20
I.6.1d 20 10
-1
1
2
-10 -20
58
2
I.6.2 Sketch the image under the map w = Log z of each of the following …gures. (a) the right halv-plane Re z > 0, (b) the half-disk jzj < 1, Re z > 0, (c) the unit circle jzjp= 1, p e), (d) the slit annulus e < jzj < e2 , z 62 ( e2 ; (e) the horizontal line y = e, (f) the vertical line x = e.
Solution (b) We have a disk with radius less then 1, this means jzj < 1, thus log jzj < 0 and is unbounded, therefore, it goes from 0 to 1. Since Re (z) > 0, the polar angle is between 2 and 2 . (d) Here, we have an annulus, the polar angle is from to , thus the argument is in this range. Since, the log function is the inverse of the exponential map, circles in z plane are stright lines in the w plane (w = log z). Therefore, the image p of this annulus under the log function is the rectangle, bounded by x = log j ej = 12 and x = log je2 j = 2.
I.6.2b z-plane
I.6.2a z-plane
I.6.2c z-plane 4
3 2
2
2
1 -3 -2 -1 -1
1
2
3
-2
2
-4
-2
2 -2
-2
-2 -4
-3
59
4
I.6.2a w-plane
I.6.2b w-plane
I.6.2c w-plane
3
π π/2
2 1 -3 -2 -1 -1
1
2
3 −π/2
-2
−π
-3
I.6.2d z-plane
I.6.2e z-plane
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8
2 4 6 8
I.6.2d w-plane
-4
-4
I.6.2f z-plane
4
4
2
2
-2
2
4
-4
-2
2
-2
-2
-4
-4
I.6.2e w-plane
I.6.2f w-plane
4
4
4
2
2
2
-2
2
4
-4
-2
2
4
-4
-2
2
-2
-2
-2
-4
-4
-4
60
4
4
I.6.3 De…ne explicitly a continuous branch of log z in the complex plane slit along the negative imaginary axis, Cn [0; i1). Solution
I.6.3
We have log z = log rei = log r + i . To avoid the negative imaginary axis we chose =2 < < 3 =2. We use the branch f rei
= log r + i ;
of log z.
61
=2 <
< 3 =2;
I.6.4 How would you make a branch cut to de…ne a single-valued branch of the function log (z + i 1)? How about log (z z0 )? Solution Any straight line cut from z0 to 1, in any direction, will do.
62
I.7.1 Find all values and plot them. (a) (1 + i)i (b) ( i)1+i (c) 2
1=2
p 1+i 3
(d)
(1 i)
Solution (a) Suppose that m; n = 0; 1; 2; : : : (1 + i)i = = ei log(1+i) = ei(logj1+ij+i Arg(1+i)+i2
m)
= ei(log
p
2+i =4+i2 m)
=e
2 m
= e2 n e
=4 i log
e
e
=4 i log
e
p
2
:
=2
:
(b) Suppose that m; n = 0; 1; 2; : : : ( i)1+i = = e(1+i) log( =e
2 m
e
i)
= e(1+i)(logj
=2 i(
e
=2+2 m)
ij+i Arg( i)+i2 m)
= [m =
= e(1+i)(
n] = e2 n e
i =2+i2 m)
=2 i(
e
=2 2 n)
=
=
=
ie2 n e
(c) Suppose that m = 0; 1; 2; : : :
2
1=2
=e
= log 2=2
=e
(logj2j+i Arg 2+i2 m)=2
=e
(log 2+i2 m)=2
=e
log 2=2
e
= (d) Suppose that n = 0; 1; 2; : : :
63
i m
= p 1= 2:
p
2
= [m =
n] =
p 1+i 3
1 i
=
= e(1 = e(1
p i) log(1+i 3)
= e(1
i)(log 2+i =3+i2 n)
p p i)(logj1+i 3j+i Arg(1+i 3)+i2 n)
= e2 n elog 2+
=3 i(
e
log 2+ =3+2 n)
= e2 n elog 2+
64
=
=3 i(
e
= log 2+ =3)
:
I.7.2 h i 2i Compute and plot log (1 + i) .
Solution We rewrite the expression, suppose that k; m; n = 0; 1; 2; : : : h i log (1 + i)2i = i h log(1+i) 2i = log e2i log(1+i) = log e2i(logj1+ij+i Arg(1+i)+i2 k) = = log e h i p = log e2i(log 2+i =4+i2 k) = log e 4 k e =2 ei log 2 = [k = m] = log e4 m e = log e4
m
e
=2 i log 2
e
+i Arg e4
m
e
=2 i log 2
e
=
I.7.2 20 10
10 -10 -20
65
e
+i2 n = 4 m =2+i log 2+i2 n = =2 + i log 2 + 4 m + 2 in:
Thus the set is a square lattice.
-10
=2 i log 2
20
=
I.7.3 Sketch the image of the sector f0 < arg z < =6g under the map w = z a for (b) a = i (c) a = i + 2 (a) a = 32 Use only the principal branch of z a .
Solution
I.7.3a z-plane
I.7.3b z-plane
5
-5
I.7.3c z-plane
5
5
-5
5
5
-5
-5
-5
-5
I.7.3b w-plane
I.7.3a w-plane
5
I.7.3c w-plane
2
5
1
-5
5
-2
2 1 -2
-5
Some points on the y-axis is marked, and the point (1; 0) too.
66
I.7.4 Show that (zw)a = z a wa , where on the right we take all possible products.
Solution Let 2 (zw)a , and k; m; n = 0; 1; 2; : : : = = ea log(zw) = ea(logjzwj+i Arg(zw)+i2
n)
= ea(logjzj+logjwj+i Arg z+i Arg w+i2
a(logjzj+i Arg z+i2 (m+n)) a(logjwj+i Arg w)
=e Conversely, if
e
m+i2 n) a
=
a
2z w :
2 z a wa , say
= = ea log z ea log w = ea(logjzj+i Arg z+i2 = ea(logjzj+logjwj+i Arg z+i Arg w+i2
(m+n))
m) a(logjwj+i Arg w+i2 n)
e
= ea(logjzwj+i Arg(zw)+i2
a(logjzwj+i Arg(zw)+i2 (k+m+n))
=e
=
k+i2 (m+n))
=
= = ea log(zw) 2 (zw)a
We have that (zw)a = z a wa as sets
67
I.7.5 i Find ii . Show that is does not coincide with ii i = i 1 . Solution We rewrite the expression ii , suppose that k = 0; 1; 2; : : :
ii = = ei log i = ei(logjij+i Arg(i)+i2
k)
= ei(i
=2+i2 k)
= =e
2 k
=2
:
i
Now we can …nd ii we use that we have ii , suppose that k; m = 0; 1; 2; : : : i
ii = = ii
i
= ei log(i ) = ei(log(e i
i( 2 k
=e
2 k
=2+i2 m)
=2
)) = ei(logje
=e
Which not coincide with ii i = i f ie2 m ; m = 0; 1; 2; : : :g.
2 k
2 m i( 2 k
e
1
= 1=i =
68
=2)
=2
j+i Arg(e
=
ie
2 k
2 m
=2
= =
)+i2 ie2
m)
m
=
:
i which is smaller than
I.7.6 Determine the phase factors of the function z a (1 z)b at the branch points z = 0 and z = 1. What conditions on a and b guarantee that z a (1 z)b can be de…ned as a (continuous) single-valued function on Cn [0; 1]? Solution Phase factors is e2 ia at 0 and e2 ib at 1. Require e2 ia e2 ib = 1, or a + b = n, where n 2 Z , to have a continuous single–valued determination of z a (1 z)b on Cn [0; 1].
69
I.7.7 Let x1 < x2 < < xn be n consecutive points on the real axis. p Describe the Riemann surface of (z x1 ) (z xn ). Show that for n = 1 and n = 2 the surface is topologically a sphere with certain punctures corresponding to the branch points and 1. What is it when n = 3 or n = 4? Can you say anything for general n? (Any compact Riemann surface is topologically a sphere with handles. Thus a torus is topologically a sphere with one handle. For a given n, how many handles are there, and where do they come from?)
Solution p p For n = 1 and n = 2, the functions f1 p = z x1 and f2 = (z x1 ) (z x2 ), and for a general n we have fn (x) = (z x1 ) (z x2 ) : : : (z xn ). To constuct the Riemann Suface we use two sheets with slits [x1 ; x2 ] ; [x3 ; x4 ] ; : : :. If n is odd, we also need a slit [xn ; +1) . Identify top edge of slit on one sheet with bottom edge of slit on other sheet. If n = 1 and n = 2 the surface is topologically a sphere with certain punctures corresponding to the branch points and 1. If n = 3 or n = 4, surface is a torus. For a general even n, fn (x) will be continuous on Cn ([x1 ; x2 ] [ [x3 ; x4 ] [ : : : [ [xn 1 ; xn ]), and its Riemann surface will be a sphere with n2 1 holes. For a general odd n though, fn (x) will be continuous on Cn ([x1 ; x2 ] [ [x3 ; x4 ] [ : : : [ [xn ; +1]), its Riemann surface will be a sphere with n 2 1 holes.
70
I.7.8 p Show that z 2 1=z can be de…ned as a (single-valued) continuous function outside the unit disk, that is, for jzj > 1. Draw branch cuts so that the function can be de…ned continuous o¤ the branch cuts. Describe the Riemann surface of the function.
Solution p The function is z 1 1=z 3 . If jzj > 1, can use the principal value of the square root to de…ne a branch of the function. There are branch points at z = 0 and z 3 = 1, that is at 0; 1; e2 i=3 and e 2 i=3 . Make two branch cuts by connecting any two pairs of point by curves; for instance, connect 0 to 1 by a straight line, and the other cube roots of unity by a straight line or arc of unit circle. The resulting two-sheeted surface with identi…cation of cuts and with points at in…nity is a torus.
71
I.7.9 q Consider the branch of the function z (z 3 1) (z + 1)3 that is positive at z = 2. Draw branch cuts so that this branch of the function can be de…ned continuously o¤ the branch cuts. Describe the Riemann surface of the function. To what value at z = 2 does this branch return if it is continued continuously once counterclockwise around the circle fjzj = 2g? Solution We have that the function is p p (z + 1) z 2 z (1
1=z 3 ) (1 + 1=z):
If jzj > 1, the second square root can be de…ned to be single-valued for jzj > 1. The value of the function at z = 2 returns to thepnegative of their initial value then we travers the circle jzj = 2 , because z does. We can construct a Riemann Surface by making cuts at 1; 0; 1; e2 i=3 ; e 2 i=3 and 1. The function for the Riemann Surface is q (z + 1) z (z 1) (z + 1) (z e2 i=3 ) (z e 2 i=3 ):
72
I.7.10 q Consider the branch of the function z (z 3 1) (z + 1)3 (z 1) that is positive at z = 2. Draw branch cut so that this branch of the function can be de…ned continuously o¤ the branch cuts. Describe the Riemann surface of the function. To what value at z = 2 does this branch return if it is continued continuously once counterclockwise around the circle fjzj = 2g? Solution We have that the function is q 4 z (1 1=z 3 ) (1 + 1=z)3 (1
1=z)
The branch that is positive for z = 2 is, it is de…ned continuously for jzj > 1. Branch return to original value around circle jzj = 2. We can construct a Riemann Surface by making cuts at 0; 1; e2 i=3 and e 2 i=3 . The function for the Riemann Surface is q (z 1) (z + 1) z (z + 1) (z e2 i=3 ) (z e 2 i=3 ):
73
I.7.11 p Find the branch points of 3 z 3 of the function.
1 and describe the Riemann surface
Solution
I.7.11a
I.7.11b 1
I.7.11c 1
-1
1
1
-1
-1
1
-1
-1
1
-1
We rewrite as follows p 3
z3
1=
q 3
(z
1) (z
e2
i=3 ) (z
e
2 i=3 ):
This equation will have 3 branch points, which are the cube root of unity. So the phase factor is e2 i=3 . The Riemann surface is obtained ty pasting three sheets with the corresponding branch cuts, we end up with a one hole torus. Make two cuts, from 1 to e 2 i=3 , on each sheet. In this case the cuts share common endpoint. We nee three shets where f0 (z), f1 (z) = e2 i=3 f0 (z) and f2 (z) = e 2 i=3 f0 (z) Note: Can use Riemann–formula to see that surface is a torus. Cheek by going around each little tip what phase change to, which of for your sheet.
74
I.8.1 Establish the following addition formulae (a) cos (z + w) = cos z cos w sin z sin w; (b) sin (z + w) = sin z cos w + cos z sin w; (c) cosh (z + w) = cosh z sinh w + sinh z sinh w; (d) sinh (z + w) = sinh z cosh w + cosh z sinh w; Solution (a) Using the de…nitions of sine and cosine functions given on page 29 in CA we have,
cos (z + w) = 1 ei(z+w) + e i(z+w) = 2ei(z+w) + 2e i(z+w) = = 2 4 1 i(z+w) i(z w) = e +e + e i(z w) + e i(z+w) + 4 + ei(z+w) ei(z w) e i(z w) + e i(z+w) = eiz + e iz eiw + e iw eiz e iz eiw e iw = = 2 2 2i 2i = cos z cos w sin z sin w: (b) Using the addition formula (a), and using that sin z = cos (z cos z = sin (z =2) we have,
sin (z + w) = cos z + w
2 = cos z cos w
=2) and
=
sin z sin w = 2 2 = cos z sin w + sin z cos w = sin z cos w + cos z sin w:
(c) Using the addition formula (a) and the formulas cos (iz) = cosh z and sin (iz) = i sinh z given on page 30 in CA we have,
75
cosh (z + w) = cos (i (z + w)) = cos (iz) cos (iw) sin (iz) sin (iw) = = cosh z cosh w i sinh z i sinh w = cosh z cosh w + sinh z sinh w: (d) Using the addition formula (b) and the formulas cos (iz) = cosh z and sin (iz) = i sinh z given on page 30 in CA we have, sinh (z + w) = i sin (i (z + w)) = i [sin (iz) cos (iw) + cos (iz) sin (iw)] = = i [i sinh z cosh w + cosh z i sinh w] = sinh z cosh w + cosh z sinh w:
76
I.8.2 Show that jcos zj2 = cos2 x + sinh2 y, where z = x + iy. Find all zeros and periods of cos z.
Solution. Use trigonometric formulas from page 29 and 30 in CA we have, cos z = cos (x + iy) = cos x cos (iy) sin x sin (iy) = cos x cosh y i sin x sinh y: Now take the modulus squared, and use cosh2 y = 1 + sinh2 y, jcos zj2 = cos2 x cosh2 y + sin2 x sinh2 y =
= cos2 x 1 + sinh2 y + sin2 x sinh2 y = = cos2 x + cos2 x + sin2 x sinh2 y = cos2 x + sinh2 y:
The identity for jcos zj2 shows that the only zeros of cos z are the zeros of cos z on the real axis, because cos z = 0 ,
cos x = 0 , x = 2 + m ; m = 0; 1; 2; : : : ; sinh y = 0 , y = 0:
Translation by any period of cos z sends zeros to zeros. Thus any period is an integral multiple of , and since odd integral multiples are not periods, the only periods of cos z are 2 n; 1 < n < 1.
77
I.8.3 Find all zeros and periods of cosh z and sinh z. Solution We have that cosh z = cos (iz), thus the zeros of cosh z are at z = i =2 +im , m = 0; 1; 2; : : :, and the periods of cosh z are 2 mi, m = 0; 1; 2; : : :. We have that sinh z = i sin (iz), thus the zeros of sinh z are at z = m i, m = 0; 1; 2; : : :, and the periods of cosh z are 2 mi, m = 0; 1; 2; : : :.
78
I.8.4 Show that tan
1
z=
1 log 2i
1 + iz 1 iz
;
where both sides of the identity are to be interpreted as subsets of the complex plane. In other words, show that tan w = z if and only if 2iw is one of the values of the logarithm featured on the right.
Solution Set z = tan w, we have that z = tan w =
eiw e sin w = cos w i (eiw + e
iw
=
iw )
e2iw 1 : i (e2iw + 1)
Solve for e2iw e2iw =
1 + iz ; 1 iz
and take logarithm and solve for w w=
1 log 2i
Because z = tan w then w 2 tan tan
1
1
z=
1 + iz 1 iz
:
z and we have the identity 1 log 2i
79
1 + iz 1 iz
:
I.8.5 Let S denote the two slits along the imaginary axis in the complex plane, one running from i to +i1, the other from i to i1. Show that (1 + iz) = (1 iz) lies on the negative real axis ( 1; 0] if and only if z 2 S. Show that the principal branch Tan
1
z=
1 Log 2i
1 + iz 1 iz
maps the slit plane CnS one-to-one onto the vertical strip fjRe wj < =2g. Solution
I.8.5 z-plane
I.8.5
I.8.5 π
i -1
-i
−π
w1 =
1+iz 1 iz
w2 = Log w1
I.8.5 w-plane
−π/2
π/2
w = 2i1 w2 = 2i1 Log 11+iz iz We begin to show that 1+iz 2 ( 1; 0] if and only if z 2 ( i1; i] [ [i; i1). 1 iz Set 1 + iz = w1 1 iz and solve for z, thus 80
z=
1 w1 i 1 + w1
1 w1 i goes from i to i1 along iR, and when As w1 goes from 1 to 1, 1+w 1 1 w1 w1 goes from 1 to 0, 1+w1 i goes from i1 to i along iR. Since we have that
1 + iz 1 w1 = w1 ) z = i 1 iz 1 + w1 the map is one to one. Remark that the map makes correspondence between the interval [ i: i1] in z plane and [ 1; 1] in w1 plane, and between interval [i; i1] in z plane and [ 1; 0] in w1 plane. We have that w2 = Log w1 maps Cn( 1; 0) onto fjIm wj < g, thus w = w2i2 maps jIm w2 j < onto jRe wj < 2 , se …gures. We have that the function Tan
1
z=
1 Log 2i
1 + iz 1 iz
maps CnS onto jRe wj < 2 , where Tan 1 z is the principal branch for tan 1 z. The other branches are given by fn (z) = Tan 1 z + n , 1 < n < 1.
81
I.8.6 Describe the Riemann surface for tan
1
z.
Solution
I.8.6 z-plane
I.8.6
I.8.5
____ ++++
+++++++++++ π
i
++++++ _ _ _ _-1 __
____ ++++
-i
_ _ _ _ −π _______
w1 =
1+iz 1 iz
w2 = Log w1
+++++++++++
___________
I.8.6 w-plane
−π/2
w=
1 w 2i 2
=
π/2
1 2i
Log
1+iz 1 iz
We have that f0 (z) = Tan
1
z=
1 Log 2i
1 + iz 1 iz
;
other branches of tan 1 z are fn (z) = f0 (z) + n , where 1 < n < 1. Use one copy of the double slit plane S for each integer n, and de…ne fn (z) = Tan 1 z + n on the nth sheet to the (n + 1) th sheet along one of the cuts, so that fn (z) and fn+1 (z) have same value at the junction. 82
Make in…nite many copies of S and call them Sn . De…ne fn (z) = Tan 1 z+n on Sn . Identify "+" side of cut on Sn to " " side of cut on Sn+1 . Then fn on Sn continuous to fn+1 on Sn+1 , and fn maps Sn onto vertical strip n 21 < Re z < n + 12 . Note that composite function is not de…ned at i and 1 (endpoints of slits), and its image omits the sequence 12 + n, 1 < n < 1.
83
I.8.7 Set w = cos z and
= eiz . Show that = w h p 1 w2 cos w = i log w
p
w2 i 1 ;
1. Show that
where both sides of the identity are to be interpreted as subsets of the complex plane.
Solution Set w = cos z and
= eiz , we have that w = cos z =
eiz + e 2
iz
+ 1= : 2
=
Solve for =w and set
p
w2
1;
= eiz and take the logarithm and solve for z z=
i log w
Because w = cos z then z 2 cos cos
1
w=
1
p
w2
1 :
w we have the identity i log w
84
p
w2
1 :
I.8.8 Show that the vertical strip jRe (w)j < =2 is mapped by the function z (w) = sin w one-to-one onto the complex z plane with two slits ( 1; 1] and [+1; +1) on the real axis. Show p that the inverse function is the branch of sin 1 z = i Log iz + 1 z 2 obtained by taking the principal value of the square root. Hint. First show that the function 1 z 2 on the slit plane omits the negative real axis, so that the principal value of the square root is de…ned and continuous on the slit plane, with argument in the open interval between =2 and =2.
Solution Set w = x + iy, by formulas on page 30 in CA we have that
sin w = sin (x + iy) = sin (x) cos (iy) + cos (x) sin (iy) = = sin x cosh (y) + i cos (x) sinh y: If jRe wj = jxj < because
=2 then sin w is mapped on Cn ( 1; 1] [ [+1; +1),
Im (sin w) = 0 ) y = 0 ) sin w = sin (x) and 1 < sin (x) < 1 for these x. Let now z = sin w. The value of w is referred to as sin complex number whose sine is z. Note that z= Now with
= eiw and 1= = e
iw
1
z, that is, the
eiw
e iw : 2i in (1), we have
1= : 2i Mulitplying the above by 2i and doing some rearranging, we …nd that z=
85
(1)
2
2iz =
1
2
or
2iz
1 = 0:
With the quadratic formula, we solve this equation for = iz +
p
1
z2
or
eiw = iz +
p
and …nd 1
z2
We now take the logarithm of both sides of the last equation and divide the result by i to obtain w= and, since w 2 sin
1
p 1 log iz + 1 i
z2
z, sin
1
z=
i log iz
86
p
1
z2 :
II 1 2 3 1 X X 2 X X X 3 X X X 4 X 5 X X X 6 7 8
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 X X X X X X X X X X X X X X X X X X X X X X X X X X X X
1
II.1.1 Establish the following: n (a) lim n+1 = 1 (c) lim n!1
(b)
lim 2n n!1 n +1
= 0 (d)
2np +5n+1 = p n!1 nn +3n+1 z lim = 0; z 2 n!1 n!
2; p > 1 C:
Solution (a) n 1 = lim = 1: n!1 n + 1 n!1 1 + 1=n lim
(b) 1=n n = lim = 0: n!1 1 + 1=n2 n!1 n2 + 1 lim
(c) 2np + 5n + 1 2 + 5n1 lim = lim n!1 np + 3n + 1 n!1 1 + 3n1
p
+n p+n
p p
= 2:
(d) Let m be a positive integer so that m > 2 jzj and let n > m, then 0
jzjm jzjn m zn = = n! m! (m + 1) (m + 2) n jzj jzj jzj jzjm jzj jzj jzj jzjm = m! m + 1 m + 2 n m! m m m m m jzj 1 1 1 jzj 1 = !0 n m! 2 2 2 m! 2 m
as n ! 1, it follows by instängning that shown.
2
zn n!
! 0 as n ! 1 as was to be
II.1.2 For which values of z is the sequence fz n g1 n=1 bounded? For which values of z does the sequence converge to 0?
Solution The sequence is bounded for jzj
1, and the series converge to 0 for jzj < 1.
3
II.1.3 Show that fnn z n g converges only for z = 0. Solution If we choose n> then
2 jzj
2 < n jzj and jnn z n j = jn zjn > 2n ! 1 when n ! 1 for z 6= 0.
4
II.1.4 Show that lim
N! k N !1 N (N k)!
= 1;
k
0.
Solution We have that
Nk
N (N 1) : : : (N k + 1) N! = = (N k)! N N ::: N
as N ! 1 (k is …xed).
5
1
1 N
1
2 N
::: 1
k
1 N
!1
II.1.5 Show that the sequence 1 1 1 + + + log n; n 1; 2 3 n is decreasing, while the sequence an = bn 1=n is increasing. Show that the sequences both converge to the same limit . Show that 1 < < 35 . Remark. The limit of the sequence is called Euler’s 2 constant. It is not known whether Euler’s constant is a rational number or an irrational number. bn = 1 +
Solution We have that 1 1 + + 2 3 1 1 = 1+ + + 2 3
bn = 1 +
+
an
+
1 n
1
+
1 n
1
1 n
log n;
log n:
We trivially have that log n =
Z
n
1 dt t
((1))
1 1 dt < t k
((2))
1
An elementary estimation gives 1 < k+1
Zk+1 k
We take the di¤erence an+1 an+1
1 an = n
an and by (1) and (2) we have that 1 log (n + 1) + log n = n
n+1 Z
dt > 0; t
n
thus an+1 > an , so the sequence an is increasing. We take the di¤erence bn+1 bn and by (1) and (2) we have that
6
bn+1
1 bn = n+1
1 log (n + 1) + log n = n+1
n+1 Z
dt < 0; t
n
thus bn+1 < bn , so the sequence bn is decreasing. We have after some investigation on the increasing sequence an that an
a7 =
49 20
1 log 7 > ; 2
n
7;
thus 21 < an , if n 7. We have after some investigation on the decreasing sequence bn that bn
b22 =
3 log 22 < ; 5
19 093 197 5173 168
n
22
thus bn 3=5, if n 22. The both sequences converge to the same limit because bn an = 1=n have the limit 0 as n ! 1. And this limit is in the interval 12 < < 53 .
7
II.1.6 For a complex number choose n" by = 1;
0
n
, we de…ne the binomial coe¢ cient " (
=
1)
( n!
n + 1)
;
n
1:
Show the following (a) The sequence (b) (c) If
is bounded if and only if Re
n
! 0 if and only if Re
n
6= 0; 1; 2; : : :, then 6=
(d) If Re
1,
(e) If Re a >
1 and
>
1.
n+1
!
n
1, then
1.
>
n+1
is not an integer, then
1. n
for all n n+1
<
0. n
for n large Solution (a) We have that
n
=
n Y
k=1
1
1 Y
+1 = = k k=1
s
2 (1 + Re ) (1 + Re )2 (1 + Im )2 + + : k k2 k2
1
Let (1 + Re )2 (1 + Im )2 tn = + k2 k2 If Re < 1, we have that tk > 0 and diverge is n unlimited. If Re
=
1, we have that tk =
(Im )2 k2
A 6= 0 becauxe for all k we have that 1 so n is limited. 8
n
2 (1 + Re ) k
! 1 then n ! 1 because and +1 k
P
tk
! A then n ! 1, and P P (Im )2 > 1 and tk = <1 k2 n
If Re > 1 we have that it exists k 2 N so that 1 < tk < 0 for all k > K. This gives that n is a decreasing series with respect to n, and so limited. The conclution is that
n
is limited if and only if Re
1.
b) We have that n ! 0 if and only if n ! 0. From (a) we see that the only possibility for this is Re > 1. For Re > 1 we can see that 0 is decreasing. Let M = lim n . Then we hav that M = lim n = n n lim = M so M = n!1 n n+1 and only if Re > 1.
c) If
6= 0; 1; 2; : : :, then
n (
n+1 n
=
n!1
n!1
M () M = 0 so
! 0, then n ! 1 if
n
6= 0 and we can divide, to get
1):::( (n+1)+1) (n+1)! ( 1):::( n+1) n!
=
1 n = n 1 ! n+1 1+ n
1
as n ! 1. d) j(n+1)j = j(n)j e) Let
j
nj n+1
=
distance( ;n) distance( 1;n)
be …xed and Re > q j(n+1)j = (Re j(n)j
> 1, if Re
1,
6=
1
1. Then n)2 +(Im )2 (n+1)2
<1
for n >
j j2 1 : 2(1+Re )
Parts (a) and (b) seems to require some facts about products: n Q
k=1 n Q
k=1 n Q
k=1
1+
k
! 1 as n ! 1,
>0
1
k
! 1 as n ! 1,
>0
1
k2
! A 6= 0 as n ! 1
9
II.1.7 De…ne x0 = 0, and de…ne by induction xn+1 = x2n + 14 for n 0. Show that xn ! 12 . Hint. Show that the sequence is bounded and monotone, and that any limit satis…es x = x2 + 41 . Solution We …rst prove by induction that for every integer n 0
xn
1 2
and
xn
For n = 0, we have x0 = 0 and x1 = x20 +
1 4
1 4
=
0 we have
xn+1 : and so
1 1 and x0 = 0 = x1 : 2 4 Hence, the assertion holds for n = 0. Now suppose that the assertion holds for n = k, that is, 0
x0
0 For 0 0 xk
x 1 2
xk
1 2
and
xk
xk+1 :
1 , 2
x2 +
it is straightforward to show that x and xk+1 = x2k + 14 we have 0
and hence 0
xk+1
xk 1 , 2
xk+1 =
x2k
1 + 4
1 2
2
+
1 4
1 . 2
Since
1 1 = 4 2
and xk+1
x2k+1 +
1 = xk+2 : 4
Hence, the inductive step holds and so the sequence fxn gn is bounded and increasing. It follows by Bounded Monotone Sequence Theorem that it converges, say xn ! x. Note that the real-valued function f given by f (t) = t2 + 41 for t 2 R is continuous and therefore we have, 1 x = lim xn+1 = lim f (xn ) = f (x) = x2 + : n!1 n!1 4 It follows that x = x2 +
1 4
and so x = 12 . Hence, xn ! 10
1 2
as required.
II.1.8 Show that if sn ! s, then jsn
sn 1 j ! 0.
Solution We use standard "=2 proof and let sn ! s. Let " > 0, choose N such that jsn
sj < "=2
sn j
jsn+1
for n > N . If n > N , then we have jsn+1
sn j = jsn+1
s+s
sj + jsn
Thus jsn+1
sn j ! 0
as n ! 1.
11
sj < "=2 + "=2 = ":
II.1.9 Plot each sequence and determine its lim inf and lim sup. (c) sn = sin ( n=4) (a) sn = 1 + n1 + ( 1)n n (b) sn = ( n) (d) sn = xn (x 2 R …xed) Solution (a) lim sup sn = 2 lim inf sn = 0 (b) lim sup sn = 1 lim inf sn = 1 (c) lim sup sn = 1 lim inf sn = 1 (d) 8 jxj > 1; < +1; 1; jxj = 1; lim sup sn = : 0; jxj < 1: and 8 1; 1 > > > x = 1; < 1; 0; jxj < 1; lim inf sn = > > 1; x = 1; > > : +1; x > 1:
1;
12
II.1.10 At what points are the following function continuous? Justify your answer. (a) z (c) z 2 = jzj (b) z= jzj (d) z 2 = jzj3 Solution (a) continuous everywhere (b) Continuous except at z = 0, where it is not de…ned (c) Continuous for jzj = 6 0. It is not de…ned at z = 0, but it has a limit 0 as z ! 0. If we de…ne the function to be 0 at z = 0, it is contionuous there. (d) Contionuous for jzj = 6 0. It has no limit as z ! 0
13
II.1.11 At what points does the function Arg z have a limit? Where is Arg z continuous? Justify your answer.
Solution Arg z have a limit at each point of Cn ( 1; 0]. It is continuous in Cn ( 1; 0]. It is discontinuous at each point of ( 1; 0]. Values of function ! f g at points of ( 1; 0], the function may take any value in intervall [ ; ] as z ! 0. The value is depending in that direktion we approach z = 0.
14
II.1.12 Let h (z) be the restriction of the function Arg z to the lower halfplane fIm z < 0g. At what points does h (z) have a limit? What is the limit?
Solution No limit at 0, has limit at all other points of closed lower half-plane, limit = at points of ( 1; 0).
15
II.1.13 For which complex values of does the principal value of z have a limit as z tends to 0? Justify your answer.
Solution Set z = rei = eLog r ei = eLog r+i and
z = eLog r+i
Re +i Im
= eLog r Re e
Im
= eLog r Re ei(Log r Im
+ i Im , then
= Re
+i Log r Im +i Re + Re )
= rRe e
Im Im
=
ei(Log r Im
+ Re )
;
thus jz j = rRe e
Im
:
Because e Im is bounded we look at rRe in three intervals If Re < 0, rRe ! 1 as r ! 0 thus jz j have no limit as z ! 0, and not z either. If Re = 0, we get z = e Im , have no limit unless Im = 0, thus we have limit if = 0. If Re > 0, rRe ! 0 as r ! 0 thus jz j have limit 0 as z ! 0, and z ! 0 as z ! 0. The conclution is z ! 0 if Re > 0, and z ! 1 if = 0, otherwise we have no limit at 0.
16
II.1.14 Let h (t) be a continuous complex-valued function on the unit interval [0; 1], and consider Z 1 h (t) dt: H (z) = z 0 t
Where is H (z) de…ned? Where is H (z) continuous? Justify your answer. Hint. Use the fact if jf (t) g (t)j < " for 0 t 1, then R1 jf (t) g (t)j dt < ". 0 Solution We have that H (z) is de…ned for z 2 Cn [0; 1]. If h (z1 ) = 0 for some z1 2 [0; 1]. Then H (z) is also de…ned for z = z1 . R1 Then H(z) = h(t) dz is continuous for z 2 Cn [0; 1]. If zn ! z 2 Cn [0; 1], t z 0
then th(t) ! h(t) uniformly for 0 zn t z H (zn ) ! H (z).
t
17
1, so the used proof shows that
II.1.15 Which of the following sets are open subsets of C? Which are close? Sketch the sets. (a) The punctured plane, Cn f0g . (b) The exterior of the open unitdisk in the plane, fjzj 1g . (c) The exterior of the closed unitdisk in the plane, fjzj > 1g . (d) The plane with the open unitinterval removed, Cn (0; 1) . (e) The plane with the closed unitinterval removed, Cn [0; 1] . (f) The semidisk, fjzj < 1; Im (z) 0g . (g) The complex plane, C. Solution (a) open (c) open (e) open (b) closed (d) neither (f) neither Note: Sketches are di¢ cult to do reasonably.
18
(g) both open and closed
II.1.16 Show that the slit plane Cn ( 1; 0] is star-shaped but not convex. Show that the slit plane Cn [ 1; 1] is not star-shaped. Show that a punctured disk is not star-shaped.
Solution (a) Cn ( 1; 0] is star–shaped, because we can see every point from z = 1. To show that Cn ( 1; 0] is not convex we may choose any two points such that the stright line segment joining them contains a point in ( 1; 0]. For example take 1 + i and 1 i. These two points are joined by a vertical line which contain 1. From this follows that Cn ( 1; 0] is not convex. (b) Cn [ 1; 1] is not star–shaped. Given any z we have that z can not bee seen from a stright line between z and z that must contain 0. We also have that if z belong to the domain then z belongs also to the domain. (c) Same argument as in (b) shows that Cn f0g is not star–shaped.
19
II.1.17 Show that a set is convex if and only if it is star-shaped with respect to each of its points.
Solution (A. Kumjian) Let D C, since the empty set trivially satis…es both conditions we will assume that D 6= ?. Suppose …rst that D is convex. We must show that D is star-shaped with respect to each of its points. So …x z0 2 D, to show that D is star-shaped with respect to z0 . We must show that for every point z 2 D the line segment connecting z0 to z is contained in D. Let z 2 D be given, then since D is convex it contains the line segment joining z0 and z. Hence, D is star-shaped with respect to z0 , then, since z0 was chosen arbitrarily, D is star-shaped with respect to each of its points. Conversely, suppose that D is star-shaped with respect to each of its points. To show that D is convex we must show that given two points in D, the line segment joining them is also contained in D. So let z0 ; z1 2 D be given. Then since D is star-shaped with respect to z0 , D contains the line segment joining z0 and z1 . Hence, D is convex.
20
II.1.18 Show that the following are equivalent for an open subset U of the complex plane. (a) Any two points of U can be joined by a path consisting of straight linesegments parallel to the coordinate axis. (b) Any continuously di¤erentiable function h (x; y) on U such that rh = 0 is constant. (c) If V and W are disjoint open subsets of U such that U = V [ W , then either U = V or U = W . Remark. In the context of topological spaces, this latter property is taken as de…nition of connectedness.
Solution Show that the following equivalent for an open subset U of the complex plane. (a) ) (b) Suppose rh = 0 in D. Fix z0 2 D. If z1 2 D, join by polygonal curve. rh = 0 ) h is constant on each segment of curve ) h (z0 ) = h (z1 ). ) h is constant in D. (b) ) (c) Suppose rh = 0, h not constant, say h (z0 ) = 0 for some z0 . Let W = fh = 0g, V = fh 6= 0g. Evidently V is open. Since rh = 0, h is constant in a neighborhood of each point, so fh = 0g is open and W is open. W 6= ?, V 6= ?, W \ V = ?, W [ V = U (c) ) (a) Suppose z0 2 U . Let D = points that can be joined to z0 by a polygonal curves, intervals parallel to coordinate axis. Evidently D is open. Also if can connected points near z0 2 U , then can connect to U , so U nD is open. By (c), U nD must be empty, some D 6= ?:
21
II.1.19 Give a proof of the fundamental theorem of algebra along the following lines. Show that if p (z) is a nonconstant polynomial, then jp (z)j attains its minimum at some point z0 2 C. Assume that the minimum is attained at z0 = 0, and that p (z) = 1 + az m + , where m 1 and a 6= 0. Contradict the minimality by showing that P "ei 0 < 1 for an appropriate choice of 0 . Solution (K. Seip) Set p (z) = an z n + an 1 z n
1
+
+ a0 ;
where we assume an 6= 0. Since jp (z)j n = jan j ; jzj!1 jzj lim
9 R > 0 such that jp (z)j > ja0 j for all jzj > R. Thus jp (z)j attains its minimum in jzj R. We may assume it is attained at z0 = 0. Suppose a0 6= 0, say a0 = 1. Then p (z) = 1 + az m + higher order
terms, a 6= 0. Choose z = " Then
p (z) = 1
1 a
1 m
(any m
th root does the job).
"m + O "m+1 :
Thus for " su¢ ciently small " we have jp (z)j < 1, which is a contradiction.
22
Follow the suggestion. Let p (z) = a0 + a1 z + am+1 z m+1 + ::: + aN z N ; aN 6= 0: Choose R so that jaN j RN > ja0 j + ja1 j R + jam+1 j Rm+1 + ::: + jaN
1j R
N 1
:
Then p (z) 6= 0 for jzj = R, the term aN z N dominates and further jp (z)j > jp (0)j for jzj = R. Let z0 be a point in the disk jzj 6 R at which the continuous function for jp (z)j attains a minimum. Then jz0 j < R, so that is a disk centred at z0 so that jp (z)j > jp (z0 )j on the disk. Suppose p (z0 ) 6= 0. We can assume that p (z0 ) = 1. Write z0 )m + O (z
p (z) = 1 + am P (z where am 6= 0. Suppose am = r0 ei 0 . Conside zm = z0 + "e i
0 =m
e
i =m
z0 )m+1 ;
; " > 0:
We have p (zm ) = 1 + r0 ei 0 em e Have p (zm ) 1 that p (z0 ) = 0.
i
0
e
i
+ O "m+1 = 1
r0 "m + O "m+1 :
r0 "m < 1 for " > 0 small. Contradiction! We conclude
23
II.2.1 Find the derivatives of the following function. n (a) z 2 1 (c) (z 2 1) (e) 1= (z 2 + 3) (b) z n 1 (d) 1= (1 z) (f) z= (z 3 5) Solution (a) 2z (b) nz n
1
(c) n (z 2 (d) 1= (1
n 1
1) z)2
(g) (az + b) = (cz + d) (h) 1= (cz + d)2 2
(e) 2z= (z 2 + 3) (f) ( 2z 3 5) = (z 3
2z
24
2
5)
(g) (ad bc) = (cz + d)2 (h) 2c= (cz + d)3
II.2.2 Show that 1 + 2z + 3z 2 +
+ nz n
1
=
zn z)2
nz n : 1 z
z n+1 ; 1 z
z 6= 1:
1 (1
Solution Use the geometric sum 1 + z + z2 + z3 +
+ zn =
1
Di¤erentiate both sides
1 + 2z + 3z 2 +
+ nz n 1 = (n + 1) ( 1) z n (1 z) + (1 z n+1 ) = = (1 z)2 nz n+1 nz n + z n+1 z n + 1 z n+1 = = (1 z)2 1 zn nz n+1 nz n 1 zn = + = (1 z)2 (1 z)2 (1 z)2
25
nz n : (1 z)
II.2.3 Show from the de…nition that the functions x = Re z and y = Im z are not complex di¤erentiable at any point.
Solution Di¤erentiation f (z) = x = Re z (from the de…nition)
lim
f (z +
z!0
z) z
f (z)
= lim
z!0
Re (z +
z) Re (z) = z Re z 1; if = = lim 0; if z!0 z
z= x : z=i y
Di¤erentiation g (z) = y = Im z (from the de…nition)
lim
z!0
g (z +
z) z
g (z)
=
lim
=
lim
Im (z +
z!0
Im z = z!0 z
z) z
Im (z) 0;
if i; if
= z= x : z=i y
The functions x = Re z and y = Im z are not complex di¤erentiable at any point, because the functions have di¤erent limit as z ! 0 through real and imaginary axis.
26
II.2.4 (+ IV.8.2) Suppose f (z) = az 2 + bz z + cz 2 , where a, b, and c are …xed complex numbers. By di¤erentiating f (z) by hand, show that f (z) is complex di¤erentiable at z if and only if bz + 2cz = 0. Where is f (z) analytic? Solution We will …nd f (z +
=
a (z +
z) z
d f dz
(z) using the limit de…nition
f (z)
z)2 + b (z +
a z 2 +2z z + =
=
z2
2
z) + c(z + z) (az 2 + bz z + cz 2 ) = z 2 + b z z + z z+z z + z z + c z 2 + 2z z + z z) (z +
z 2 2az z + a z + bz z + bz z + b z z + 2cz z + c z = = z 2 z z z = 2az + a z + bz + bz + b z + 2cz +c = z z z z = 2az + a z + bz + b z + c z + (bz + 2cz) z 2
We know z is not analytic at any open set of C and lim
4z!0
z z
z z
does not ex-
d ist. Therefore, unless bz + 2cz = 0, the derivative dz f (z) would not exist. Therefore f (z) is analytic on C if b = c = 0, otherwise is the function not analytic on any open set.
27
(az 2 + bz z + cz 2 )
II.2.5 Show that if f is analytic on D, then g (z) = f (z) is analytic on the re‡ected domain D = fz : z 2 Dg, and g 0 (z) = f 0 (z). Solution (A. Kumjian) Let f be analytic on the domain D and de…ne g on D as above (note that D is also a domain). For a complex-valued function ' de…ned near a point z0 it is easy to show that lim ' (z) exists i¤ lim ' (z) exists;
z!z0
z!z0
and if either exists, then the two limits are complex conjugates of each other. Let z 2 D be given. We …rst show that g is di¤erentiable at z and that g 0 (z) = f 0 (z). 1 lim (g (z + h) h!0 h
1 f z+h h!0 h
g (z)) = lim
1 f (z) = lim (f (z + k) k!0 k
f (z)) = f 0 (z)
where equation ( ) follows using the substitution k = h and basic algebraic properties of conjugation, note that conjugation is a continuous function so h ! 0 i¤ h ! 0. Since f is analytic on the domain D, its derivative f 0 is continuous on D. Further, since g 0 (z) = f 0 (z) for all z 2 D , g 0 is the composite of continuous functions and therefore is itself continuous. Hence, g is analytic on D .
28
II.2.6 Let h (t) be a continuous complex-valued function on the unit interval [0; 1], and de…ne Z 1 h (t) dt; z 2 Cn [0; 1] : H (z) = z 0 t
Show that H (z) is analytic and compute its derivative. Hint. Differentiate by hand, that is, use the de…ning identity (2.4) of complex derivative.
Solution We use de…nition (2.4) of derivative to …nd H 0 (z).
H 0 (z) =
Z 1 h (t) h (t) dt dt = = lim z!0 z z z 0 t 0 t Z 1 1 1 1 = lim h (t) = z!0 z t z z t z 0 Z 1 Z 1 h (t) h (t) dt = = lim = z!0 z z) (t z) z) (t z) 0 (t 0 (t Z 1 h (t) = dt: z)2 0 (t H (z +
z) z
H (z)
1 = lim z!0 z
Z
1
H 0 (z) is continuous in z, also by uniform convergence of integrand H (z) is analytic for z 2 Cn [0; 1].
29
II.3.1 Find the derivatives of the following functions. sin z sinh z (a) tan z = cos (b) tanh z = cosh (c) sec z = 1= cos z z z Solution (a) d d sin z cos z cos z sin z ( sin z) cos2 z + sin2 z 1 tan z = = = = : 2 2 dz dz cos z cos z cos z cos2 z (b) cosh2 z sinh2 z 1 d d sinh z cosh z cosh z sinh z sinh z = = : tanh z = = 2 2 dz dz cosh z cosh z cosh z cosh2 z (c) d d 1 sec z = = dz dz cos z
1 sin z ( sin z) = = tan z sec z: cos2 z cos2 z
30
II.3.2 Show that u = sin x sinh y and v = cos x cosh y satisfy the CauchyRiemann equations. Do you recognize the analytic function f = u + iv? (Determine its complex form)
Solution We have @ @ @v @u = sin x sinh y = cos x sinh y = cos x cosh y = ; @x @x @y @y @u @ @ @v = sin x sinh y = sin x cosh y = cos x cosh y = : @y @y @x dx Hence, u and v satisfy Cauchy-Riemann equations, and now we calculate f (z).
f (z) = e ix ey e y eix + e ix ey + e y +i = 2i 2 2 2 eix+y + eix y + e ix+y + e ix eix+y eix y e ix+y + e ix iy +i 4 4 ix y ix+y i(x+iy) i(x+iy) iz e +e e +e e + e iz =i =i =i = 2 2 2 = i cos z:
= u + iv = =
i
eix
31
y
=
II.3.3 1 2 3 P
L K LLL Show that if f and f are both analytic on a domain D, then f is constant.
Solution (A. Kumjian) Let D be a domain and let f be a complex valued function such that both f and f are analytic on D. Then both Re f = 21 f + f and Im f = 2i1 f f must also be analytic on D. Since Re f and Im f are also both real-valued, it follows by the theorem on page 50, that both are constant. Hence, f = Re f + i Im f is constant.
32
II.3.4 Show that if f is analytic on a domain D, and if jf j is constant, then f is constant. Hint. Write f = jf j2 =f . Solution (with use of hint) If f (z) = 0 for some z 2 D. Then f 0 on D, since jf j is constant. Assume that f (z) 6= 0 for every z 2 D. Since f is analytic in D we have that 1=f and f = jf j2 =f (where jf j is constant) are analytic. From this follows that Re f and Im f are analytic and real-valued and therefore constant. So f = Re f + i Im f is constant.
33
II.3.4 Show that if f is analytic on a domain D, and if jf j is constant, then f is constant. Hint. Write f = jf j2 =f . Solution Set f = u + iv, where u and v are realvalued functions. Now suppose that jf j = k is constant. We have that u2 + v 2 = k 2 . Di¤erentiate both sides with respect to x respectively y 0
0
uux + vvx = 0; 0 0 uuy + vvy = 0: 0
0
(1) 0
We use Cauchy-Riemanns equations ux = vy and uy = 0
0
vx in (1) and have
0
uux vuy = 0; 0 0 uuy + vux = 0:
(2)
In the simultaneusly systems of equations in (2) we …rst …rst multiply the …rst row with u and second row with v in (2) and add them together, and 0 have (u2 + v 2 ) ux = 0. And if we in the same system multiply the …rst row 0 with v and the second row with u and add them we have (u2 + v 2 ) uy = 0. Thus 0
(u2 + v 2 ) ux = 0; 0 (u2 + v 2 ) uy = 0:
(3)
If in (3) u2 + v 2 = 0, and we have that u = v = 0 and thus f is constant. 0 0 Suppose that u2 + v 2 6= 0. We have that (3) gives that ux = uy = 0, with says that u is constant. Cauchy-Riemanns equations gives 0
0
0 = ux = v y ; 0 0 0 = uy = v y ; thus by (4) f is constant.
34
(4)
II.3.5 If f = u + iv is analytic, then jruj = jrvj = jf 0 j. Solution (K. Seip) Because f = u + iv is analytic. Then ru =
@u @u ; @x @y
=
@u ; @x
@v @x
=
from which it follows that jruj = jf 0 (z)j = jrvj :
35
@v ; @y
@v @x
II.3.6 If f = u + iv is analytic on D, then rv is obtained by rotating ru by 90 . In Particular, ru and rv are orthogonal. Solution (A. Kumjian) Let 2 R then rotation of a vector (x; y) 2 R2 by the angle origin) is given by matrix multiplication: x y
x y
7! A
cos sin
=
sin cos
x y
(about the
x cos y sin x sin + y cos
=
:
Hence, rotation by 90 is given by the transformation (x; y) 7! ( y; x). By the Cauchy-Riemann equations we have rv =
@v @v ; @x @y
@u @u ; @y @x
=
=A
=2 ru:
Thus, rv is obtained by rotating ru by 90 . It follows that ru and rv are orthogonal. Solution (D. Jakobsson) Given since f is analytic, then (0.6)
@v @u = @x @y
and
@v @v Since ru = @u ; @u and rv = @x ; @y @x @y rv rv = 0. To prove that we use (0.6).
@u = @y
@v @x
all we need to check is whether
@v @u @v @u @v @v @v @v + = =0 @x @x @y @y @x @y @y @x We conclude that ru and rv are orthogonal. Notice that the operation that sends ru ! rv is given by the counter-clockwise rotation matrix with = 2 . Keeping in mind (0.6), one can write the following. ru rv =
rv =
vx vy
=
0 1 1 0
vy vx
= 36
cos 2 sin sin 2 cos 2
2
vy vx
=R
2
ru
Remark. We have that
vx =
@v ; @x
vy =
@v ; @y
ux =
@u ; @x
and R ( =2) = A
37
=2 :
uy =
@u ; @y
II.3.7 Sketch the vector …elds ru and rv for the following functions f = u + iv. (a) iz (b) z 2 (c) 1=z Solution
II.3.7a
II.3.7a y
II.3.7a y
4
-2
2
4
-2
-4
x
-2
2
4
-2
-4
4 2
2
2 -4
y
4
-4
x
-2
2 -2
-4
-4
(a) f (z) = iz =
u= y ) v=x
y + ix )
(b) f (z) = z 2 = x2
ru = (0; 1) rv = (1; 0)
u = x2 y 2 ) v = 2xy
y 2 + 2ixy )
ru = (2x; 2y) rv = (2y; 2x)
(c) f (z) =
cos x u = x2 +y 1 z x iy 2 = r = 2 = 2 ) y 2 v = = z x + y 2 2 jzj x +y 8 2 2 2 2 @u 2 > = (xy2 +yx2 )2 = sin r2cos = cos > @x r2 > > 2xy 2 cos sin < @u = = sin2 2 2 = 2
> > > > :
@y @v @x @v @y
=
=
(x2 +y 2 ) 2xy (x2 +y 2 )2 y 2 x2 (x2 +y 2 )2
ru = rv =
1 r2 1 r2
=
=
r 2 cos sin r2 sin2 cos2 r2
r sin 2 2 r 2 = cos r2
=
( cos 2 ; sin 2 ) ( sin 2 ; cos 2 )
38
sin r
4
x
II.3.8 (see III.4.2) (36) Derive the polar form of the Cauchy-Riemann equations for u and v, @u @u ; = r @v : = 1r @v @ @ @r @r Check that for any integer m, the functions u rei = rm cos (m ) and v rei = rm sin (m ) satisfy the Cauchy-Riemann equations.
Solution Set x = r cos and y = r sin , then take the partial derivative, @x @r @y @r
= cos = sin
@x @ @y @
= r sin = r cos
Use Cauchy-Riemanns equations and the derivative to get the …rst equation @u @u @x @u @y @u @u @v @v = + = cos + sin = cos sin = @r @x @r @y @r @x @y @y @x 1 @v @v 1 @v @x @v @y 1 @v = r sin + r cos = + = ; r @x @y r @x @ @y @ r@ Use Cauchy-Riemanns equations and the derivative to get the second equation @u @x @u @y @u @u @v @v @u = + = r sin + r cos = r sin r cos = @ @x @ @y @ @x @y @y @x @v @v @v @x @v @y @v = r cos + sin = r + = r : @x @y @x @r @y @r @r we get @u @r @u @
= =
1 @v ; r@ r @v : @r
Set u rei = rm cos (m ) and v rei Cauchy-Riemann’s equations, 39
= rm sin (m ) and use it in the
@u @ m = (r cos (m )) = mrm 1 cos (m ) = @r @r @u @ m = (r cos (m )) = mrm sin (m ) = @ @
40
1 m 1 @ m 1 @v r m cos (m ) = (r sin m ) = ; r r@ r@ @ @v rmrm 1 sin (m ) = r (rm sin (m )) = r : @r @r
II.4.1 Sketch the gradient vector …elds ru and rv for (a) u + iv = ez (b) u + iv = Log z Solution a) We have u + iv = ez = ex cos y + iex sin y and get ru = (ex cos y; ex sin y) = ex (cos y;
sin y)
(in black), and rv = (ex sin y; ex cos y) = ex (sin y; cos y) (in red). b) We have u + iv = Log z = log r + i and get 1! ur ; r 1! rv = u : r
ru =
FIGURE II.4.1a NF FIGURE II.4.1b NF
41
II.4.2 Let a be a complex number a 6= 0, and let f (z) be an analytic branch of z a on Cn ( 1; 0]. Show that f 0 (z) = af (z) =z. (Thus f 0 (z) = az a 1 , where we pick the branch of z a 1 that corresponds to the original branch of z a divided by z.)
Solution We may write the branch as follows f (z) = z a = ea log z = ea(Log z+2 where c = e2
iam
im)
= ea Log z+2
iam
= cea Log z ;
for some m 2 Z. It follows by the chain rule that f 0 (z) = cea Log z
So, the desired result follows.
42
af (z) a = : z z
II.4.3 p Consider the branch of f (z) = z (1 z) on Cn [0; 1] that has positive imaginary part at z = 2. What is f 0 (z)? Be sure to specify the branch of the expression for f 0 (z). Solution p Set f (z) = z (1
f 0 (z) =
z) (principal branch) and di¤erentiate 1 1 p 2 z (1
d (z (1 z) dz
z)) =
1 1 2z p : 2 z (1 z)
Branch of f (z) is i increasing for x > 0 large, thus f 0 (z) is i positive for x > 0 large. p Use branch of f 0 (z) = z (1 z) that it is i positive, i.e., same on f (z). f 0 (z) =
1 1 2z : 2 f (z)
Solution (D. p Jakobsson) Given f (z) = z (1 z), we can de…ne w = f (z) and get w2 = z (1 d (1 dz 1
z) z =
2z = 2w
z).
d 2 w dz
dw dz
One gets dp z (1 dz
1 2z z) = p 2 z (1 z)
and is de…ned on Cn [0; 1]. Take the principal branch with the positive imaginary part.
43
II.4.4 Recall that the principal branch of the inverse tangent function was de…ned on the complex plane with two slits on the imaginary axis by Tan
1
=
1 2i
Log
1+iz 1 iz
z 62 ( i1; i] [ [i; i1) :
;
Find the derivative of Tan 1 z. Find the derivative of tan 1 z for any analytic branch of the function de…ned on a domain D.
Solution We have that d Tan dz
1
z=
1 1 i 2i 1 + iz
Any two branches of tan
1 1 1
iz
( i) =
1 1 iz + 1 + iz 1 : = 2 (1 + iz) (1 iz) 1 + z2
z di¤er by a constant, so derivatives are same.
44
II.4.5 p Recall that cos 1 (z) = i log z z 2 1 . Suppose g (z) is an ana1 lytic branch of cos (z), de…ned on a domain D. Find g 0 (z). Do di¤erent branches of cos 1 (z) have the same derivative?
Solution We have that d cos dz
1
z= i 1 2 1=2 1 z 1 2z = 2 2 z z 1 i z i p = 1 p =p = 2 2 2 z z 1 z 1 z 1 =
p
=p
Derivatives of branches of cos
1
z are not always the same.
45
1 1
z2
:
II.4.6 Suppose h (z) is an analytic branch of sin 1 (z), de…ned on a domain D. Find h0 (z). Do di¤erent branches of sin 1 (z) have the same derivative?
Solution (A. Kumjian) Regard h (z) as a local inverse of f (z) where f (z) = sin z. Then f 0 (z) = cos z, so using the formula given in the statement of theorem on page 51 we have 1 1 = h0 (z) = 0 f (h (z)) cos h (z) if cos h (z) 6= 0. One local inverse of f satis…es h1 (0) = 0, while another local inverse of f satis…es h2 (0) = . Using the formula above we see 0
h1 (0) =
1 =1 cos 0
Hence, di¤erent branches of sin ative.
and 1
0
h2 (0) =
1 cos
=
1:
(z) do not necessarily have the same deriv-
46
II.4.7 Let f (z) be a bounded analytic function, de…ned on a bounded domain D in the complex plane, and suppose that f (z) is one-oneone. Show that the area of f (D) is given by ZZ 2 Area (f (D)) = jf 0 (z)j dxdy: D
Solution (K. Seip) We have f : D ! C, jf (z)j M for some M < 1 when z 2 D. Since f is assumed to be one-to-one, we may compute ZZ A (f (D)) = dudv f (D)
by the change of variables (u (x; y) ; v (x; y)) ! (x; y), and since det Jf = jf 0 (z)j2 , we get ZZ 2 jf 0 (z)j dxdy: Area (f (D)) = D
47
II.4.8 Sketch the image of the circle fjz Compute the area of the image.
1g under the map w = z 2 .
1j
Solution
II.4.8 z-plane y
II.4.8 w-plane y
4
4
2
-4
-2
2
2 -2
4
-4
x
-2
2
4
-2
-4
x
-4
We have that f (z) = z 2 , that gives f 0 (z) = 2z and Jf (z) = 4 jzj2 . The area of the image is ZZ ZZ Jf dxdy = 4 x2 + y 2 dxdy: (x 1)2 +y 2 61
Set t = x
4
ZZ
t2 +y 2 61
1, get
t2 + 2t + 1 + y 2 dtdy = 2
6 = 44
ZZ
t2 + y 2 dtdy + 0 +
t2 +y 2 61
ZZ
t2 +y 2 61
3
7 dtdy 5 = =4
h
2
+
i
=6 :
Solution (D. Jakobsson) d Given w = f (z) = z 2 , one can di¤erentiate f (z) and get dz f (z) = 2z = 2 (x + iy), we can use maple to integrate and get the following 48
Area (f (D)) = ZZ Z 2 0 = jf (z)j dxdy = =
Z
0
p 2Z 1 p
D
(x 1)
0
2
Z p1 p
2
4 x2 + y 2 dx dy =
1 (x 1)2
Z
(x 1)2
1 (x 1) 2
2
4 jx + iyj2 dx dy =
p 8x2 2x
x2 +8=3 2x
x2
3=2
dx =
0
=6 Change of variables x = 1 + sin t dx = cos tdt Z
=2
8 (1 + sin t)2
=2
Z
=2
p
1
sin2 t +
8 1 3
sin2 !
3=2
cos tdt =
8 4 2 2 8 1 + 2sin |{z}t + sin t cos + 3 cos tdtdt = =2 odd Z =2 2 = 16 2 cos2 cos4 tdt = 3 0 Z =2 1 =8 2 (1 + cos 2t) 1 + 2 cos 2t + cos2 2t dt = 3 0 Z =2 5 4 1 =8 + cos 2t (1 + cos 4t) dt = 3 3 3 0 5 1 9 =8 =4 =6 : 2 3 6 6 =
49
II.4.9 Compute ZZ
D
2
2
jf 0 (z)j dxdy;
for f (z) = z and D the open unit disk fjzj < 1g. Interpret your answer in terms of areas.
Solution By the result above ZZ ZZ 2 2 0 jf (z)j dxdy = jf 0 (z)j dx dy = ; jzj<1 x>0
jzj<1 x<0
thus ZZ
2
jf 0 (z)j dx dy = 2 :
jzj<1
Remark. The function maps the top and bottom halves of the unit disk one-to-one onto the unit disk, so by Exersice II.4.7 the integrals on top and bottom halves are each .
50
II.4.10 1 2 3 P L K For smooth functions g and h de…ned on a bounded domain U , we de…ne the Dirichlet form DU (g; h) by ZZ
@g @h @g @h + dx dy @y @y U @x @x Show that if z = f ( ) is a one-to-one analytic function from the bounded domain V onto U , then DU (g; h) =
DU (g; h) = DV (g f; h f ) : Remark. This shows that the Dirichlet form is a "conformal invariant". Solution RR Du (g; h) = rgrhdxdy Assume h; g ???? U RR Du (g; h) = r (g f ) r (h f )dudv = U RR g (u (x; y) ; v (x; y)) h (u (x; y) ; v (x; y)) dudv
U @g(u(x;y);v(x;y)) @h(u(x;y);v(x;y)) + @x @x @g(u(x;y);v(x;y)) @h(u(x;y);v(x;y)) = @y @y @g @u @g @v @g @u @g @v @h @u @v @h @u @v + @v + @h + @u + @v + @h @u @x @x @u @x @v @x @y @y @u @y @v @y @g @u @h @v @g @v @h @u @g @h @v 2 @g @h @u 2 + @u + @v + @v + @u @u @x @x @v @x @x @u @x @v @x 2 2 @g @h @u @g @v @h @u @g @h @v @g @u @h @v + @v + @v + @u = @u @u @y @y @v @y @y @u @y @v @y 2 2 @g @h @v 2 @g @h @u @g @h @v @g @h @u 2 + + + = @u @u @x @v @v @x @v @v @y ! @u @u @y ! 2 2 2 2 @u @u @v @v @g @h @g @h + + @v + + @u @u @v
|
@x
{z
jf 0 (z)j2
@y
}
|
@x
{z
jf 0 (z)j2
@y
}
@g @u @h @v @g @v @h @u @g @u @h @v @g @v @h @u + + + @u @x @v @x @v @x @u @x @u @y @v @y @v @y @u @y | {z } Use C-R equations RR RR rgrh jf 0 (z)j2 dxdy = rgrhdudv U
U
51
=
II.5.1 Show that the following functions are harmonic, and …nd its harmonic conjugates: (a) x2 y 2 (c) sinh x sin y (e) tan 1 (y=x) ; x > 0 2 2 (b) xy + 3x2 y y 3 (d) ex y cos (2xy) (f) x= (x2 + y 2 ) Solution (a) Set u (x; y) = x2
y2 )
0
ux = 2x ) 0 uy = 2y
00
uxx = 2 ) 4u = 0; 00 uyy = 2
thus is the function u (x; y) is harmonic. Cauchy Riemann give us 0
0
0
vx = uy = 2y ) v (x; y) = 2xy + g (y) ) vy = 2x + g 0 (y) 0 0 vy = ux = 2x Thus we have that g 0 (y) = 0 ) g (y) = C; thus v (x; y) = 2xy + C: (b) Set
u (x; y) = xy+3x2 y y 3 )
0
ux = y + 6xy 0 uy = x + 3x2
00
3y 2
)
uxx = 6y ) 4u = 0; 0 uyy = 6y
thus is the function u (x; y) is harmonic. Cauchy Riemann give us 0
0
vx = uy = x 3x2 + 3y 2 ) v (x; y) = 0 0 vy = ux = y + 6xy Thus we have that
52
x2 2
0
x3 + 3xy 2 + g (y) ) vy = 6xy + g 0 (y)
y2 g (x) = y ) g (y) = + C; 2 0
thus x2 y2 + 3xy 2 x3 + C: 2 2 In paranthesis we remark that the analytic function f (z) = u + iv is given by v (x; y) =
f (z) = xy +3x2 y y 3 +i
y2 + 3xy 2 2
x2 2
x3 + C
=
iz 2 (z + 1=2)+iC:
(c) Set 0
u (x; y) = sinh x sin y )
ux = cosh x sin y ) 0 uy = sinh x cos y
00
uxx = sinh x sin y ) 4u = 0; 0 uyy = sinh x sin y
thus the function u (x; y) is harmonic. Cauchy Riemann give us 0
0
vx = uy = sinh x cos y ) v (x; y) = 0 0 vy = ux = cosh x sin y
0
cosh x cos y + g (y) ) vy = cosh x sin y + g 0 (y)
Thus we have that g 0 (y) = 0 ) g (y) = C; thus v (x; y) =
cosh x cos y + C
In paranthesis we remark that the analytic function f (z) = u + iv is given by
53
f (z) = sinh x sin y i (cosh x cos y + C) = eiy e iy ex + e x eiy + e iy ex e x i + iC = = 2 2i 2 2 i = ex e x eiy e iy + ex + e x eiy + e iy + iC = 4 ex+iy + e (x+iy) + iC = i cosh z + iC: = i 2 (d) Set
u (x; y) = ex 00
)
2
y2
0
cos (2xy) )
2
2
2
2
2
ux = 2xex y cos (2xy) 2yex y sin (2xy) ) 2 2 2 2 0 uy = 2yex y cos (2xy) 2xex y sin (2xy)
2
uxx = 2ex y (cos 2xy + 2x2 cos 2xy 2y 2 cos 2xy 4xy sin 2xy) ) 4u = 0; 0 2 2 uyy = 2ex y (cos 2xy + 2x2 cos 2xy 2y 2 cos 2xy 4xy sin 2xy)
thus the function u (x; y) is harmonic. Cauchy Riemann give us 8 0 0 2 2 2 2 2 < vx = uy = 2yex y cos (2xy) + 2xex y sin (2xy) ) v (x; y) = ex 0 2 2 2 2 ) v = 2yex y sin (2xy) + 2xex y cos (2xy) + g 0 (y) : 0 y 0 2 2 2 2 vy = ux = 2xex y cos (2xy) 2yex y sin (2xy)
y2
sin (2xy) + g (y) )
Thus we have that
g 0 (y) = 0 ) g (y) = C; thus v (x; y) = ex
2
y2
sin (2xy) + C
In paranthesis we remark that the analytic function f (z) = u + iv is given by f (z) = ex
2
y2
cos (2xy) + i ex
2
(e) 54
y2
2
sin (2xy) + C = ez + iC:
Set
1
(y=x) ; x > 0 ) 8 1 0 > < ux = 1 + (y=x)2 =) x > : u0y = 2 x + y2
u (x; y) = tan
y x2
)
=
x2
y + y2
=)
8 00 > > < uxx = 0 > > : uyy
2xy (x2 + y 2 )2 ) 4u = 0; 2xy = (x2 + y 2 )2
thus the function u (x; y) is harmonic. Cauchy Riemann give us 8 0 > < vx =
x ) v (x; y) = a x2 + y 2 y 0 0 > : v y = ux = 2 x + y2 0
uy =
1 2
0
log (x2 + y 2 ) + g (y) ) vy =
y + g 0 (y) 2 2 x +y
Thus we have that
g 0 (y) = 0 ) g (y) = C; thus v (x; y) =
1 log x2 + y 2 + C: 2
(f) Set
u (x; y) =
x2
8 0 > > < ux =
x ) 0 > + y2 > : uy =
8 3y 2 x2 x2 y 2 00 > > u = 2x < xx (x2 + y 2 )3 (x2 + y 2 )2 ) ) 4u = 0; 2xy 3y 2 x2 0 > > : uyy = 2x 2 (x2 + y 2 )2 (x + y 2 )3
thus is the function u (x; y) is harmonic. Cauchy Riemann give us
55
8 0 > > < vx =
y 2xy 0 uy = + g (y) ) vy = 2 ) v (x; y) = 2 2 2 2 (x + y ) (x + y ) 2 2 x y 0 0 > > : v y = ux = (x2 + y 2 )2 0
x2 y 2 0 2 + g (y) 2 2 (x + y )
Thus we have that
g 0 (y) = 0 ) g (y) = C; thus v (x; y) =
(x2
y +C + y2)
In paranthesis we remark that the analytic funktion f (z) = u + iv is given by
f (z) =
x +i x2 + y 2
y +C (x2 + y 2 )
56
=
z z 1 + iC = + iC: 2 + iC = zz z jzj
II.5.2 Show that if v is a harmonic conjugate for u, then conjugate for v.
u is a harmonic
Solution If f (z) = u (x; y) + iv (x; y) is analytic then if (z) = v (x; y) analytic. Thus u is a harmonic conjugate of v.
57
iu (x; y) is
II.5.3 De…ne u (z) = Im (1=z 2 ) for z 6= 0, and set u (0) = 0. (a) Show that all partial derivatives of u with respect to x exist at all points of the plane C, as do all partial derivative of u with respect to y. 2 2 (b) Show that @@xu2 + @@yu2 = 0. (c) Show that u is not harmonic on C. @2u (d) Show that @x@y does not exist at (0; 0). Solution (a) We …rst de…ne 8 <
z2 (x iy)2 2xy 1 = (x; y) 6= (0; 0) Im 2 = Im 4 = Im 2 2 2 2 u (z) = z jzj (x + y ) (x + y 2 )2 : 0 (x; y) = (0; 0)
(x;y) For all (x; y) 6= (0; 0), the n th derivative of u (z) will be (x2f+y 2 )n+2 with f (x; y) a polynomial of degree n + 2 in (x; y). At (0; 0), the function itself is 0, thus u (x; y) 2 C1 (x). By symmetry u (x; y) 2 C1 (y). (b) We have
@u @2u @2u + 2 = 2 @x @y @x
2y
3x2 y 2 (x2 + y 2 )3 =
3y 2 x2 = (x2 + y 2 )3 x2 y 2 x2 y 2 + 24xy =0 24xy (x2 + y 2 )4 (x2 + y 2 )4
+
@u @y
2x
(c) Notice that the function u (x; y) is not harmonic because the function itself, its …rst and second derivative are not continuous at the origin. The limit of u (x; y) as (x; y) ! (0; 0) does not exist, for instance u n1 ; n1 goes to 1 as n ! 1 whereas we de…ned the function u to be 0 at the origin. lim u
n!1
1 1 ; n n
= lim
n!1
58
n2 = 2
1:
(d) @u = @y
2
2x3 + 6xy 2 2x (x2 + y 2 ) + 2xy 2 (x2 + y 2 ) 2y = (x2 + y 2 )4 (x2 + y 2 )3
thus 3
2
@2u ( 6x2 + 6y 2 ) (x2 + y 2 ) ( 2x3 + 6xy 2 ) 3 (x2 + y 2 ) = @x@y (x2 + y 2 )6
2x
=
6x4
We have @2u = @x@y
(
6x4 36x2 y 2 +6y 4 (x2 +y 2 )4
0
@2u (x; 0) = @x@y
(x; y) 6= (0; 0) (x; y) = (0; 0)
6=x4 ; x 6= 0 : 0; x=0
This partial derivative is not continuous at x = 0, so (0; 0). .
59
@2u @x@y
does not exist at
36x2 y 2 + 6y 4 : (x2 + y 2 )4
II.5.4 Show that if h (z) is a complex-valued harmonic function (solution of Laplace’s equation) such that zh (z) is also harmonic, then h (z) is analytic. Solution First partial derivative for zh (z) is, @z @h @ (zh) = h+z ; @x @x @x @z @h @ (zh) = h+z : @y @y @y Second derivative for zh (z) is, @ 2 (zh) @2z @z @h @z @h @2h = h + + + z ; @x2 @x2 @x @x @x @x @x2 @2z @z @h @z @h @2h @ 2 (zh) = h + + + z : @y 2 @y 2 @y @y @y @y @y 2 We have z = x+iy, thus @z=@x = 1, @z=@y = i, and @ 2 z=@x2 = @ 2 z=@y 2 = 0, therefore @ 2 (zh) @h @2h = 2 + z ; @x2 @x @x2 @ 2 (zh) @h @2h = i2 + z : @y 2 @y @y 2 Add the equations @h @h @ 2 (zh) @ 2 (zh) + =2 + 2i +z 2 2 @x @y @x @y
@2h @2h + @x2 @y 2
:
Because h (z) and zh (z) is harmonic, then both @ 2 (zh) =@x2 +@ 2 (zh) =@y 2 = 0 and @ 2 (h) =@x2 + @ 2 (h) =@y 2 = 0, we have @h = @x 60
i
@h : @y
Write h = u + iv, get @v @u +i = @x @x
i
@u @v +i @y @y
=
@v @y
i
@u : @y
Now we take real and imaginary parts and get @u @x
=
@v @y
and
@u @y
=
@v : @x
Thus yields Cauchy-Riemann’s equations for u + v, then h (z) is analytic as was to be shown.
61
II.5.5 Show that Laplace’s equation in polar coordinates is 1 @2u @ 2 u 1 @u + = 0: + @r2 r @r r2 @ 2 Solution Let x = r cos and y = r sin , and get partial derivatives @x = cos ; @r @y = sin ; @r Taking the …rst derivative of u with
@x = r sin ; @ @y = r cos : @ respect to r
@u @u @x @u @y @u @u @u @u = + = cos + sin = cos + sin : @r @x @r @y @r @x @y @x @y Taking the second derivative of u with respect to r @2u = @r2 @ 2 u @x @ 2 u @y @ 2 u @x @ 2 u @y cos + cos + sin + sin = @x2 @r @y@x @r @x@y @r @y 2 @r @ 2 u @x @2u @2u @ 2 u @y = cos + sin cos + cos sin + sin = @x2 @r @y@x @x@y @y 2 @r 2 @2u @2u 2 @ u = cos2 + 2 sin cos + sin : @x2 @x@y @y 2 =
Taking the …rst derivative of u with respect to @u @u @x @u @y @u @u = + = ( r sin ) + r cos : @ @x @ @y @ @x @y Taking the second derivative of u with respect to
62
@2u = @ 2 @ 2 u @y @u @ 2 u @x ( r sin ) + ( r sin ) + ( r cos ) + 2 @x @ @y@x @ @x @ 2 u @x @ 2 u @y @u + (r cos ) + 2 (r cos ) + ( r sin ) = @x@y @ @y @ @y @2u @u @2u = (r cos ) ( r sin ) + ( r cos ) + ( r sin ) ( r sin ) + 2 @x @y@x @x @2u @2u @u + ( r sin ) (r cos ) + 2 (r cos ) (r cos ) + ( r sin ) = @x@y @y @y 2 @2u @2u @u @u 2 2 2 @ u = r2 sin2 2r sin cos + r cos r cos r sin : 2 2 @x @x@y @y @x @y =
Combining these partial derivatives, one gets @ 2 u 1 @u 1 @2u = + + @r2 r @r r2 @ 2 @2u = cos2 + 2 sin cos @x2 1 @2u + 2 r2 sin2 2r2 sin r @x2 @2u + 2 sin cos = cos2 @x2 @2u + sin2 2 sin cos @x2 = sin2
@2u @2u 1 @u @u + sin2 + cos + sin + 2 @x@y @y r @x @y @2u @2u @u @u cos + r2 cos2 r cos r sin 2 @x@y @y @x @y 2 2 @ u @ u 1 @u 1 @u + sin2 + cos + sin + 2 @x@y @y r @x r @y @2u @2u 1 @u 1 @u + cos2 cos sin = 2 @x@y @y r @x r @y @2u @2u @2u @2u 2 2 + sin + cos = + : + cos2 @x2 @y 2 @x2 @y 2
Because u is a harmonic function we have @ 2 u=@x2 + @ 2 u=@y 2 = 0 and we have, @ 2 u 1 @u 1 @2u @2u @2u + + = + = 0: @r2 r @r r2 @ 2 @x2 @y 2
63
=
II.5.6 Show using Laplace’s equation in polar coordinates that log jzj is harmonic on the punctured plane Cn f0g. Solution Set z = rei , we have u (r; ) = log jzj = log r. We can compute the Laplacian of log jzj without worrying about the origin, i.e. r 6= 0 because this case is taken out, thus we have log r =
1 @u2 @ 2 u 1 @u + + = @r2 r @r r2 @ 2
64
1 11 + + 0 = 0: r2 r r
II.5.7 Show that log jzj has no conjugate harmonic function on the punctured plane Cn f0g, though it does have a conjugate harmonic function on the slit plane Cn ( 1; 0). Solution (A. Kumjamin) Set D := Cn ( 1; 0] and let u : D ! R be given by u (z) := log jzj, it is easily veri…ed that u (z) = Re Log z and this shows that u is harmonic on D by the main theorem in section II.5. Moreover, it follows by Cauchy-Riemann equations that a harmonic conjugate v is given by v (z) := Im Log z = Arg z
for
z 2 D:
Any other harmonic conjugate di¤ers from v by a constant. Now if we set ~ := Cn f0g and de…ne u~ : D ! R by u~ (z) := log jzj, it can easily be D ~ shown by an argument similar to the one above that u~ is harmonic on D even though it cannot be expressed as the real part of an analytic function ~ (the property of being harmonic is local). Suppose that u~ does have a on D ~ Then its restriction to D would have to harmonic conjugate, say v~, on D. be a harmonic conjugate of u and so it would be of the form z 7 ! Arg z + c ~ for some c 2 R. But such a function has no continuous extension to D. ~ (since this implies v~ is This contradicts the fact that v~ is harmonic on D ~ continuous on D). Therefore, log jzj has no conjugate harmonic function on the punctured plane Cn f0g. Solution (K. Seip) In Cn ( 1; 0], we know that any harmonic conjugate of log jzj must have the form v (x; y) = Arg z + c: Such a function can not be made continuous in C since for x < 0, we have lim v (x; y) =
y!00+
+ c;
and lim v (x; y) =
y!00
65
+ c:
II.5.8 Show using Laplace’s equation in polar coordinates that u rei = log r is harmonic. Use the polar form of the Cauchy-Riemann equations. (Exercise 3.8) to …nd a harmonic conjugate v for u. What is the analytic function u + iv? Solution Set u rei
= log r )
The function u rei
(
0
ur =
)
r u = log r 0
(
00
urr = u
00
=0
r2
= log r is harmonic because
1 1 u = urr + ur + 2 u = + = 0: r r r2 r2 Cauchy Riemann equations in polar form (page 50) gives us 8 <
1 0 vr = u = : v 0 = rur0 = r 0
log r ) v (r; ) = r
(log r)2 0 + g ( ) ) v = g0 ( ) 2
Thus we have that
2
g0 ( ) =
) g( ) =
2
+ C;
thus 2 (log r)2 + + C: 2 2 We have that the analytic function f (z) = u + iv is given by
v (r; ) =
2
f (z) = log r+i
2
(log r)2 +C 2
!
=
66
i (log r + i )2 +C = 2
i (log z)2 +iC: 2
II.6.1 Sketch the families of level curves of u and v for the following functions f = u + iv. (a) f (z) = 1=z (b) f (z) = 1=z 2 (c) f (z) = z 6 Determine where f (z) is conformal and where it is not conformal.
Solution (a) We have that 8 < u=
x 2 z x iy 1 x + y2 ) f (z) = = 2 = 2 y : v= z x + y2 jzj 2 x + y2
Conformal except at z = 0, where it is not de…ned. (Conformal everywhere if we view it as a mapping of the sphere). We have 1 2 c 2 c2 1 (x + y 2 ) = x; x + y2 = ; u= ; c c 2 4 which is circles centred on x axis and tangent to y axis. We have 1 2 a 2 a2 1 (x + y 2 ) = y; x2 + y + v= ; = ; a a 2 4 which is circles centred on y axis and tangent to x axis. (b) We have 1 z2 (x iy)2 x2 y 2 2ixy f (z) = 2 = 4 = = ) z jzj (x2 + y 2 )2 (x2 + y 2 )2 ( 2 2 2 2 u = (xx2 +yy2 )2 = cos r2sin = =) 2xy 2 cos sin v = (x2 +y = 2 )2 = r2 and
67
cos(2 ) r2 sin(2 ) r2
1 1 u= ; r2 = c cos (2 ) ; v= ; c c Conformal everywhere except at z = 0. (c) We have that
r2 =
f (z) = z 6 = r6 (cos (6 ) + i sin (6 )) ) Figure repeats itself. Rotate of fu = 0g by repeats itself every =6. Conformal everywhere except at z = 0.
c sin (2 ) :
u = r6 cos (6 ) v = r6 sin (6 )
=12 is fv = 0g. Each grind
FIGURE II.6.1a NF FIGURE II.6.1b NF FIGURE II.6.1c NF
68
II.6.2 Sketch the families of level curves of u and v for f (z) = Log z = u+iv. Relate your sketch to a …gure in Section I.6. Solution
II.6.2 3π/4
π/2
π/4
0
−3π/4
−π/2
−π/4
We have
f (z) = Log z = log jzj+i Arg z = log r+i )
u = log jzj ) v = Arg z
u = log r : v=
If u = constant and r = constant we have circles centred at 0, and if v = constanat and = constanat we have rays issuing from 0, < < . Refer to the Figure for the map w = Log z in section I.6. The sketch is of the inverse images of horizontal and vertical lines in that Figure.
69
II.6.3 Sketch the families of level curves of u and v for the function f = u + iv given by (a) f (z) = ez , (b) f (z) = e z , where is complex. Determine where f (z) is conformal and where it is not conformal.
Solution f (z) = ez = ex (cos y + i sin y) ) u = ex cos yv = ex sin y u = const, x = log (c cos y) = log c log (cos y) v = const, x = log (c sin y) = log c log (sin y) Level curves are invariant under ????? in x–direction. Level curves are invariant under ????? in y–direction by . The function f (z) is conformal everywhere. f (z) = e z = e(a+ib)(x+iy) = eax by (cos (ay + bx) + i sin (ay + bx)) ) u = eax by cos (ay + bx) v = eax by sin (ay + bx) Figure repeat themselves u = 0 at ay + bx = n + 2 , y = ab x + na + 2a . v = 0 at ay + bx = n , y = ab x + na FIGURE II.6.3a NF FIGURE II.6.3b NF
70
II.6.4 Find a conformal map of the horizontal strip f A < Im z < Ag onto the right half-plane fRe w > 0g. Hint. Recall the discussion of the exponential function, or refer to the preceding problem.
Solution (A. Kumjian) FIGURE II.6:4az FIGURE II.6:4aw It is clearly intended that A > 0 so we will assume this. We de…ne the map z h : C ! C by h (z) = e 2A for all z 2 C. Then h0 (z) =
z
e 2A
2A for all z 2 C. Hence, h is analytic with a nowhere vanishing derivative and so h is conformal. Note that the exponential functions maps the horizontal strip fz 2 C : =2 < Im z < =2g onto the right half-plane fw 2 C : Re w > 0g = fw 2 C : Hence, h maps the horizontal strip fz 2 C : half-plane fw 2 C : Re w > 0g.
71
=2 < Arg w < =2g : A < Im z < Ag onto the right
II.6.5 Find a conformal map of the wedge f B < arg z < Bg onto the right half-plane fRe w > 0g. Assume 0 < B < . Solution Try w = z , multiply angels by
2B
, so eiB ! ei
=2
,w=z
FIGURE II.6.5az FIGURE II.6.5aw
72
=(2B)
does the trick
II.6.6 Determine where the function f (z) = z +1=z is conformal and where it is not conformal. Show that for each w, there are at most two values z for which f (z) = w. Show that if r > 1, f (z) maps the circle fjzj = rg onto an ellipse, and that f (z) maps the circle fjzj = 1=rg onto the same ellipse. Show that f (z) is one-to-one on the exterior domain D = fjzj > 1g. Determine the image of D under f (z). Sketch the images under f (z) of the circles fjzj = rg for r > 1, and sketch also the images of the parts of the rays farg z = g lying in D. Solution f (z) = z + 1=z, f 0 (z) = 1 1=z 2 , f 0 (z) = 0 at z 2 = 1, z = at z = 0, not conformal at z = 1. If f (z) = , then z =
1.pNot de…ned 2 4 =2.
If z = ei , then w = u + iv = cos + i sin + cos = i sin = , which can be written u2 = ( + 1= )2 + v 2 = ( 1= )2 = 1, i.e. image is an ellipse, replace by 1= and get same ellipse. Since f (D) = f (D), and f (z) is at most two–to–one, f (z) is one–to–one on D. Since f (z) maps @D onto the interval [ 2; 2], and f (z) maps onto C, the image of D is Cn [ 2; 2].
73
II.6.7 For the function f (z) = z + 1=z = u + iv, sketch the families of level curves of u and v. Determine the images under f (z) of the top half of the unit disk, the bottom half of the unit disk, the part of the upper half-plane outside the unit disk, and the part of the lower half-plane outside the unit disk. Hint. Start by locating the images of the curves where u = 0, where v = 0, and where v = 1. Note that the level curves are symmetric with respect to the real and imaginary axis, and they are invariant under the inversion z 7! 1=z in the unit circle. Solution
74
II.6.8 ( From Hints and Solutions) 1 2 3 P L K Consider f (z) = z + ei =z, where 0 < < . Determine where f (z) is conformal and where it is not conformal and where it is not conformal. Sketch the images under f (z) of the unit circle fjzj = 1g and the intervals ( 1; 1] and [+1; +1) on the real axis. Show that w = f (z) maps fjzj > 1g conformally onto the complement of a slit plane in the w plane. Sketch roughly the images of the segments of rays outside the unit circle farg z = ; jzj 1g under f (z). At what angels do they meet the slit, and at what angles do they approach 1? Solution f (z) = z + ei =z , f 0 (z) = 1 ei =z 2 , f 0 (z) = 0 at z 2 = ei . Not de…ned at z = 0, not conformal at z = ei =2 . The expression f (z) = ei =2 e i =2 z + 1= e i =2 z , shows that f (z) is a composition of the rotation by =2, the function of Exercises 6 and 7, and a rotation by =2. Thus f (z) maps fjzj > 1g one–to–one onto the complement rotate of [ 2; 2] by =2.
75
II.6.9 Let f = u + iv be a continuously di¤erentiable complex-valued function on a domain D such that the Jacobian matrix of f does not vanish at any point of D. Show that if f maps orthogonal curves to orthogonal curves, then either f or f is analytic, with nonvanishing derivative. Solution Suppose f (0) = 0 and ux 6= 0 at 0. The tangents in the orthogonal directions (1; t) and ( t; 1) are mapped to the tangents in the directions (ux ; vx ) + t (uy ; vy ) and t (ux ; vx ) + (uy ; vy ). The orthogonality of these directions for all t gives h(ux ; vx ) + t (uy ; vy ) ; t (ux ; vx ) + (uy ; vy )i = 0;
can be rewritten
h(ux ; vx ) ; (uy ; vy )i+t (h(uy ; vy ) ; (uy ; vy )i
h(ux ; vx ) ; (ux ; vx )i) t2 h(uy ; vy ) ; (ux ; vx )i = 0:
The systems of equations (1) (2)
u2y
ux u y + v x v y = 0 + vy2 (u2x + vx2 ) = 0
most hold to get orthogonality. Put (1) into (2) and get u2x u2y + vy2
u2x
vx2 = vx2 vy2 + u2x vy2
= vx2 vy2
u2x + u2x vy2
u4x
u2x vx2 =
u2x = vx2 + u2x
vy2
u2x = 0:
If ux 6= 0 we are lead to ux = vy and uy = vx . Tangent to curve s ! (s; ts) is tangent to image curve s + (u (s; ts) ; v (s; ts)) (1; 0) ! (ux ; vx ), (0; 1) ! (uy ; vy ), orthogonal ) (uy ; vy ) = C ( vx ; uy ) for some. A ????? gives C = 1, same sign holds in ?????. Either ux = vy , uy = vx ) conformal in neighborhood of 0, or ux = vy , uy = vx ) anti conformal in neighborhood of 0. ux = vy and uy = vx . So Cauchy Riemann is satis…ed for either f or f .
76
II.7.1 Compute explicitly the fractional linear transformations determined by the following correspondences of triples (a) (1 + i; 2; 0) 7 ! (0; 1; i 1) (e) (1; 2; 1) 7 ! (0; 1; 1) (b) (0; 1; 1) 7 ! (1; 1 + i; 2) (f) (0; 1; i) 7 ! (0; 1; 1) (c) (1; 1 + i; 2) 7 ! (0; 1; 1) (g) (0; 1; 1) 7 ! (0; 1; i) (d) ( 2; i; 2) 7 ! (1 2i; 0; 1 + 2i) (h) (1; i; 1) 7 ! (1; 0; 1) Solution (w (w
(z w2 ) = w0 ) (z
w0 ) (w1 w2 ) (w1
z0 ) (z1 z2 ) (z1
z2 ) z0 )
(a) We have the points z0 = 1 + i z1 = 2 z2 = 0
w0 = 0 w1 = 1 w2 = i 1
The mapping is (w (w
w0 ) (1 w2 ) (1
w2 =w1 ) (z = w0 =w1 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
take limit as w1 ! 1, to obtain w w
0 (i
1)
=
(z (z
(1 + i)) (2 0) 2iz + (2 2i) )w= : 0) (2 (1 + i)) z 2
(b) We have the points z0 = 0 z1 = 1 z2 = 1
w0 = 1 w1 = 1 + i w2 = 2
The mapping is (w (w
w0 ) (w1 w2 ) (w1
w2 ) (z z0 ) (z1 =z2 1) = ; w0 ) (z=z2 1) (z1 z0 )
take limit as z2 ! 1, to obtain 77
(w (w
1) ((1 + i) 2) ((1 + i)
2) z = 1) 1
0 2z )w= 0 z
(1 + i) : (1 + i)
(c) We have the points z0 = 1 z1 = 1 + i z2 = 2
w0 = 0 w1 = 1 w2 = 1
The mapping is (z=z0 1) (z1 z2 ) (w w0 ) (w1 =w2 1) = ; (w=w2 1) (w1 w0 ) (z z2 ) (z1 =z0 1) take limit as z0 ; w2 ! 1, to obtain w 1
0 (1 + i) 2 i = )w= 0 z 2 z
1 : 2
(d) We have the points z0 = 2 w0 = 1 2i z1 = i w1 = 0 z2 = 2 w2 = 1 + 2i The mapping is (w (w
w0 ) (w1 w2 ) (w1
w2 ) (z = w0 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
obtain (w (w
(1 2i)) (0 (1 + 2i)) (0
(1 + 2i)) (z = (1 2i)) (z
( 2)) (i 2) ) w = iz + 1: 2) (i ( 2))
(e) We have the points z0 = 1 z1 = 2 z2 = 1
w0 = 0 w1 = 1 w2 = 1 78
The mapping is (w w0 ) (w1 =w2 1) (z z0 ) (z1 =z2 1) = ; (w=w2 1) (w1 w0 ) (z=z2 1) (z1 z0 ) take limit as z2 ; w2 ! 1, to obtain w 1
0 z = 0 2
1 )w=z 1
1:
(f) We have the points z0 = 0 w0 = 0 z1 = 1 w1 = 1 z2 = i w2 = 1 The mapping is (w w0 ) (w1 =w2 1) (z = (w=w2 1) (w1 w0 ) (z
z0 ) (1 z2 ) (1
z2 =z1 ) ; z0 =z1 )
take limit as z1 ; w2 ! 1, to obtain w 1
z 0 = 0 z
0 z )w= i z i
(g) We have the points z0 = 0 z1 = 1 z2 = 1
w0 = 0 w1 = 1 w2 = i
The mapping is (w (w
w0 ) (1 w2 ) (1
w2 =w1 ) (z z0 ) (z1 =z2 1) = ; w0 =w1 ) (z=z2 1) (z1 z0 )
take limit as z2 ; w1 ! 1, to obtain w w
0 z = i 1
0 iz )w= : 0 z 1
(h) 79
We have the points z0 = 1 w0 = 1 z1 = i w1 = 0 z2 = 1 w2 = 1 The mapping is (w (w
w0 ) (w1 w2 ) (w1
(z w2 ) = w0 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
obtain (w (w
1) (0 ( 1)) (z = ( 1)) (0 1) (z
1) (i ( 1)) iz + 1 )w= : ( 1)) (i 1) z+i
80
II.7.2 Consider the fractional linear transformation in Exercise 1a above, which maps 1 + i to 0, 2 to 1, and 0 to i 1. Without referring to an explicit formula, determine the image of the circle fjz 1j = 1g, the image of the disk fjz 1j < 1g, and the image of the real axis. Solution It is obvious that the tripple (1 + i; 2; 0) lies on the circle jz 1j = 1 in the z plane. From exercise II.7.1a we know that points in the w plane the tripple is mapped on, i.e, z0 = 1 + i z1 = 2 z2 = 0
w0 = 0 w1 = 1 w2 = i 1
The circle jz 1j = 1 is mapped to the straight line through 0 and i 1, i.e Im w = Re w since tree points determines the "circle". By preservation of orientation the disk goes to the half-plane to lower left of the straight line through 0 and i 1. Image of the real axis is "circle" orthogonal to image of jz 1j = 1, that must be the straight line through i 1, with slope 1, i.e. the line Im w = Re w + 2.
II.7.2 z-plane
II.7.2 w-plane
81
II.7.3 Consider the fractional linear transformation that maps 1 to i, 0 to 1 + i, 1 to 1. Determine the image of the unit circle fjzj = 1g, the image of the open unit disk fjzj < 1g, and the image of the imaginary axis. Illustrate with a sketch.
Solution We have the points z0 = 1 w0 = i z1 = 0 w1 = 1 + i z2 = 1 w2 = 1 The mapping is (w (w
w0 ) (w1 w2 ) (w1
(z w2 ) = w0 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
obtain (w (w
i) ((1 + i) 1) ((1 + i)
1) (z = i) (z
1) (0 ( 1)) i 1 )w= ( 1)) (0 1) z+i
By our mapping w = (i 1) = (z + i), we have that i on the unitcircle in the z plane is mapped to 1 in the w plane. The circle jzj = 1 is mapped to the straight line through i and 1, i.e Im w = Re w + 1 since tree points determines the "circle". By preservation of orientation the disk goes to the half-plane to upper left of the straight line through i and 1. The tripple (1; 0; 1) on the real axis is mapped to the points (i; 1 + i; 1) in the w plane, it must be so that plane is mapped p the real axis in the z 2 1 1 + i = in the w plane. onto the circle w 2 2 2 Imaginary axis must be mapped to a circle through 1+i, that is orthogonal to image of real line, this is the straight line through 1+i and 0, i.e Im w = Re w.
82
II.7.3 z-plane
II.7.3 w-plane
83
II.7.4 Consider the fractional linear transformation that maps 1 to i, 1 to 2i, and i to 0. Determine the image of the unit circle fjzj = 1g, the image of the open unit disk fjzj < 1g, and the image of the interval [ 1; +1] on the real axis. Illustrate with a sketch. Solution We have the points z0 = 1 w0 = i z1 = 1 w1 = 2i z2 = i w2 = 0 The mapping is (w (w
w0 ) (w1 w2 ) (w1
(z w2 ) = w0 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
obtain (w (w
( i)) (2i 0) (z = 0) (2i ( i)) (z
( 1)) (1 i) ( 6 2i) z 2 + 6i )w= i) (1 ( 1)) 5z 3 + 4i
It is obvious that the tripple ( 1; 1; i) lies on the circle jzj = 1 in the z plane. And we can see that thiese three points are mapped to three points on the imaginary axis in the w plane, i.e Re w = 0, since tree points determines the "circle". By preservation of orientation the disk goes to the right half-plane. The interval [ 1; 1] must be mapped to a circle through i and 2i that is orthogonal to image of the unit circle, this is an arc of the circle w 21 i = 32 . We have that w (0) = 65 25 i must be an point on the arc, thus the arc is the circle w 21 i = 23 in the right half-plane from i to 2i.
84
II.7.4 z-plane
II.7.4 w-plane
85
II.7.5 What is the image of the horizontal line through i under the fractional linear transformation that interchanges 0 and 1 and maps 1 to 1 + i? Illustrate with a sketch.
Solution We have the points z0 = 0 w0 = 1 z1 = 1 w1 = 0 z2 = 1 w2 = 1 + i The mapping is (w (w
w0 ) (w1 w2 ) (w1
(z w2 ) = w0 ) (z
z0 ) (z1 z2 ) (z1
z2 ) ; z0 )
obtain (w (w
1) (0 (1 + i)) (z = (1 + i)) (0 1) (z
0) (1 ( 1)) iz )w= ( 1)) (1 0) z
i i
The tripple (0; 1; 1) on the real axis is mapped to the points (1; 0; 1 + i) in the w plane, it must be so pthat the real axis in the z plane is mapped onto 1 + 12 i = 22 in the w plane, since tree points determines the circle w 2 the "circle". The real axis and the line through i have one common point at in…nity. The common point at in…nity for both real axis and the line through i must be mapped to the same point in the w plane, we have that the point is w (1) = i. The two lines in the z plane is parallell, thus the images of the lines must go through i in the w plane and be parallell in that point, thus the mapping of the horizontal line through i must be the line through i, which is tangent line to circle at i, i.e. Im z = Re z + 1. Remark that the image can not be a circle because w (i) = 1, so the image of the horizontal line must contain the point at in…nity.
86
II.7.5 z-plane
II.7.5 w-plane
87
II.7.6 Show that the image of a straight line under the inversion z 7! 1=z is a straight line or circle, depending on whether the line passes through the origin.
Solution The image of the straight line ax+by = c is (kontrollera x iy i lösningshäftet) w= where u = x=r2 and v = au
bv =
1 z 1 = = 2 = (x z zz jzj
iy)
1 r2
y=r2 . Get
ax + by c c = 2 = 2 = c u2 + v 2 2 r r x + y2
The image is the solution of c (u2 + v 2 ) If c = 0 we have that
au + bv = 0.
au + bv = 0; this is a straight line through 0. If c 6= 0 we have c u2 + v 2
au + bv = 0;
which can be rewritten as
u
a 2c
2
it is a circle through 0.
+ v+
b 2c
0s
2
=@
a 2c
2
+
b 2c
2
12
A ;
Solution (K. Seip) Set f (z) = z1 . Then 1 2 f (S) , 0 2 S, which proves the claim.
88
II.7.7 Show that the fractional linear transformation f (z) = (az + b) = (cz + d) is the identity mapping z if and only if b = c = 0 and a = d 6= 0. Solution We have that az + b = z , az + b = cz 2 + dz , cz 2 + (d a) z b = 0: cz + d Put in values on z for intance 0 and 1 the expession becomes (1) c (d a) = 0; (2) b=0 (3) c + (d a) = 0;
z = 1: z=0 z = 1:
From (2) we have b = 0 and if we add (1) and (3) we have c = 0. Thus the fractional linear transformation f (z) is the identity mapping if and only if b = c = 0, a = d 6= 0. Solution maps R to R . Then 0 7! db 2 R , Set R = R [ f1g. Suppose f (z) = az+b cz+d 1 7! ac 2 R , 0 - ab 2 R , 1 - dc 2 R . At least one of theese rations is di¤erent from 0 and 1, and since one of a; b; c; d can be chosen freely, the result follows.
89
II.7.8 Show that any fractional linear transformation can be represented in the form f (z) = (az + b) (cz + d), where ad bc = 1. Is this representation unique?
Solution If w = ( z + )( z + ) divide each coe¢ cient by the square root of
to obtain representation with ad
bc = 1:
Representation is not unique, as can multiply all coe¢ cients by
90
1.
II.7.9 Show that the fractional linear transformations that are real on the real axis are precisely those that can be expressed in the form (az + b) = (cz + d), where a; b; c;and d are real.
Solution (K. Seip) maps R to R . Then 0 7 ! db 2 R , Set R = R[f1g. Suppose f (z) = az+b cz+d 1 7! ac 2 R , 0 - ab 2 R , 1 - dc 2 R . At least one of these is di¤erent from 0 and 1, and since one of a; b; c; d can be chosen freely, the result follows.
91
II.7.9 Show that the fractional linear transformations that are real on the real axis are precisely those that can be expressed in the form (az + b) = (cz + d), where a; b; c;and d are real.
Solution (A. Kumjian) Set R = R [ f1g, note that a fractional linear transformation f : C ! C is real on the real axis i¤ f (R ) R (since f is continuous). For A = ac db with det A 6= 0 we de…ne the fractional linear transformation fA : C ! C by the formula fA (z) = (az + b) = (cz + d). It remains to prove that given a fractional linear transformation f , we have f (R ) R i¤ f = fA for some 2 2 matrix A with real entries. Let f : C ! C be a fractional linear transformation. Suppose that f = fA where A is a 2 2 matrix with real entries a; b; c and d as above. Then for x 2 R we have ax + b 2R : f (x) = fA (x) = cx + d This is clear for x 2 R but requires a little work for x = 1. We have az + b = z!1 cz + d
f (1) = lim
a=c if 1 if
c 6= 0; c = 0:
Hence, f (R ) R . Conversely, suppose that f (R ) R , we must show that f = fA for some 2 2 matrix with real entries. If A is a 2 2 matrix (with non-zero determinant), then fA 1 = (fA ) 1 . Hence, f = fA for some 2 2 matrix A with real entries i¤ its inverse f 1 has the same property. By the unlikeness part of the theorem on page 64, f is completely determined by the three extended real numbers x0 = f (0), x1 = f (1) and x1 = f (1) (recall that we have assumed f (R ) R ). So it su¢ ces to show that the fractional linear transformation g for which g (x0 ) = 0, g (x1 ) = 1 and g (x1 ) = 1 has the requisite properties, because we must have g = f 1 (again by the uniqueness part of the theorem). There are four cases to consider. The …rst case is when x0 ; x1 ; x1 6= 1 and the remaining three cases are x0 = 1; x1 = 1 and x1 = 1. The formula for g (z) in the four cases is given as follows (in the …rst case k = xx11 xx10 2 R): 92
x x0 k if x0 ; x1 ; x1 6= 1; x x1 x x0 if x1 = 1; x x1
x1 x x x1
x1 if x0 = 1; x1 x0 if x1 = 1: x0
In each case g (z) has the desired form. Hence, f = fA for some 2 A with real entries as required.
93
2 matrix
II.7.10 Suppose the fractional linear transformation (az + b) = (cz + d) maps R to R, and ad bc = 1. Show that a; b; c; and d are real or they are all pure imaginary.
Solution Because R is mapped on R and the orientation is perserved, f (x) will be either increasing or decreasing. Case 1 : Suppose f (x) is increasing, then f 0 (x) = 1= (cx + d)2 > 0 for all x ) c; d are real. f (z) = (az + b) = (cz + d), then ax + b is real for all x 2 R, so a; b are also real. Case 2 : Suppose f (x) is decreasing, then f 0 (z) = 1= (cz + d)2 < 0 for all x ) c; d are pure imaginary. f (z) = (az + b) = (cz + d), then ax + b are pure imaginary for all x 2 R, so a; b are also pure imaginary.
94
II.7.11 Two maps f and g are conjugate if there is h such that g = h f h 1 . Here the conjugating map h is assumed to be one-to-one, with appropriate domain and range. We can think of f and g as the "same" map, after the change of variable w = h (z). A point z0 is a …xed point of f if f (z0 ) = z0 . Show the following. (a) If f is conjugate to g, then g is conjugate to f . (b) If f1 is conjugate to f2 and f2 to f3 , then f1 is conjugate to f3 . (c) If f is conjugate to g, then f f is conjugate to g g, and more generally, the m fold composition f f (m times) is conjugate to g g (m times). (d) If f and g are conjugate, then the conjugating function h maps the …xed points of f to …xed points of g. In particular, f and g have the same number of …xed points.
Solution (a) If f is conjugate to g then g=h f
h
1
)f =h
thus g is conjugate to f (b) If f1 = h f2 h 1 and f2 = g f3 g f1 = h g f3 g
1
h
1
1
1
g h
we get 1
= (h g) f3 (h g)
:
(c) m times
g=h f
(d) g=h f
}| { z h 1 theng g : : : g = =h f h 1 h f h
1
::: h f
h
z =h f h 1 , f (z0 ) = z0 , and w0 = h (z0 ), then 95
1
=
m times
f
}| { ::: f h
1
g (w0 ) = h f h
1
(w0 )
= h (f (z0 )) = h (z0 ) = w0 :
We have that f and g have the same number of …xed points since every point of g is mapped to a …xed point of f by the function h 1 .
96
II.7.12 Classify the conjugacy classes of fractional linear transformations by establishing the following (a) A fractional linear transformation that is not the identity has either 1 or 2 …xed points, that is, points satisfying f (z0 ) = z0 . (b) If a fractional linear transformation f (z) has two …xed points, then it is conjugate to the dilation z 7! az with a 6= 0, a 6= 1, that is, there is a fractional linear transformation h (z) such that h (f (z)) = ah (z). Is a unique? Hint. Consider a fractional linear transformation that maps the …xed points to 0 and 1. (c) If a fractional linear transformation f (z) has exactly one …xed point, then it is conjugate to the translation 7! + 1. In other words, there is a fractional linear transformation h (z) such that h (f (h 1 ( ))) = + 1, or equivalently, such that h (f (z)) = h (z) + 1. Hint. Consider a fractional linear transformation that maps the …xed point to 1. Solution = z , az + b = cz 2 + dz , cz 2 + (d a) z b = 0. If this is = 0 a) az+b cz+d for every z, we get w (z) = z. It is not identically equall to zero, so it has either 1 or 2 …nite solutions. If c 6= 0, it has 2 solutions possibly one with multiplicity and 1 is not a …xed point. If c = 0, it has one …nite solution z1 = b= (d a), and z2 = 1 is a …xed point, so they are two. b) Make a change of variables h (z) = ( z + ) = ( z + ) that maps …xed points to 0 and 1 . Then h f h 1 is FLT with …xed points at 0 and 1. (h f h 1 ) (w) = aw for some a 6= 0, a 6= 1. f is conjugate to a dilation Suppose aw is conjugate to Aw, i.e. 9h, (h (ah 1 )) (w) = Aw, h (az) = Ah (z). h must map …xed points to …xed points, so either h (z) = cz or h (z) = c=z. So either h (z) = cz ) h (az) = caz = Ah (z) = Acz ) A = a or h (z) = c=z ) h (az) = c= (az) = Ah (z) = Ac=z ) A = 1=a. Thus a is not unique. c) Suppose h maps the …xed points to 1, then h f h 1 has only one …xed point at 1, so (h f h 1 ) (w) = aw+b. Since this has no …nite …xed points, aw +b = w has no solutions and a = 1, b 6= 0. Thus (h f h 1 ) (w) = w +b. 97
Now w + b is conjugate to w + 1, by a dilation g (w) = Aw. w ! w=A ! w=A + b ! w + Ab, take A = 1=b. So g
h f
h
1
()
g
1
(w) = w + 1 ()
(g h) f
(g h) 1 (w) = w + 1 () () ((g h) f ) (z) = (g h) (z) + 1:
98
III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8
1
III.1.1 R 2 Evaluate y dx + x2 dy along the following paths from (0; 0) to (2; 4), (a) the arc of the parabola y = x2 , (b) the horizontal interval from (0; 0) to (2; 0), followed by the vertical interval from (2; 0) to (2; 4), (c) the vertical interval from (0; 0) to (0; 4), followed by the horizontal interval from (0; 4) to (2; 4).
Solution (a) Z
2
2
y dx+x dy =
x=t y = t2
dx = dt; dy = 2t dt
0
dx = dt dy = 0 dt
0
t
2
=
Z
2
2 2
t
0
Z dt+
2
t2 2t dt =
0
72 : 5
(b) Z
Z
2
2
y dx + x dy = 1
2
2
x=t y=0 x=2 y=t
dx = 0 dt dy = dt
t
2
=
Z
2 2
0 dt +
0
0
t
Z
4
Z
t2 0 dt = 0
0
4 2
y dx + x dy = = t 0 dt + 0 2 Z Z Z 2 2 2 2 y dx + x dy + y 2 dx + x2 dy = 16: y dx + x dy = 1
2
Z
4
22 dt = 16
0
2
(c) Z
Z
2
2
y dx + x dy = 1
x=0 y=t
dx = 0 dx dy = dt
0
t
4
=
Z
Z
0 2
4 2
y 0 dt +
x=t dx = dt 0 t 2 y 2 dx + x2 dy = = 42 dt + y = 4 dy = 0dt 0 2 Z Z Z y 2 dx + x2 dy = y 2 dx + x2 dy + y 2 dx + x2 dy = 32: 1
2
2
Z
0
Z
4
02 dt = 0
0
2
t2 0 dt = 32
III.1.2 R Evaluate xy dx both directly and using Green’s theorem, where R 1), is the boundary of the square with vertices at (0; 0), (1; 0), (1; and (0; 1).
Solution (A. Kumjian)
VII.1.2 γ 1
γ
γ
4 3 2 1 γ
1
Denote the square by D and note that xy dx = P dx + Q dy where P = xy and Q = 0. Then P and Q are continuously di¤erentiable on D and = @D, hence by Green’s Theorem we have, Z
xy dx =
Z
P dx+Q dy =
@D
ZZ
D
@Q @x
@P dy
dx dy =
Z
0
1
Z
1
x dx dy =
0
Observe that = 1 + 2 + 3 + 4 where 1 and 3 are the bottom and top of the square while 2 and 4 are the last two sides taken in the order indicated by the order of the vertices in the statement of the problem (so the boundary is oriented counter clockwise). Note that the path integrals on 2 and 4 are zero because the edges are vertical. Z
xy dx =
Z
xy dx + 1
Z
xy dx = 3
Z
1
x 0 dx
0
3
Z
0
1
x 1 dx = 0
1 = 2
1 : 2
1 : 2
III.1.3 R Evaluate @D x2 dy both directly and using Green’s theorem, where D is the quarter-disk in the …rst quadrant bounded by the unit circle an the two coordinate axes.
SolutionR Evaluate @D x2 dy directly, set @D =
Z
Z
2
x dy = 1
2
x dy = 2
Z
x=t y=0
x = cos t y = sin t
1
dx = dt dy = 0 dt dx = sin t dt dy = cos t dt
+
2
0
3.
+ t
1
=
Z
1
t2 0 dt = 0
0
0
t
=2
=
Z
Z
cos2 t cos t dt =
0 1
2 3
dx = 0 dt 0 t 1 02 dt = 0 = t dy = dt 0 3 Z Z Z Z 2 2 2 2 x dy = x dy + x dy + x2 dy = 3 @D 1 2 3 R Now we evaluate @D x2 dy, this time using Green’s theorem. In this case, P (x; y) = 0 and Q (x; y) = x2 . Z
x2 dy =
2
ZZ
x=0 y=1
=2
x dy = 2x dxdy = D ZZ x = r cos x dxdy = =2 y = r sin D @D
dx dy = r dr d
=2
Z
0
4
0 0
r
1 2
r dr
1 Z
0
=2 =2
=2
Z
0
1
Z
2 cos d = : 3
0
=2
r cos rdrd =
III.1.4 R Evaluate y dx both directly and using Green’s theorem, where is the semicircle in the upper half-plane from R to R. SolutionR Evaluate y dx directly, set Z
Z
y dx = 1
y dx = 2
x=t y=0
=
1
+
dx = dt dy = 0 dt
2.
R
t
R
=
x = R cos t dx = R sin t dt 0 t y = R sin t dy = R cos t dt Z Z Z 2 2 x2 dy = x dy + x dy = @D
R
0 dt = 0 R
=
Z
R sin t R sin t dt =
0
R 2
2
1
Z
2
R
Now we evaluate y dx, this time using Green’s theorem. In this case, P (x; y) = y and Q (x; y) = 0. Z
ZZ
y dx = ZZ dx dy =
@D
=
D
dx dy =
D
x = r cos y = r sin
dx dy = r dr d
=
Z
0 0
0
5
r
R
r dr
R Z
0
=
Z
0
d =
R
Z
0 2
R : 2
r dr d =
R2 2
III.1.5 R R Show that @D x dy is the area of D, while @D y dx is minus the area of D. Solution Using R in the two cases For @D x dy, we have P = 0 and Q = x, using Green’s theorem we have Z Z Z x dx = dx dy = Area D: For
R
@D
@D
D
y dx, we have P = y and Q = 0, using Green’s theorem we have Z Z Z y dx = dx dy = Area D: @D
D
6
III.1.6 Show that if P and Q are continuous complex-valued functions on a curve , then R
R Qdy P dx + ; (z = x + iy) z w z w is analytic for w 2 Cn . Express F 0 (w) as a line integral over . Solution Di¤erentiate by hand, use uniform convergence of 1 w z as
1 (w +
1 w)
z
w
=
1 (z
w) (z
(w +
w))
!
w ! 0, uniformly for z 2 . Get F 0 (w) =
R
R Qdx P dx ; 2 + (z w) (z w)2
7
z = x + iy
1 (z
w)2
III.1.7 Show that the formula in Green’s theorem is invariant under coordinate changes, in the sense that if the theorem holds for a bounded domain U with piecewise smooth boundary, and if F (x; y) is a smooth function that maps U one-to-one onto another such domain V and that maps the boundary of U one-to-one smoothly onto the boundary of V , then Green’s theorem holds for V . Hint. First note the change of variable formulae for line and area integrals, given by
ZZ
Z
Pd
=
@V
Rd d
=
V
Z
(P
Z@U Z
@ @ dx + dy ; @x @y
F)
(R F ) det JF dx dy;
U
where F (x; y) = ( (x; y) ; (x; y)), and where JF is the Jacobian matrix of F . Use these formulae, with R = @P=@ . The summand R Q d is treated similarly. Solution ZZ
@P d d = @
V
ZZ
@P @
( (x; y) ; (x; y)) det (JF ) dx dy =
U
F (x; y) = ( (x; y) ; (x; y)) @F @x @F @y
det JF (x; y) =
@ @ @x @y =
= =
@ @x @ @y
+ +
@ @x @ @y
@ @ = @y @x ZZ @P ( (x; y) ; (x; y)) @ U
Using Green’s theorem 8
@ @ @x @y
@ @ @y @x
dx dy
Z
Pd =
@V
Z
Z @ @ @ @ (P F ) P ( (x; y) ; (x; y)) = dx + dy = dx + dy = @x @y @x @y @U @U ZZ @ @ @ @ = P + P dx dy = @y @x @x @y U ZZ @ @ @2 @ @ @2 = P ( (x; y) ; (x; y)) P + P ( (x; y) ; (x; y)) +P dx dy = @y @x @y@x @x @y @x@y U ZZ @P @ @P @ @P @ @P @ @ @ = + + dx dy = @ @y @ @y @x @ @x @ @x @y U ZZ @P @ @ @P @ @ + = dx dy = @ @y @x @ @x @y U ZZ @P JF dx dy: = @ U
R Similar argument as above hold for Qd Here we use that Z Z @ @ Qd = (Q F ) dx + dy @x @y @V @U and replace R =
@P=@ by R=@Q=@ to conclude that Z ZZ @Q Qd = d d : @V V @
9
III.1.8 Prove Green’s theorem for the rectangle de…ned by x0 < x < x1 and y0 < y < y1 (a) directly, and (b) using the result for triangles. Solution
VII.1.8a
VII.1.8b III
IV
III II
R
IV
S
I
T
II
I
(a) Make …gure and set the linesegment between (x0 ; y0 ) and (x1 ; y0 ) to proceed in this way in counterclockwise direction. We have Z Z Z Z Q dy = P dx = Q dy = P dx = 0: 1
2
3
Integrate around the rectangle
10
4
1
and
ZZ
R
@Q @x Z
y1
@P dx dy = @y Z y1 Z x1 @Q = dx dy x0 @x y0
Z
x1
Zx0x1
Z
y1
y0
@P dy dx = @y
P (x; y0 )] dx = Z x1 Z y1 Z y1 P (x; y0 ) dx = P (x; y1 ) dx + Q (x0 ; y) dy Q (x1 ; y) dy = x0 x0 y0 y0 Z x1 Z y1 Z x1 Z y1 = P (x; y0 ) dx+ Q (x1 ; y) dy P (x; y1 ) dx Q (x0 ; y) dy = x0 y0 x0 y0 Z Z Z Z Q dy = P dx+ Q dy + P dx+ = 4 3 2 1 Z = (P dx + Q dy) : =
[Q (x1 ; y)
Q (x0 ; y)] dy
[P (x; y1 )
Zx0x1
y0
@R
(b) Write R = S [ T . The integrals on the diagonal are in di¤erent directions and will cancel Z
@R
(P dx + Q dy) = Z Z (P dx + Q dy) = (P dx + Q dy) + = @T @S ZZ ZZ @Q @P @Q @P = dx dy + dx dy = @x @y @x @y S T ZZ @Q @P = dx dy: @x @y R
11
III.2.1 Determine whether each of the following line integrals is independent of path. If it is, …nd a function h such that dh = P dx + Q dy. If it is not, …nd a closed path around which the integral is not zero. (a) x dx + y dy, (b) x2 dx + y 5 dy, (c) y dx + x dy, (d) y dx x dy.
Solution We compute the partial derivatve because if h exist we have that dh = P dx + Qdy where @P=@y = @Q=@x. (a) P =x Q=y @Q @P =0 =0 @y R R@x x2 P dx = 2 + g (y) Qdy = x2 +y 2 )h= 2
y2 2
+ f (x)
y6 6
+ f (x)
(b) P = x2 Q = y5 @Q @P =0 =0 @x R@y R 3 P dx = x3 + g (y) Qdy = 2x3 +y 6 )h= 6 (c) P =y Q=x @Q @P =1 =1 R@y R@x P dx = xy + g (y) Qdy = xy + f (x) ) h = xy (d) P =y Q= x @P = 1 @Q = 1 @y @x The line integral is not independent of path, we use Greens’theorem Z ZZ (y dx x dy) = 2 dx dy = 2 : jzj=1
jzj<1
12
III.2.2 Show that the di¤erential ydx + xdy ; x2 + y 2
(x; y) 6= (0; 0) ;
is closed. Show that it is not independent of path on any annulus centered at 0.
Solution P = @P @y
y x2 +y 2 (x2 +y2 )+y(2y)
=
(x2 +y 2 )2
Q= =
y 2 x2 (x2 +y 2 )2
@Q @x
x x2 +y 2 (x2 +y2 ) x(2x)
=
(x2 +y 2 )2
=
y 2 x2 (x2 +y 2 )2
Because @P=@y = @Q=@x;the di¤erential is closed and we can use Green’s Theorem I
P dx + Q dy =
jzj=r
=
I
jzj=r
I x 1 y dx+ 2 dy = 2 y dx + x dy = x2 + y 2 x + y2 r jzj=r ZZ 1 1 = 2 (1 + 1) dx dy = 2 2 r2 = 2 6= 0 r r jzj
Thus the line integral is not independent of path.
13
III.2.3 Suppose P and Q are smooth functions on the annulus fa < jzj < bg that Hsatisfy @P=@y = @Q=dx. Show directly using Green’s theorem that jzj=r P dx + Qdy is independent of the radius r, for a < r < b. Solution (A. Kumjian)
III.2.3 γ1 γ2 D
Let r1 ; r2 be given so that a < r1 < r2 < b. We must prove that I I P dx + Qdy = P dx + Qdy: jzj=r1
jzj=r2
Let D denote the annulus fz 2 C : r1 < jzj < r2 g. Observe that D is a bounded domain with piecewise smooth boundary @D = 1 [ 2 where 1 denotes the circle fz 2 C : jzj = r1 g with clockwise orientation while 2 denotes the circle fz 2 C : jzj = r2 g with counter clockwise orientation. Moreover, both P and Q are continuously di¤erentiable on D = D [ @D. Hence, Green’s Theorem applies and we obtain Z ZZ @Q @P dxdy = 0; P dx + Qdy = @x @y @D D and since, as noted above, @D consists of the two circles with opposite orientation, Z I I P dx + Qdy = P dx + Qdy P dx + Qdy: @D
jzj=r2
Hence, equation ( ) holds and r, for a < r < b.
H
jzj=r
jzj=r1
P dx + Qdy is independent of the radius
14
III.2.4 Let P and Q be smooth functions on D satisfying @P=@y = @Q=@x. Let 0 and 1 be two closed paths in D such that the straight line segment Rfrom 0 (t) to 1 R(t) lies in D for every parameter value t. Then P dx + Q dy = P dx + Q dy. Use this to give another 0 1 solution to the preceding exercise. Solution Use the theorem on page 81. Use the straight lines to deform 0 to 1 . De…ne s) 0 (t), 0 6 s 6 1, a1 6 t 6 b1 . The theorem on page s (t) = s 1 (t) + (1 81 applies. Obtain the result for the annulus above by parameterizing the circles jzj = r0 and jzj = r1 by 0 (t) = r0 e2 it , 1 (t) = r1 e2 it , 0 6 t 6 1. The straight line segments joining 0 (t) to 1 (t) are radical are in the annulus, so the …rst part applies.
15
III.2.5 Let 0 (t) and 1 (t), 0 t 1, be paths in the slit annulus fa < jzj < bg n ( b; a) from A to B. Write down explicitly a family of paths s (t) from A to B in the slit annulus that deforms continuously to 1 . Suggestion. Deform separately the modulus and the principal value of the argument.
Solution Write 0 (t) = r0 ei 0 (t) , 0 6 t 6 1 , when continuous. (Use the fact that (t) = Arg annulus) Also 1 (t) = r1 ei 1 (t) . Then consider s
s) r0 (t)] ei[s
(t) = [sr1 (t) + (1
< 0 (t) < , r0 (t), 0 (t) (t) is continuous on the slit 0
1 (t)+(1
s)
0 (t)]
;
0
s
1:
This does the trick, it deform 0 to 1 continuously in D, and each s is a path in D from A to B. We see that s (t) belongs to the slit annulus for 0 s 1 and 0 t 1 since for every 0 5 1 we have a < min (r0 (t) ; r1 (t))
sr1 (t) + (1
s) r0 (t)
max (r0 (t) ; r1 (t)) < b
and < min (
0
(t) ;
1
(t))
s
1
(t) + (1
16
s)
0
(t)
max (
0
(t) ;
1
(t)) <
III.2.6 Show that any closed path (t), 0 t 1, in the annulus fa < jzj < bg can be deformed continuously to the circular path (t) = (0) e2 imt , 0 t 1, for some integer m. Hint. Reduce to the case where j (t)j j (0)j is constant. Then start by …nding a subdivision 0 = t0 < t1 < < tn = 1 such that arg (t) has a continuous determination on each interval tj 1 t tj . Solution We …rst note that there are two cases. For the closed path (t) enclosing the origin we can deform (t) to a curve with constant modulus. Suppose j (t)j = r for 0 6 t 6 1 and (0) = (1) = 1 where a < r < b. Follow hint, write (t) = rei j (t) , tj 1 6 t 6 tj . Note j (tj ) j+1 (tj ) is a integral multiple of 2 , since (t) is continuous. Add multiples of 2 to the j ‘s, obtain (t) continuous for 0 6 t 6 1 such that (0) = 0 and (t) = rei (t) . Note (1) = 2 m for some integer m. Deform by s (t) = rei[(1 s) (t)+2 sm] . The second case is when (t) does not enclose the origin. For this case (t) the counter of a region that lies entirely inside the annulus. Therefrom the curve (t) may be deformed to any point of this region. In particular it can be continuously deformed to the point (0) = (t) = (0) e2 imt when m = 0.
17
III.2.7 Show that if 0 and 1 lie in di¤erent connected components of the complement C nD of D in the extended complex plane, then there R is a closed path in D such that d 6= 0. Hint. The hypothesis means that there are > 0and a bounded subset E of CnD such that 0 2 E; and every point of E has distance at least 5 from every point of CnD not in E. Lay down a grid of squares in the plane with side length ;and let F be the union of the closed squares in the grid that meet E or that border on a square meeting R E. Show that @F is a …nite union of a closed paths in D, and that @F d = 2 . Solution Proceed as in the hint. R Assume that 0 Ris the centre of one of the squares S0 in the grid. Then @S0 d = 2 , while @S d = 0 for any other square in PR d = 2 , where we sum over squares. S0 ; S1 ; : : : ; Sn the grid. Thus @Sj in F . If two squares in F are ?????, the corresponding integrals along the edge cancel. Then this is the sum over the integrals over the edges of the Sj ´s that have F on oneRside and CnF on the other, that is the edges that form @F . We have thus @F d = 2 . Note that by construction, @F D. Now note how edges @F can meet. Either a vertex in @F have one edge coming in and one out, or it have two coming in and two out. By starting on following edges, and making a left turn at vertices where 4 edges meet, we must eventually end where we started, with a closed path. Three paths must be disjoint, P R except for vertices with Rfour edges. Call the paths 1 ; : : : ; m . d = 2 , we must have d 6= 0 for one of the j ‘s. (In fact, Since j R we will have d = 2 on the j ´s we get by moving from 0 to @F .) j
18
III.3.1 For each of the following harmonic functions u, …nd du, …nd dv, and …nd v, the conjugate harmonic functions of u. (a) u (x; y) = x y (c) u (x; y) = sinh x cos y y (b) u (x; y) = x3 3xy 2 (d) u (x; y) = x2 +y 2 Solution (a) Use Cauchy-Riemanns equations page 83. u (x; y) = x y du = dx dy dv = dx + dy v (x; y) = x + y (b) Use Cauchy-Riemanns equations page 83. u (x; y) = x3 3xy 2 du = (3x2 3y 2 ) dx 6xy dy dv = 6xy dx + (3x2 3y 2 ) dy v (x; y) = 3x2 y y 3 (c) Use Cauchy-Riemanns equations page 83. u (x; y) = sinh x cos y du = cosh x cos y dx sinh x sin y dy dv = sinh x sin y dx + cosh x cos y dy v (x; y) = cosh x sin y (d) Use Cauchy-Riemanns equations page 83. y x2 +y 2 2xy du = (x2 +y 2 )2 dx 2 2 dv = (xy2 +yx2 )2 dx x v (x; y) = x2 +y 2
u (x; y) =
+
x2 +y 2 y(2y) dy (x2 +y 2 )2 2xy dy (x2 +y 2 )2
19
=
2xy (x2 +y 2 )2
dx +
x2 y 2 dy (x2 +y 2 )2
III.3.2 Show that a complex-valued function h (z) on a star-shaped domain D is harmonic if and only if h (z) = f (z) + g (z), where f (z) and g (z) are analytic on D.
Solution Write h = u + iw where u, w are harmonic. By the Theorem on page 83 both u and w have harmonic conjugate in D, (since D is star-shaped), so thus there are ', analytic such that u = Re ', w = Re . u = (' + ') =2;
w=
+
=2;
gives h=
1 2
'+'+i +i
= 12 [' + i ] +
1 2
'+i
:
Take g = 12 ['
f = 12 [' + i ] ;
i ];
then h (z) = f (z) + g (z). To show the oposite direction. Assume that h (z) = f (z) + g (z) where f (z) and g (z) are analytic on D. Then h = Re f + Re g + i (Im f + Im g) and from Cauchy-Riemanns equations follows that @ 2 (Re f ) @ 2 (Re g) @ 2 (Im f ) @ 2 (Im g) + +i @x2 @x2 dx2 @x2 @ 2 (Im f ) @ 2 (Re f ) @ 2 (Re g) + + i + @y 2 @y 2 @y 2 @ 2 Im f @ 2 Im g @ 2 Re f @ 2 Re g = + +i + @x@y @x@y @x@y @x@y +i so h (z) is harmonic.
20
+ @ 2 (Im g) = @y 2 @ 2 Im f @ 2 Im g + @y@x @y@x @ 2 Re f @ 2 Re g =0 @y@x @y@x
III.3.3 Let D = fa < jzj < bg n ( b; a), an annulus slit along the negative real axis. Show that any harmonic function on D has a harmonic conjugate on D. Suggestion. Fix c between a and b, and de…ne v (z) explicitly as a line integral along the path consisting of the straight line from c to jzj followed by the circular arc from jzj to z. Or map the slit annulus to a rectangle by w = Log z.
Solution Integrate the C–R equations we have Z z @u 1 @u dr + r d = v (z) = r@ @r z0
Z
@u dx + @y
Z
@u dy @x
integrate along r interval, then interval. Perception is same as getting from z0 to z1 , then from z1 to little disk. Integral is well-de…ned, continuous, harmonic and is harmonic conjugate for u.
21
III.3.4 (From Hints and Solutions) Let u (z) be harmonic on the annulus fa < jzj < bg. Show that there is a constant C such that u (z) C log jzj has a harmonic conjugate on the annulus. Show that C is given by Z 2 @u r rei d ; C= 2 0 @r where r is any …xed radius, a < r < b.
Solution u has harmonic conjugate v1 on the annulus slit along ( b; a), and also a harmonic conjugate v2 on the annulus slit (a; b). Since v1 v2 is constant above the slit ( b; a), and also constant below the slit, v1 jumps by a constant across the slit. Arg z also jumps by a constant across the slit. By appropriate choice of C, v1 C Arg z is continuous across the slit ( b; a), and u C log jzj has a harmonic conjugate v1 C log jzj on the annulus. For the identity, use the polar form of the Cauchy-Riemann equations to convert the r–derivative of u to a –derivative of v.
22
III.3.5 The ‡ux of a function u across a curve is de…ned to be Z Z @u ds = ru n ds; @n where n is the unit normal vector to and ds is arc length. Show that if a harmonic function u on a domain D has a conjugate harmonic function v on D, then the integral giving the ‡ux is independent of path in D. Further, the ‡ux across a path in D from A to B is v (B) v (A).
Solution ! ds ds t = dx n = dy ; dy , ! ; dx , ds ds R R R @u @u dy @u dx ru! n ds = ds = dx + @u dy |{z} = @x ds @y ds @y @x R @v R @v dx + @y dy = dv @x ( ) C–R equations. If goes from P to Q, this is v (Q) v (P ), which shows that the integral is independent of path in D.
23
III.4.1 Let f (z) be a continuous function on a domain D. Show that if f (z) has the mean value property with respect to circles, as de…ned above, then f (z) has the mean value property with respect to disks, that is if z0 2 D and D0 is a disk RR centered at z0 with area A and 1 contained in D, then f (z0 ) = A D0 f (z) dx dy. Note: This MVP for area is not exactly what is de…ned as the MVP for disks. What is then is that if h (z0 ) is the average of h (z) for any circle fjz z0 j = rg, 0 < r < 1. Then h (z) is the average of h (z) over any disk fjz z0 j = r0 g, 0 < r < r0 . Express dx dy in polar coordinates at z0 and integrate …rst with respect to . Solution Suppose that f satis…es the mean value property, Z 2 1 f z0 + rei d f (z0 ) = 2 0
for all z0 2 D and r > 0 such that Br (z0 ) D. Let D0 = fz 2 C : jz z0 j Rg D. We have, A = R2 and parametrizing in polar coordinates, and use the mean value property with respect to circles,
ZZ 1 x = x0 + r cos dx dy = r dr d 0 r R f (z) dx dy = f (z) dx dy = 2 y = y0 + r sin 0 2 R D0 D0 Z 2 Z R Z R Z 2 Z R 1 1 1 = f z0 + rei r dr d = f z0 + rei d r dr = 2 f (z0 ) r d 2 2 R 0 R 0 R2 0 0 0 Z R 1 r2 2f (z0 ) R = r dr = 2 f (z0 ) = f (z0 ) : R2 R 2 0 0 1 A
ZZ
Another note: It’s better to use something other than 1=A as A was already used as the circle average.
24
III.4.2 Derive (4.2) from the polar form of the Cauchy-Riemann equations (Exercise II.3.8).
Solution The polar form of Cauchy-Riemann’s equations (from Exercise II.3.8) 1 @v @u = ; @r r@
@u = @
r
@v : @r
Now we have r
Z
0
2
@u z0 + rei @r
d =
Z
0
25
2
@v z0 + rei @
d =0
III.4.3 A function f (t) on an interval I = (a; b) has the mean value property if f
s+t 2
=
f (s) + f (t) ; 2
s; t 2 I:
Show that any a¢ ne function f (t) = At + B has the mean value property. Show that any continuous function on I with the mean value property is a¢ ne.
Solution (A. Kumjian) Let I := (a; b) and let f : I ! R be given. Suppose …rst that f is an a¢ ne function, that is, there are A; B 2 R such that f (t) = At + B for all t 2 I. Now let s; t 2 I be given, then f
s+t 2
=A
s+t 2
+B =
1 1 f (s) + f (t) (As + B) + (At + B) = : 2 2 2
Hence, f has the mean value property. Next suppose that f is continuous and that it has the mean value property. We must prove that f is a¢ ne. We claim that for any c; d 2 I, with c < d, and t 2 [0; 1] we have f (tc + (1
t) d) = tf (c) + (1
t) f (d) :
So let c; d 2 I be given with c < d. We will …rst prove that equation ( ) holds for t = k=2m where k; m are nonnegative integers with k 2m . We do this by induction on m. For m = 0, we have t = k = 0; 1 and the assertion is obvious. Now suppose that the assertion holds for some integer m 0 (that is, equation ( ) holds for t = k=2m for all k = 0; : : : ; 2m ). This inductive step entails showing that equation ( ) holds for t = k=2m+1 for any nonnegative integer k with k 2m+1 . By the inductive hypothesis we may assume k = 2j + 1 for some integer j = 0; : : : ; 2m 1 (for if k were even we could rewrite t as j=2m for some integer j with 0 j 2m ). Then setting t0 = j=2m and t1 = (j + 1) =2m , we have t = (t0 + t1 ) =2 and 1 t = ((1 t0 ) + (1 t1 )) =2, thus,
26
tc + (1
t) d =
1 ((t0 c + (1 2
t0 ) d + (t1 c + (1
t1 ) d))
hence,
f (tc + (1
t) d) = =f
1 ((t0 c + (1 2
t0 ) d + (t1 c + (1
t1 )d)
=
1 (f (t0 c + (1 t0 ) d) + f (t1 c + (1 t1 ) d)) = by the mean value property 2 1 = (t0 f (c) + (1 t0 ) f (d) + t1 f (c) + (1 t1 ) f (d)) = by inductive hypothesis 2 = tf (c) + (1 t) f (d) : =
So we have proved that equation ( ) holds for t = k=2m where k; m are nonnegative integers with k 2m . Since such numbers are dense in [0; 1] the claim follows by the continuity of f . It is now straightforward to verify that for any c; d 2 I, with c < d, there are unique constants A and B such that f (x) = Ax + B for all x 2 [c; d] (take A = (f (c) f (d)) = (c d) and B = f (c) Ac ). The desired result now follows by observing that A and B do not depend on c; d.
27
III.4.4 Formulate the mean value property for a function on a domain in R3 , and show that any harmonic function has the mean value property. Hint. For A 2 R3 and r > 0, let Br be the ball of radius r centered at A, with volume element d , and let @Br be its boundary sphere, with area element d and unit outward normal vector n. Apply the Gauss divergence theorem ZZZ ZZ r Fd F nd = Br
@Br
to F = ru.
Solution Prove MV theorem for harmonic functions in Rn , as follows u (R! x) R R @u ! ! ! u 0 = 0 @r (r x ) dr. Let d = area measure, x = unit vector, get R @u ! R @u ! R R R @2u ! (R x ) d (r x ) d = (r x ) drd (! x ). Get directional @r
@r
0
0
S @r2
derivative of Strokes formula. Better: Apply stroke’s theorem to shell f 0 < < get R R! R R ! ::: V ! n d surface = : : : r V volume. Apply to ru = ! v, Volume R R R R get : : : ru! n d surface = : : : r2 ! n d volume = 0, R R R @u R @u ::: @ d ::: @ d ! ! j x x0 j= 0 j x x0 j=R 1 R @u : : : ud _ is constant. @ ! x x = j 0j
28
1 g,
III.5.1 Let D be a bounded domain, and let u be a real-valued harmonic function on D that extends continuously to the boundary @D. Show that if a u b, then a u b on D.
Solution Since D [ @ is a compact set we have that u attains both maximum and minimum values. Assume that u (z0 ) = c > b fore some z0 2 D. Then it follows by the maximum principle that u c on D, but this is a contradiction to the asumption that u is continuously extended to @D. Thereby u b on D. For u we have that b u a on @D. Assume that u (z0 ) = c0 > a for seme z0 2 D. Then by the maximum principle u c0 on D, but this contradicts that u is continuously extended to @D with u a on @D. So u a on D, equivalently a u b on D. This show that a u b on D.
29
III.5.2 Fix n 1, r > 0 and = ei' . What is the maximum modulus of z n + over the disk fjzj rg? Where does z n + attain its maximum modulus over the disk?
Solution (A. Kumjian) We claim that the maximum modulus of z n + over the disk fz 2 C : jzj rg is rn + and this is attained at = rei where = (2k + ')=n for k = 0; 1; : : : ; n 1. By the maximum modulus principle the maximum modulus occurs on the boundary fz 2 C : jzj = rg. For z 2 C with jzj = r we have by the triangle inequality jz n + j
jz n j + j j = rn +
so the claim is proven.
30
= (rn + ) ei' = j
n
+ j:
III.5.3 Use the maximum principle to prove that the fundamental theorem of algebra, that any polynomial p (z) of degree n 1 has a zero, by applying the maximum principle to 1=p (z) on a disk of large radius.
Solution It is su¢ cient to show that any p (z) has one root, for by division we can then write p (z) = (z z0 ) g (z), with g (z) of lower degree. Note that if p (z) = an z n + an 1 z n
1
+
+ a0 ;
then as jzj ! 1, jp (z)j ! 1. This follows as a0 an 1 + + n : z z 1 Assume p (z) is non-zero everywhere. Then p(z) is bounded when jzj R. 1 1 Also, p (z) 6= 0, so p(z) is bounded for jzj R by continuity. Thus, p(z) is a bounded entire function, which must be constant. Thus, p (z) is constant, a contradiction which implies p (z) must have a zero. p (z) = z n an +
Solution (K. Seip) If p (z) has no zeros, then 1=p (z) is an entire function. Also, if we denote by m (R) the maximum of 1=p (z) on the circle jzj = R, then m (R) ! 0 when R ! 1, unless p (z) is a constant. By the maximum principle j1=p (z)j m (R) when jzj R, which means 1=p (z) = 0. This is impossible, and so 1=p (z) is not an entire function.
31
III.5.4 Let f (z) be an analytic function on a domain D that has no zeros on D. (a) Show that if jf (z)j attains its minimum on D, then f (z) is constant. (b) Show that if D is bounded, and if f (z) extends continuously to the boundary @D of D, then f (z) attains its minimum on @D.
Solution (a) If jf (z)j attains its minimum on D, then j1=f (z)j attains its maximum on D, which can only happens if f (z) is a constant, by the maximum principle. (b) Since D [ @D is compact and f (z) is continuous it follows that jf (z)j attains both maximum and minimum value on D [ @D. Assume that jf (z)j does not attain its minimum on @D. Then jf (z)j attains its minimum on D and it follows by (a) that f is constant so jf (z)j attains its minimum on @D.
32
III.5.5 Let f (z) be a bounded analytic function on the right half-plane. Suppose that f (z) extends continuously to the imaginary axis and satis…es jf (iy)j M for all points iy on the imaginary axis. Show that jf (z)j M for all z in the right half-plane. Hint. For " > 0 small, consider (z + 1) " f (z) on a large semidisk. Solution Now we consider the function (z + 1) " f (z) in the right half-plane, because f (z) is bounded here, we assume that jf (z)j C in this domain. On a semidisk of radius r in the right half-plane we have (z + 1)
"
C
f (z)
jr
For r = R su¢ cient large, we have (z + 1)
"
1j"
C
f (z)
(R
1)"
M
for all jzj R, when Re z 0. On the imaginary axis we have that (z + 1)
"
f (z)
M
because jf (iy)j M for all points iy on the imaginary axis. By the maximum principle, we have (z + 1)
"
f (z)
M
for jzj R, when Re z 0. Now let " ! 0, we have jf (z)j
M
for all z in the right half-plane. Solution (K. Seip) We assume that jf (z)j
C for Re z > 0 and jf (iy)j F" (z) = (1 + z) 33
"
f (z) :
M . For " > 0 set
Then jF" (iy)j
M
and F" Rei
C R
"
M
for su¢ ciently large R. Thus for arbitary z, Re z > 0 we have jF" (z)j
M;
or jf (z)j
M (1 + jzj)" :
This holds for every " > 0, thus jf (z)j
34
M.
III.5.6 Let f (z) be a bounded analytic function on the right half-plane. Suppose that lim sup jf (z)j M as z approaches any point of the imaginary axis. Show that jf (z)j M for all z in the half-plane. Remark. This is a technical improvement on the preceding exercise for students who can deal with lim sup (see Section V.1).
Solution Set g (z) = (z + 1)
"
f (z) :
Then jg (z)j 6 jf (z)j becauce jz + 1j > 1 for all z in the right half-plane. Take R large so that jg (z)j 6 M for jzj > R. The lim sup condition implies that there is > 0 such that jg (z)j 6 M + ";
jzj < R; 0 < Re z 6 :
Apply the maximum principles to the domain fzj jzj < R; Re z > g, to obtain jg (z)j 6 M + ". Then g (z) ! f (z) as " ! 0.
35
III.5.7 Let f (z) be a bounded analytic function on the open unit disk D. Suppose there are a …nite number of points on the boundary such that f (z) extends continuously to the arcs of @D separating the points and satis…es f ei M there. Show that jf (z)j M on D. Hint. In the case that there is only one exceptional point z = 1, consider the function (1 z)" f (z).
Solution Let the points be a1 ; : : : ; an . Then (z aj )" , where 1 j n is analytic " (take analytic branch). We have that jz aj j ! 0 as z ! aj for 1 j n. Q Then [ (z aj )" ] f (z) = f" (z) satis…es f" (z) is continuous in D [ @D. jf" (z)j 6 M on @D. jf" (z)j 6 M on D. Let " ! 0, obtain jf (z)j 6 M on D. Solution Let C = supz2D jf (z)j. By de…nition jf (z)j C. Let z1 ; : : : ; zn 2 @D be all the points for which f (z) does not extend continuously. Now suppose that jf (z)j M for all z 2 @D except fz1 ; : : : ; zn g and to the contrary M < C, then there is a z0 2 D such that jf (z0 )j = M + with 0< C M. Let g (z) = (z
z1 )" : : : (z
zn )" f (z)
where " is chosen so that 8z 2 D [ @D. We have that M + =2 M+
j(z
z1 )" : : : (z
zn )" j
and
j(z
z1 )" : : : (z
zn )" j
Note that g (z) extends continuously to @D and that g (z) is analytic in D. By the maximum modulus principle, and since jg (z0 )j = j(z0
z1 )" : : : (z0
zn )" j f (z0 )
we have that 36
(M + )
M + =2 M+
M + =3 M
M + =2 (M + ) M+
jg (z0 )j
sup j(z
z2@D
which is a contradiction.
37
z1 )" : : : (z
zn )" j jf (z)j
M + =3 M M
III.5.8 Let f (z) be a bounded analytic function on a horizontal strip in the complex plane. Suppose that f (z) extends continuously to the boundary lines of the strip and satis…es jf (z)j M there. Show that jf (z)j M for all z in the strip. Hint. Find a conformal map of the strip onto D and apply Exercise 7.
Solution < Im z < 2 , so it is mapped to right half-plane by Assume strip is 2 w = i log z, z = eiw . Then g (w) = f (eiw ) hold on right half–plane, 6 M at all point of iR except at w = 0. Modify proof idea of Exercise III.5.5 to get jgj 6 M on the half–plane.
38
III.5.9 Let D be an unbounded domain, D 6= C, and let u (z) be a harmonic function on D that extends continuously to the boundary @D. Suppose that u (z) is bounded below on D, and that u (z) 0 on @D. Show that u (z) 0 on D. Hint. Suppose 0 2 @D, and consider functions of the form u (z) + log jzj on D \ fjzj > "g. Solution (From Hints and Solutions) Let " > 0. Take > 0 such that u (z) > " for > 0. Take R > 0 so large that u (z) + log jzj maximum principle, u (z) + log jzj > " + log hence for all z 2 D such that jzj > . Let ! 0, u (z) > 0 on D.
z 2 D, 0 < jzj < . Let > 0 for jzj > R. By the for z 2 D, < jzj < R, then let " ! 0, to obtain
Solution (K. Seip) We may assume 0 2 @D. By continuity of u at 0, we can …nd ( ; > 0) u (z) + log jzj
";
and
so that
z 2 D \ fjzj = g
for arbitary " > 0. Since u is bounded below, there is also R0 > 0 such that for R > R0 u (z) + log jzj
z 2 D \ fjzj = Rg ;
0;
Applying the maximum principle to u (z) log jzj in D \ f < jzj < Rg, we get u (z) " log jzj. Since " and can be chosen arbitarily small, the result follows.
39
III.5.10 Let D be a bounded domain, and let z0 2 @D. Let u (z) be a harmonic function on D that extends continuously to each boundary point of D except possibly z0 . Suppose u (z) is bounded below on D, and that u (z) 0 for all z 2 @D, z 6= z0 . Show that u (z) 0 on D.
Solution D hold, z0 2 @D, u > M on D, u > 0 (@D) n fz0 g. The map ' (z) = 1= (z z0 ) maps D onto an unbounded domain, to which the result of Exercise 9 applies. Or: consider w (z) = u (z) + log (1= jz z0 j). Assume jz z0 j 6 R on D, then take " > 0 small, assume R > 1. Then w (z) > " log (1=R) at all points of (@D) n fz0 g, w (z) ! +1 as z ! z0 . By the maximum principle, w (z) > " log (1=R) on D. Let " ! 0, get w (z) > 0 on D.
40
III.5.11 Let E be a bounded set of integer lattice points in the complex plane. A point m + ni of E is an interior point of E if its four immediate neighbors m 1 + ni, m + ni i belong to E. Otherwise m + ni is a boundary point of E. A function E is harmonic if its value at any interior point of E is the average of its values at the four immediate neighbors. Show that a harmonic function on a bounded set of lattice points attains its maximum modulus on the boundary of the set.
Solution Suppose u (z) attains its maximum at m + ni, call it M . Then u (m + ni) is the average of its values at its four neighbors, so all their values must coincide with M. Consider fm + ni : u (m + ni) = M g. This is a …nite set. It have a point with largest m. This point have a neighbor in the boundary of E. u attains value M at that point. u = c at some point of boundary of E. For complex–valued case, follow same approach as in book. Remark: Should just consider real-valued u, and show it attains its maximum and minimum at boundary points of E.
41
III.6.1 Consider the ‡uid ‡ow with constant velocity V = (2; 1). Find the velocity potential (z), the stream function (z), and the complex velocity potential f (z) of ‡ow. Sketch the streamlines of the ‡ow. Determine the ‡ux of the ‡ow across the interval [0; 1] on the real axis and across the interval [0; i] on the imaginary axis.
Solution V = (2; 1), V = r , = 2x + y, @@x = @@y , @@y = @@x ) = 2y x f (z) = + i = 2z iz = (z i) z Streamlines are straight line with slope 1=2, ‡ux across [0; 1] = (1) (0) = 1, ‡ux across [0; i] = (i) (0) = 2.
42
III.6.2 Fix real numbers and polar coordinates by
, and consider the vector …eld given in
V (r; ) =
ur + u ; r r where ur and u are the unit vectors in r and directions, respectively. (a) Show that V (r; ) is the velocity vector …eld of a ‡uid ‡ow, and …nd the velocity potential (z) of the ‡ow. (b) Find the stream function (z) and the complex velocity potential f (z) of the ‡ow. (c) Determine the ‡ux of the ‡ow emanating from the origin. When is 0 a source and when is 0 a sink? (d) Sketch the streamlines of the ‡ow in the case = 1 and = 1.
Solution a) V (r; ) = r ur + r u r = @@r ur + 1r @@ u = r ur + r u Get @@r = r , = log r + h ( ) 1@ = h0 ( ) = r , h0 ( ) = , h ( ) = )V =r r@ = log r + is harmonic locally. is the velocity vector …eld of a ‡ow b) = Re ( log z + arg z) stream function = = Im ( log z + arg z) = arg z log jzj complex velocity of a point f (z) = ( i ) log z c) Flux = of around a circle centred at 2 , sin u : > 0, sin u : > 0. d) = 1, = 1. Note: and are only locally de…ned.
43
III.6.3 Consider the ‡uid ‡ow with velocity V = r , where (r; ) = (cos ) =r. Show that the streamlines of the ‡ow are circles and sketch them. Determine the ‡ux of the ‡ow emanating from the origin.
Solution = cosr =
r cos r2
= Re
1 z
= Re
z
R jzj
= Im
2
1 z
=
sin r
Since is single-valued, ‡ux = jzj=" d = 0, and there is no source or sink at 0. Set w = 1=z, z = 1=w, these streamlines are the curves Im (1=z) = constant, which correspond to the curves Im w = constant get modi…ed to tangent to real line at 0 = z (1), and real axis. r = @@r ur + 1r @@ u = cos ur sin u. r2 r2
44
III.6.4 Consider the ‡uid ‡ow with velocity V = r , where (z) = log
z 1 : z+1
Show that the streamlines of the ‡ow are arcs of circles and sketch them. Determine the ‡ux of the ‡ow emanating from each of the singularities at 1.
Solution is the composition of an analytic function w = (z 1) = (z + 1) and the harmonic function log jwj , so is harmonic, and = Re log ((z 1) = (z + 1)). The stream function is = ImH(log (z 1) = H(z + 1)) = arg ((z 1) = (z + 1)) , then is not single-valued. jz 1j=" d = jz 1j=" d arg (z 1) = 2 = ‡ux H H emanating from +1. jz 1j=" d = jz 1j=" d arg (z + 1) = 2 = ‡ux emanating from 1. z = 1 is a source, z = 1 is a sink. Set = (z 1) = (z + 1), w = log . The line Im w = constant in the w plane correspond to rays issuing from 0 in the plane. (since = ew ), and this correspond to arcs of circles from 1 to 1 in the z–plane, since FLT´s maps circles to circles. Hence streamlines are arcs of circle from 1 to 1, including the circle through 1.
45
III.6.5 Consider the ‡uid ‡ow in the horizontal strip f0 < Im z < g with a sink at 0 and equal sources at 1. Find the stream function (z) and the velocity vector …eld V (z) of the ‡ow. Sketch the streamlines of the ‡ow. Hint. Map the strip to a half-plane by = ez and solve a Dirichlet problem with constant boundary values on the three intervals in the boundary separating sinks and sources.
Solution Consider ‡uid ‡ow in a horizontal strip. f0 < Im z < g with a sink 0, but no source at 1 . Find the stream function (z). Sketch the ‡ow lines. Find the complex vector …eld V (z). Hint : Map the half-plane by. Map to half-plane = ez . A arg + B arg ( 1) + C, arg = arg ez = y, Ay+C+B arg (ez 1), c3 = 0, requires A+ B+C = 0 ) C = (A + B), A (y ) + B [arg (ez 1) ]. C1 = C2 require A + B B = z [ A B ], A [ (y ) + 2 [arg (e 1) ]] = A [ ( + y) + 2 arg (ez
46
1)]
III.6.6 For a ‡uid ‡ow with velocity potential (z), we de…ne the conjugate ‡ow to be the ‡ow whose velocity potential is the conjugate harmonic function (z) of (z). What is the stream function of the conjugate ‡ow? What is the complex velocity potential of the conjugate ‡ow?
Solution Stream function of conjugate ‡ow in direction r is potential is g (z) = i = if (z) = i ( + i )
47
. Complex velocity
III.6.7 Find the stream function and the complex velocity potential of the conjugate ‡ow associated with the ‡uid ‡ow with velocity vector ur =r. Sketch the streamlines of the conjugate ‡ow. Do the particles near the origin travel faster or slower than particles on the unit circle?
Solution Flow ur =r, = log jzj, = arg z. Complex velocity potential of conjugate ‡ow is i log z. Streamlines of the conjugate ‡ow are = log jzj = constant , r = jzj = constant, which are circles centered in origin. Speed = jV (z)j = jf 0 (z)j = 1= jzj. Particles travel faster near z = 0.
48
III.6.8 Find the stream function of the conjugate ‡ow of 1 ( ur + u ) : r Sketch the streamlines of both the ‡ow and the conjugate ‡ow on the same axis. (See Exercise 2d.) V (r; ) =
Solution This is the vector …eld from 2d. Hence V = r ( log r + ). The velocity potential is = log r + . The conjugate ‡ow has velocity potential = log r = Re ((i 1) log z). The stream function of the conjugate ‡ow is then Im ((i 1) log z) = + log r = . (The steam function of the conjugate is always the negative of the velocity potential of the ‡ow. See Ex.6) Streamline of conjugate ‡ow are orthogonal to the stream of the ‡ow (Exercise 2d) and velocity vector …eld. Conjugate ‡ow is the rotation by 90 of the velocity vector …eld of the ‡ow, so the origin is a sink of both ‡ows in this case.
49
III.7.1 Find the steady-state heat distribution u (x; y) in a laminar plate corresponding to the half disk fx2 + y 2 < 1; y > 0g, where the semicircular top edge is held at temperature T1 and the lower edge ( 1; 1) is held at temperature T2 . Find and sketch the isothermal curves for the heat distribution. Hint. Consider the steady-state heat distribution for the full unit disk with the top held at temperature T1 and the bottom temperature T3 , where T2 = (T1 + T3 ) =2. Solution (z) = 2 (Arg (1 + z)
z)) has temp +1 on top, 1 on bottom T1 = A + B half and is 0 on the interval ( 1; 1). Take u = A + B, ) T2 = B A = T/ 1 T2 , get u = (T1 T2 ) (z) + T2 B = T2 u = (T1 T2 ) 2 [Arg (1 + z) Arg (1 z)] + T2 The isothermal curves are the curves where u = konstant. these are the curves Arg (1
Arg
1+z 1 z
= konsant
then (1 + z) (1
z) = 1 + 2i Im z
and Im z = konstant. 1 jzj2
50
jzj2
III.7.2 Find the potential function (x; y) for the electric …eld for a conducting laminar plate corresponding to the quater-disk fx2 + y 2 < 1; x > 0; y > 0g, where the two edges on the coordinate axis are grounded (that is, = 0 on the edges), and the semicircular edge is held at constant potential V1 . Find and sketch the equipotential lines and the lines of force for the electric …eld. Hint. Use the conformal map = z 2 and the solution to the preceding exercise.
Solution (z) = 2 (Arg (1 + w (z))
Arg (1
w (z))),
51
1
= V1 2 (Arg (1 + z 2 )
Arg (1
z 2 ))
III.7.3 Find the potential function (x; y) for the electric …eld for a conducting laminar plate corresponding to the unit disk where the boundary quater-circles in each quadrant are held at a constant voltages V1 ; V2 ; V3 and V4 . Hint. Map the disk to the upper halfplane by w = w (z) and consider potential functions of the form Arg (w a).
Solution Let w = i (1 z) = (1 + z), w (1) = 0, w ( 1) = 1, w (i) = 1, w ( i) = 1. z ! w (z) maps unit disk to upper half–plane. Su¤ers to …nd a harmonic function u (w) in upper half-plane with boundary values constant on each interval, as above. Any such function is a linear combination of u 1 (w) = 1 Arg (w + 1), u0 (w) = 1 Arg w, u1 (w) = 1 Arg (w 1), and if = A 1 u 1 + A0 u0 + A1 u1 + C, we get a system we can back solve. 8 8 A + A + A + C = V C = V2 > > 1 0 1 3 > > < < A0 + A1 + C = V3 A1 = V1 V2 ) A1 + C = V1 A0 = V4 V1 > > > > : : C = V2 A 1 = V3 V4 That does it-just plug in. Example: V2 = 1, V1 = V3 = V4 = 0, this solution is and we P can get other elementary solutions, 1 (z), (z) = vj j (z).
52
2
(z) = 1 1 Arg i 11+zz , 3 (z), 4 (z), and then
III.7.4 Find the steady-state heat distribution in a laminar plate corresponding to the vertical half-strip fjxj < =2; y > 0g, where the vertical sides at x = =2 are held at temperature T0 = 0 and the bottom edge ( =2; =2) on the real axis is held at temperature T1 = 100. Make a rough sketch of the isothermal curves and the lined of heat ‡ow. Hint. Use w = sin z to map the strip to the upper half-plane, and make use of harmonic functions of the form Arg (w a).
Solution Try
= a Arg (w + 1) + b Arg (w
8 < a = 50 b = 50 : c=0 (z) = 50 Arg (sin z + 1)
8 < a+b+c=0 a b + c = 100 1) + c, want : a b+c=0
50 Arg (sin z
53
1).
)
III.7.5 Find the steady-state heat distribution in a laminar plate corresponding to the vertical half-strip fjxj < =2; y > 0g, where the side x = =2 is held at constant temperature T0 , the side x = =2 is held at constant temperature T1 , and the bottom edge ( =2; =2) on the real axis is insulated, that is, no heat passes through the bottom edge, so the gradient ru of the solution u (x; y) is parallel there to the x axis. Hint. Try linear functions plus constants.
Solution =0 u (x; y) = ax + b is harmonic. @u @y T1 T0 a + b = T0 a= 2 ) 1 a + b = T1 b = T0 +T 2 2 1 u = 1 (T1 T0 ) x + T0 +T . 2
54
III.7.6 Find the steady-state heat distribution in a laminar plate corresponding to the upper half-plane fy > 0g, where the interval ( 1; 1) is insulated, the interval ( 1; 1) is held at temperature T0 , and the interval (1; 1) is held at temperature T1 . Make a rough sketch of the isothermal curves and the lines of heat ‡ow. Hint. Use the solution to Exercise 5 and the conformal map from Exercise 4.
Solution Let z = z (w) map the vertical half-strip to the upper half-plane with z ( =2) = 1 and z ( =2) = 1, w = w (z) : upper half–plane ! vertical semi strip. Take z = sin w, this has desired mapping property, w = sin 1 (z). 1 . Then solution is composition of sine function with 2 (T1 T0 ) Re w + T0 +T 2 T0 +T1 1 1 2 Get u = (T1 T0 ) Re sin (z) + 2 , where we take branch sin z with values in the vertical semi-strip. Note : This conformal map was used in Exercise 4.
55
III.7.7 The gravitational …eld near the surface of the earth is approximately constant, of the form F = ck, where k is the unit vector in the z direction in (x; y; z) space and the surface of the earth is represented by the plane where z = 0 (the ‡at earth theory). Show that F is conservative, and …nd a potential function for F .
Solution ! (x; y; z) = cz, r = c k Since r
F =
@ @ @ ; ; @x @y @z
we have that F is conservative.
56
(0; 0; c) = (0; 0; 0) ;
III.7.8 Show that the inverse square force …eld F = ur =r2 on R3 is conservative. Find the potential function for F , and show that is harmonic.
Solution ! ! r = ur2r = r ,
(x; y; z) =
1 r
57
III.7.9 For n 3, show that the function 1=rn 2 is harmonic on Rn n f0g. Find the vector …eld F that has this function as its potential.
Solution 1 1 2 = n 2 = u = r rn 2 2 +:::+x2 2 x (1 n) @u n = 1 2 (x21 + : : : + x2n ) @xj @2u @x2j 2 n rn
= (2
n) r
(2 n)(
n
+ (2
n n 2
n) xj
1
n 2
n) xj r
n
= (x21 + : : : + x2n ) 2xj = (2 n 2
(x21 + : : : + x2n )
n)x2j
n 2
1
2xj =
+ r n+2 (2 n)n P 2 2 n u = n rn xj = 0 this shows that u is harmonic on Rn n f0g r n+2 P @u P x F = u= e = (2rnn) xj ej = r2n n1 r j ej = (2rnn)~1ur @xj j !
F = (2r) rrn = rcnu 1 R P F d = rnc 1 Area (s) = (n
2) Area(S) rn 1
58
IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8
1
IV.1.1 1 2 3 P L
K F KKK FFF Let be the boundary of the triangle f0 < y < 1 x; 0 < x < 1g, with the usual counterclockwise orientation. EvaluateR the following integrals R R (a) Re z dz (b) Im z dz (c) z dz Solution a) We begin do split the contur into three parts so that And integrate along each side in the triangle
= We parametrizize and dz = dt Z
1
+
2
+
= [0; 1], thus
1
(t) = t, for 0
Re z dz = 1
1
Z
1
Z
0
1
t
2
+
3.
1 t dt = : 2
We parametrizize 3 = [i; 0], thus 3 (t) = i (1 t), for 0 Re (i (1 t)) = 0 and dz = i dt, then Z Z 1 Re z dz = (0) ( i) dt = 0: 0
2
1, then Re (t) = t
We parametrizize 2 = [1; i], thus 2 (t) = 1 t + it, for 0 Re (1 t + it) = 1 t and dz = ( 1 + i) dt, then Z Z 1 1 1 Re z dz = (1 t) ( 1 + i) dt = + i: 2 2 0 2
3
+
3
Re (t) dz = 1
=
t
t
1, then
1, then,
And now we take the parts in the integral together Z Z Z Z i Re z dz = Re z dz + Re z dz + Re z dz = 2 1 2 3 b) For details see a)
Z
Z
Im z dz = 1
Z
Im z dz = Z2
Z
1
0 dt = 0;
0
1
t ( 1 + i) dt =
0
Im z dz =
Z
1
(1
t) ( i) dt =
0
3
1 1 + i; 2 2 1 i: 2
thus Z
Im z dz =
1 : 2
c) For details see a)
Z
Z z dz = 2
Z
z dz =
Z 11 0
z dz = 3
Z
0
(1
Z
1
1 t dt = ; 2
t + it) ( 1 + i) dt =
1
(i (1
0
1 t)) ( i) dt = : 2
thus Z
z dz = 0:
3
1
IV.1.2 1 2 3 P L K 111 LLL Absolutbeloppet i c) nagot a •r f el Let be the unit circle fjzj = 1g, with the usual counterclockwise orientation. following Rintegrals, for m = 0; 1; 2; : : :. R m EvaluateR the (a) z dz (b) z m dz (c) z m jdzj
Solution Set z = ei , where 0 (a) Z
m
z dz =
Z
2 , and thus dz = iei d .
2 im
e
i
ie d = i
0
Z
2
ei(m+1) d = m+1
i(m+1)
e
0
=
1 ei2 m+1
(m+1)
1 =
2
= 0
0; m 6= 2 i m=
1 1:
(b) Z
m
z dz =
Z
2 im
e
i
ie d = i
0
Z
2
ei(1
m)
d =
ei2
(m 1)
0
ei(1 m) = 1 m
2
= 0
1 1
m
1 =
0; m 6= 1 2 i m = 1:
1 =
0; 2
(c) Z
m
z jdzj =
Z
0
2 im
e
eim d = im
2
= 0
4
1 ei2 im
m
m 6= 0 m = 1:
IV.1.3 1 2 3 P L K 111 Let be the circle fjzj = Rg, with the usual counterclockwise orientation. integrals, for m = 0; 1; 2; : : :. R mEvaluate theR following R m (a) jz j dz (b) jz m j jdzj (c) z dz Solution Set z = Rei , where 0 (a)
2 , and thus dz = iRei d and jzj = R.
Z
i
m
jz j dz =
Z
2 m im
R e
iRe d = iR
m+1
0
Z
2
ei d = Rm+1 ei
0
2 0
= Rm+1 ei2
1 = 0:
(b) Z
m
jz j jdzj =
Z
2 m im
R e
i
iRe
d =R
m+1
0
Z
2
d = Rm+1 [ ]20 = Rm+1 (2
0
0) = 2 Rm+1 :
(c) Z =
m
z dz =
Z
2 m
R e
im
i
iRe d = iR
Z
2 i(1 m)
e
0
0
Rm+1 i2 e 1 m
m+1
(1 m)
1 =
0; m 6= 1; = m+1 2 iR ; m = 1;
5
d =R
m+1
ei(1 m) 1 m
0; m 6= 1; 2 2 iR ; m = 1:
2
= 0
IV.1.4 1 2 3 P
L K LLL Show that if D is a bounded domain with smooth boundary, then Z z dz = 2i Area (D) : @D
Solution Substitute z = x
iy, dz = dx + idy, and apply Green‘s theorem.
6
IV.1.5 1 2 3 P L K 111 KKK Show that I
138
jz 1j=1
ez dz z+1
2 e2 :
Solution We begin with parametrizize the circle jz 1j = 1, as z = 1+eit = 1+cos t+ i sin t where 0 t 2 . For the nominator in the integrand ez = (z + 1) we have jez j = eRe z+i Im z = eRe z ei Im z = eRe z
e2 ;
since Re z 2 on the circle. And for the denominator in the integrad we have j1 + zj = j2 + cos t + i sin tj
Re (2 + cos t + i sin t) = 2 + cos t
1;
Thus M=
jez j ez = 1+z jz + 1j
e2 = e2 : 1
Becauce the integrationcontour is a circle with radius 1 we have that L = 2 . M L estimate gives I ez 2 e2 : dz jz 1j=1 z + 1
7
IV.1.6 1 2 3 P L K 111 KKK Show that there is a strict inequality I p log R Log z ; dz 2 2 z2 R jzj=R
R>e :
Solution For the integrand Log z=z 2 , with the nominator Log z = log R + i with the principalargument we have Log R = R2
p log2 R + R2
2
6
p
log2 R + R2
2
6
p
log2 R + log2 R = R2
p
2 log R : R2
there the last inequalitys follows by given fact R > e that gives 2 < log2 R. Becauce the integrationcontour is a circle with radius R we have that L = 2 R. M L estimate gives p p log R H 2 log R Log z dz 6 2 R = 2 2 ; R>e : jzj=R z2 R2 R
8
IV.1.7 1 2 3 P L K Show that there is a strict inequality I zn 2 Rn+1 dz ; R > 1; m m 1 Rm 1 jzj=R z
1; n
1:
Solution If jf (z)j 6 m0 < m on an arc R
f (z) dz +
0
R
0
of the circle,
=
0 [ 1,
then
f (z) dz 6 m0 length ( 0 )+mlength ( 1 ) = (m0
R
f (z) dz =
m) length ( 0 )+
1
mlength ( ) < mlength ( ) R Thus if f (z) dz = ML, then jf (z)j = M on . In this case, jz m j = Rn , 1 zm 1
6
1 , Rm 1
L = 2 R, get
R
ik=m with strict inequality unless z = R e|2 {z } . )M= th
zn
jzj=R z m 1
dz <
Rn
2 R=
Rm 1
9
2 Rn+1 , Rm 1
Rn , Rm 1
root of w
R > 1, m > 1, n > 0.
IV.1.8 1 2 3 P
L K LLL Suppose the continuous function f ei on the unit circle satis…es R f ei M and jzj=1 f (z) dz = 2 M . Show that f (z) = cz for some constant c with modulus jcj = M . Solution H Multiply f by a unimodular constant, assume jf (z)j 6 M , jzj=1 f (z) dz = R2 R2 i f Re f ei iei d = 2 M , e d = 2 M . Take real parts, have 0 0 i i i 6 M . If have strict inequality somewhere, Re f e ie d 6 f e R2 then 0 < 2 M . We conclude that Re f ei iei d M . Since f ei iei i i i i M , we have Im f e ie d 0. ) f e ie M , f ei ie i M , f (z) iM z. ) f (z) = cz for a constant, jcj = M .
10
6
IV.1.9 1 2 3 P L K Suppose h (z) is a continuous function on a curve . Show that Z h (z) dz; w 2 Cn ; H (w) = z w
is analytic on the complement of , and …nd H 0 (w).
Solution For the analyticity, di¤erentiate h by hand. (See i Exercise III.1.6). The DerivR h(z) h(z) ative is H 0 (w) = lim 1w dz = z (w+ w) z w w!1 R R h(z) w lim 1w (z (w+ dz = (zh(z)dz w))(z w) w)2
w!1
11
IV.2.1 1 2 3 P L K K 111 LLL KKK 246 Evaluate the following integrals, for a path that travels from i to i in the right half-plane, and also for a path from i to i in theR left half-plane.R R R (a) z 4 dz (b) ez dz (c) cos z dz (d) sinh z dz
Solution The integrals are all independent of path and can be evaluated by …nding a primitive. (a) Z
i
z5 5
z 4 dz =
= i
( i)5 5
i)5
( 5
=
2 5i : 5
(b)
(c) Z
Z
ez dz = [ez ]
cos z dz = [sin z] = sin ( i)
i i
i i
=e
i
i
e
=
1
( 1) = 0:
=
sin (
i) =
ei(
i)
e 2i
i( i)
=
ei(
i)
2 (e 2i
e i( i) = 2i e ) =i e e
:
(d) Z
sinh z dz = [cosh z]
i i
= cosh ( i)
cosh ( e i+e = 2
12
i) = i
e
i
+e 2
(
i)
= 0:
IV.2.2 1 2 3 P L K K 111 LLL 246 R Using an appropriate primitive, evaluate 1=z dz for a path that travels from i to i in the right half-plane, and also for a path from i to i in the left half-plane. For each path give a precise de…nition of the primitive used to evaluate the integral.
Solution In right half–plane use f (z) = Log z = log jzj + i , where < < primitive, get Z 1 dz = [Log z] i i = log j ij + i log j i j i = i: z 2 2
as a
In left half–plane use f (z) = log0 z = log jzj + i , where 0 < < 2 as a primitive, get Z 1 3 dz = [log0 z] i i = log j ij + i log j i j + i = i: z 2 2
13
IV.2.3 1 2 3 P L K 111 LLL KKK Show that if m 6= 1, then z m has a primitive on Cn f0g. Solution The primitive is
z m+1 , m+1
which is analytic for z 6= 0 and m 2 Z, m 6=
14
1.
IV.2.4 1 2 3 P L K p Let D = Cn ( 1; 1], and consider the branch of z 2 1 on D that is positive on the interval (1; 1). (a) p Show that z + z 2 1 omits the negative real axis, that is, the range of the function on D does not include any values in the interval ( 1; 0] on the real axis. (b) p p Show that Log z + z 2 1 is a primitive for 1= z 2 1 on D. (c) Evaluate Z dz p ; z2 1 where is the path from 2i to +2i in D counterclockwise around the circle jzj = 2. (d) Evaluate the integral above in the case is the entire circle jzj = 2, oriented counterclockwise. (Note that the primitive is discontinuous at z = 2.) Solution p D = Cn ( 1; 1], z 2 1 a) p p z + z 2 1 = t (t > 0) ) z 2 1 = t z, z 2 1 = t2 2z + z 2 , 2z = t2 + 1, z = (t2 + 1) =2. But values of the branch are > 0 on (1; 1) = D \ R. ) It assumes no negative values. b) p p Log z + z 2 1 is the composition of z+ z 2 1 and Log w on Cn ( 1; 0], p d so it’s analytic. By the chain role, dz Log z + z 2 1 = z+p1z2 1 1 + 21 (z 2 1) p1 z+ z 2 1 p of z 2
c)
1+
p z z2 1
=
p 1 , z2 1
where we use everywhere the constant branch
1, by ????? it on (1; 1) or using
15
d a z dz
=
az a z
1=2
2z =
R
p p p 2i = Log z + z 2 1 2i = Log 2i + i 5 Log 2i i 5 = i p Use fact that p z 2 1 is positive imaginary on positive imaginary axis, and p also Log 2i + i 5 Log 2i i 5 ?????. d) p R dz 2+i0 p = Log z + z 2 1 2+i0 2 z 1 p At 2 + 0t , haveLog z + z 2 1 ! p p log 2 + 3 + i Arg z + z 2 1 at z = 2. As z ! 2 + i0, have p p p p 3, arg z 2 p1 ! + , jz 2 1j ! 3. ) z + z 2 1 !p 2+( 1) 3 = 2 2 2 2 1 ! . As z ! 2+i0, have arg z 1! , jz 1j ! arg z + z p p p p 2 2 3. ) z + z 1 ! 2 + ( 1) 3 = 2 3, arg z + z 1 ! . Values coincide ) p R dz 2+i0 p = Log z + z 2 1 2+i0 = log jj log jj + i ( i) = 2 i z2 1 p dz z2 1
16
IV.2.5 1 2 3 P L K Show that an analytic function f (z) has a primitive in D if and R only if f (z) dx = 0 for every closed path in D. Solution (A. Kumjian) Let f be an analytic function de…ned on a domain D. Suppose …rst that f has a primitive in D. Then by the Fundamental Theorem of Calculus for R Analytic Functions (Part I) we have f (z) dx = 0 for every closed path 2 D (since for anyRsuch path A = B). Conversely, suppose f (z) dx = 0 for every closed path 2 D. By formula (1.1) on p. 102, we have Z Z f (z) dx + if (z) dy = f (z) dx = 0 for every close path 2 D. It follows that the integral on the left is path independent for paths which are not necessarily closed and hence, by the Lemma on page 77, there is a continuously di¤erentiable function F on D @ @ F = f and @y F = if . It follows that such that dF = f dx + if dy, that is, @x the real and imaginary parts of F satisfy the Cauchy-Riemann equations: @F @F @ @ Re F = Re = Re f = Im if = Im = Im F; @x @x @y @y @ @F @F @ Re F = Re = Re if = Im f = Im = Im F: @y @y @x @x Moreover, F 0 (z) = f (z) for all z 2 D. Hence, f has a primitive in D, namely F.
17
IV.3.1 1 2 3 P
L K LLL 2 By integrating e z =2 around a rectangle with vertices R; it R;and sending R to 1, show that Z 1 1 2 2 p e x =2 e itx dx = e t =2 ; 1 < t < 1: 2 1 Use the known value of the integral for t = 0. Remark. This 2 shows that e x =2 is an eigenfunction of the Fourier transform with eigenvalue 1. For more see the next exercise.
Solution R RR R t ( R+iy)2 =2 R t (R+iy)2 =2 RR 2 z 2 =2 x2 =2 e dz = e dx e idy+ e idy+ e (x+it) =2 dx = @D R 0 0 R 0 p R1 2 2 e ( R+iy) =2 , 0, uniformly for 0 6 y 6 1 as R ! 1. 2 = 1 e x =2 dx = R1 R1 2 2 (x+it)2 =2 e dx = e (x +2ixt t )=2 dx = 1R 1 2 2 1 et =2 1 e x =2 e ixt dx R1 2 2 ) p12 e x =2 dx = e t =2 1
18
IV.3.2 1 2 3 P L K We de…ne the Hermite polynomials Hn (x) and Hermite orthogonal functions 0 by n (x) for n Hn (x) = ( 1)n ex
2
dn e dxn
x2
;
n
(x) = e
x2 =2
Hn (x) :
(a) Show that Hn (x) = 2n xn + is a polynomial of degree n such that is even when n is even and odd when n is odd. (b) By integrating the function e(z
dn 2 e z n dz R; it R and sending R to 1, show that
it)2 =2
around a rectangle with vertices Z 1 1 itx p dx = ( i)n n (t) ; 1 < t < 1: n (x) e 2 1 Hint. Use the identity from Exercise 1, and also justify and use the identity dn e dxn (c) Show that (d) R 00 Using n
00
n
x
m
dx =
n
(x+it)2
+ (2n + 1)
R
Z
00
n n 1 n
n
=
1 dn e in dtn
(x+it)2
:
= 0.
dx and (c), show that (x)
m
(x) dx = 0;
1
n 6= m:
Remark. This shows that the n ’s form an orthogonal system of eigenfuctions for the (normalized) Fourier transform operator F with eigenvalues 1 and i. Thus F extends to a unitary operator on square-integrable functions. Further, F 4 is the identity operator, and the inverse Fourier transform is given by (F 1 f ) (x) = (Ff ) ( x). Solution 19
PartRA 2 dn z2 dz 0 = @DR e(z it) =2 dz ne R 1 (x it)2 =2 dn x2 R1 2 2 dn x2 ! 1e dx = 1 e t =2 e itx ex =2 dx dx = ne ne dx R n n^ t2 =2 1 itx t2 =2 e e ( 1) (x) dx = e ( 1) n (t) 1 R R 1 x2 =2n dn (x+it)2 R 1 x2 =2 dn 2 ! e e dx = e e (x+it) dx = n dx d(it)n (x+it)\@DR 1 1 2 n R1 n R1 2 2 2 ( i) dtd n 1 ex =2 e (x+it) dx = ( i) dtd n 1 ex =2 e 2itx et dx = Z 1 p p 2 n n 2 2 2 ( i) dtd n e t e x =2 e 2itx dx = 2 ( i)n dtd n e t = 2 e t =2 in n (t) {z } | 1 p
2 e
(2t)2 =2
If we set these equal,pwe get 2 2 ( i)n e t =2 ^ n (t) = 2 e t =2 in n (t), p12 ^ n (t) = ( i)n n (t) PartRB 2 dn z2 dz 0 = @DR ez =2 dz ne R 1 x2 =2 d z2 R 1 x2 =2 dn (x+it)2 n 2 nR1 e e dx = e e dx = ( i) ez =2 dtd n e n dx dx 1 1 1 R 1 (x it)2 =2 d x2 R1 2 2 e e dx = 1 ex =2 dzdn e (x+it) dx = dz n 1 R1 R1 2 d 2 2 2 2 e t =2 1 e itx ex =2 n e x dx + ( i)n 1 ex =2 dxdn e (x+it) dx dx | {z } ( 1)n
e
t2 =2
dn
^ (t) = in n e n dt
+ ( i)n p (t) + ( i)n 2
^ n (t) = ( i)n n
(x) = e
x2 =2
H2 (x) = e
2
d dxn
et
2
d dxn
R1
1
et e
Hn (x), Hn (x) = ( 1)n ex
H0 (x) = 1, H1 (x) = e x2
n (x)
t2
d2 dx2
e
x2
x2 d dx
e x2
=e
x2
=
4x2 e
x2
2
e
x2 =2
e
dn dxn x2
e
x2
( 2x) e
2e
x2
dx
2ixt
dx p = ( i)n 2
2t2
e
(x+it)2
d t2 e dxn
= x2
= 4x2
= 2x 2
R1 2 2 dn Hn (x) = 2n xn + lower order polynomial. 1 n (x) e itx dx = 1 e x =2 ( 1)n ex dx e n R R 2 2 2 2 1 1 ( 1)n 1 Dxn ex ex =2 ixt dx = 1 e x =2 Dxn ex =2 ixt dx = R1 x2 n (x it)2 =2 t2 =2 R1 2 2 2 e Dx e e dx = (i)n 1 e x Dtn e(x it) =2 et =2 dx = 1 R1 R1 2 2 2 2 2 2 et =2 in Dtn 1 e x e(x it) =2 dx = et =2 in Dtn 1 e x =2 e ixt e t =2 dx = p 2 2 2 2 et =2 in Dtn e t =2 e t =2 R1
20
x2
e
ixt
d
IV.3.3 1 2 3 P L K Let f (z) = c0 + c1 z + + cn z n be a polynomial. (a) If the ck ’s are real, show that Z
1
2
f (x) dx 1
Z
2 i
f e
0
d = 2
n X
c2k :
k=0
Hint. For the …rst inequality, apply Cauchy’s theorem th the function f (z)2 separately on the top half an the bottom half of the unit disk. (b) If the ck ’s are complex, show that Z
1 1
2
jf (x)j dx
Z
2 i
f e
0
d = 2
n X k=0
jck j2 :
(c) Establish the following variant of Hilbert’s inequality, that n X
cj ck j+k+1 j;k=0
n X k=0
jck j2 ;
with strict inequalityRunless the complex numbers c0 ; : : : ; cn are all zero. Hint. 1 Start by evaluating 0 f (x)2 dx.
Solution (a) R1 R R 2 f (x)2 dx = f (z)2 dz 6 f ei d 1 0 1 R1 R R2 2 f (x)2 dx = f (z)2 dz 6 f ei d 1 1 R1 R 2 2 Add, get 2 1 f (x)2 dx =6 f ei d n n n R2 P 2 P P 2 2 i i ik f e d = 2 c , by from f e = c e cm e im k k 0 k=0 k=0 k=0 P P P P 2 2 2 (b) f (x)2 = ak x k + i b k x k = ak x k + bk xk R1 P R1 R1 P P P 2 k 2 k 2 k f (x) dx = a x dx+ b x dx = a x + bk x k = k k k 1 1 1 P k ck x 21
hR
1 1
P
d 2
ak eik
+
R1 P 1
bk eik
d 2
i
n R1 P R1 P 2 cj ck xj+k dx = (c) 0 ( cj xj ) dx = 0
R1
j;k=0
R1
"
n P
j;k=0
P
xj+k+1
cj ck j+k+1
#1 0
=
n P
j;k=0
cj ck j+k+1
jj 6 0 jf (x)j dx 6 0 jf (x)j dx 6 jck j , for equality, must have R1 R1 2 2 jf (x)j = 0 jf (x)j , so f (x) = 0, for 1 6 x 6 0, and then f (x) 0 0 Suppose P (z) = a0 + a1 z + ::: + an z n is a polynomial. By integrating P (z)2 log z, for an appropriate branch of log z, around a keyhole contour, n n R1 R P P aj ak 2 show that P (x) dx 6 jak j2 . P (z)2 log z dz = 0, = j+k+1 R1
2
j;k=0 2
2
k=0
0
2
@D
R1 R2 2 gives 0 P (z) log x dx 0 P (z)2 (log x + 2 i) dx+ 0 P ei iei d = 0. R1 R R 2 Get 2 i 0 P (x)2 dx = i 0 P ei ei d . Note that P ei d , so latter R2 R1 R2 2 integral is i 0 P ei ei ( ) d . Thus 0 P (x)2 dx = 0 P ei ei ( n R1 R2 P 2 d i P e ) 0 P (x)2 dx 6 = jak j2 2 0 k=0
22
) 2d
IV.3.4 1 2 3 P L K Prove that a polynomial in z without zeros is constant (the fundamental theorem of algebra) using Cauchy’s theorem, along the following lines. If P (z) is a polynomial that is not a constant, write P (z) = P (0) + zQ (z), divide by zP (z), and integrate around a large circle. this will lead to a contradiction if P (z) has no zeros. Solution P (z) is a polynomial with no zeros. Assume P (0) = 1 . Write p (z) = 1 + zQ (z). H P (z) H H H Q(z) 1 1 1 dz = dz = + dz = dz2 = i zP (z) zP (z) P (z) zP (z) jzj=R z jzj=R
jzj=R
! 0 by ML-estimate if deg P > 1. 1 , L R, when n = det P M = max zP1(z) Rn+1 ML
z=R 1 Rn
! 0, if R ! 1, if n > 1.
23
jzj=R
IV.3.5 1 2 3 P L K Suppose that D is a bounded domain with piecewise smooth boundary, and that f (z) is analytic on D [ @D. Show that sup jz
f (z)j
z2@D
2
Area (D) : Length (D)
Show that this estimate is sharp, and that R in fact there exist D and f (z) for which equality holds. Hint. Consider @D jz f (z)j dz, and use Exercise 4 in section 1. Solution R R [z f (z)] dz = zdz = 2iArea (D) @D @D R [z f (z)] dz 6 sup jz f (z)j Length (@D) @D ) sup jz z2@D
f (z)j >
z2@D Area(D) 2 Length(@D)
To see its sharp, take D = D = unit disk, get sup jz z2@D
Area(D) 2 Length(@D) =2
2
=1
24
f (z)j = 1
IV.3.6 1 2 3 P L K Suppose f (z) is continuous in the closed disk H fjzj Rg and analytic on the open disk fjzj < Rg. Show that jzj=R f (z) dz = 0. Hint. Approximate f (z) uniformly by fr (z) = f (rz). Solution (A. Kumjian) H H To prove jzj=R f (z) dz = 0 it su¢ ces to show that jzj=R f (z) dz ", for every " > 0. So let " > 0 be given. Since f is continuous in the closed disk fjzj Rg, it must be uniformly continuous there (since the close disk is both closed and bounded). Hence, there is a > 0 such that for all z1 ; z2 in the closed disk, we have jf (z1 )
f (z2 )j <
" 2 R
if
jz1
z2 j < :
For r 2 (0; 1), de…ne fr on the disk fz : jzj < R=rg by fr (z) = f (rz) for all z such that jzj < R=r. Then since H fr is analytic on a domain containing the closed disk fjzj Rg, we have jzj=R f (z) dz = 0 by Cauchy’s theorem. Now let r be chosen such that 1 =R < r < 1. Then for all z with jzj = R we have jz zrj = R (1 r) < and so by the choice of we have jf (z)
fr (z)j = jf (z)
f (rz)j <
" : 2 R
I
j(f (z)
Hence we have I
jzj=R
f (z) dz =
I
(f (z)
fr (z)) dz
jzj=R
jzj=R
by the ML-estimate theorem on p. 105.
25
fr (z))j jdzj
" 2 R = ": 2 R
IV.4.1 1 2 3 P L K 111 LLL Evaluate H then following integrals, Husingezthe Cauchy integral formula: (a) jzj=2 zz 1 dz; n 0 (e) jzj=1 zm dz; 1
Solution We rearragnde the Cauchy integral formula. Let D be bounded domain with piecewise smooth boundary @D and let f be an analytic function de…ned in a domain which contains D. Then for any z0 2 D and positive integer m we have, Z dm f (z) 2 i f (z) : dz = z0 )m (m 1)! dz m @D (z z=z0
(a) The nominator has one zero at z = 1 inside jzj = 2, thus by by Cauchy´s theorem, H
zn jzj=2
z
1
dz =
2 i n [z ]z=1 = 2 i: 0!
(b) The nominator has no zeros inside jzj = 1, thus by by Cauchy´s theorem, H
zn jzj=1
z
2
dz = 0:
(c) The nominator has one zero at z = 0 inside jzj = 1, thus by by Cauchy´s theorem, H
jzj=1
sin z 2 i dz = [sin z]z=0 = 0: z 0!
(d) The nominator has one tripel zero at z = 0 inside jzj = 1, thus by by Cauchy´s theorem,
26
H
cosh z 2 i d2 dz = cosh z jzj=1 z3 2! dz 2
= i: z=0
(e) The nominator has one tripel zero at z = 0 inside jzj = 1, thus by by Cauchy´s theorem. If m 0, then the integrand z m ez de…nes an analytic function on C, hence, by Cauchy’s Theorem the integral must be zero. If m 1, the nominator has a zero of order m at z = 0 inside jzj = 1, hence by by Cauchy´s theorem. I 2 i dm 1 z 2 i ez dz = e : = m m 1 (m 1)! dz (m 1)! jzj=1 z z=0 Thus
I
jzj=1
ez dz = zm
0 2 i (m 1)!
if m if m
0; 1:
(f) Let f be the principal branch of the logarithm de…ned on the slit plane, that is, f (z) = Log z for all z 6= 0 with Arg z 6= . Then f is analytic on a domain jz 1 ij 5=4. The nominator has one double zero at z = 1 inside jz 1 ij < 5=4, thus by by Cauchy´s theorem. H
2 i d Log z (Log z) 2 dz = 1! dz (z 1)
jz 1 ij=5=4
= 2 i: z=1
(g) The nominator has one double zero at z = 0 inside jzj = 1, thus by by Cauchy´s theorem, H
dz jzj=1 2 z
(z 2
4) ez
=
2 i d 1! dz
e z2
z
4
= z=0
i : 2
(h) The nominator has one zero at z = 0, and one zero at z = 2 inside jz thus by by Cauchy´s theorem, I
jz 1j=2
dz z
(z 2
4) ez
=
2 i 1 2 0! (z 4) ez 27
+ z=0
2 i 1 0! z (z + 2) ez
1j = 2,
= z=2
i i + 2: 2 4e
IV.4.2 1 2 3 P
L K LLL Show that a harmonic function is C 1 , that is, a harmonic function has partial derivatives of all orders.
Solution @ @v @ u = Re f 0 , @y u = @u = @x = Re (f 0 ). Write u = Re f , and note that @x @y @v v = Im f = Re ( if ), @x = Re ( if ). Now take successive derivatives, get m @ u 0 a multiple of Re f is Re (if 0 ). @xp @y q
28
IV.4.3 1 2 3 P
L K LLL Use the Cauchy integral formula to derive the mean value property of harmonic functions, that Z 2 d ; z0 2 D; u (z0 ) = u z0 + ei 2 0 whenever u (z) is harmonic in a domain D and the closed disk jz z0 j is contained in D. Solution H Write f (z0 ) = jz
f (z) dz z0 j= z z0
=
H2 0
f z0 + ei
29
d 2
, which is the MVP.
IV.4.4 1 2 3 P L K Let D be a bounded domain with smooth boundary @D, and let z0 2 D. Using the Cauchy integral formula, show that there is a constant C such that jf (z0 )j
D sup fjf (z)j : z 2 @Dg
for any function f (z) analytic on D [ @D. By applying this estimate to f (z)n , taking nth roots, and letting n ! 1, show that the estimate holds with C = 1. Remark. This provides an alternative proof of the maximum principle for analytic function. Solution Write f (z0 ) =
1 2 i
R
f (z) dz @D z z0
. Letkf k = max jf (z)j, z2@D
1 jz z0 j
6
1 dist(z0 ;@D) n
= d1 .
By ML–estimate, L = length @D, jf (z0 )j 6 2 1 d kf k L. Apply to f , get jf (z0 )jn 6 2 1 d kf n k L. Use kf n k = kf kn , take nth roots, get jf (z0 )j 6 kf k
L 1=n . 2 d
Let n ! 1, get jf (z0 )j 6 kf k.
30
IV.5.1 1 2 3 P L K 111 LLL Denna beh• over N og f ixas till Lite Show that if u is a harmonic function on C that is bounded above, then u is constant. Hint. Express u as the real part of an analytic function, and exponentiate.
Solution Let u be the real part of the analytic function f = u + iv, where we know that u C, for some constant C. Then set g = eu+iv where g is an entire function. Now we have that jgj = eu+iv = eu eiv = jeu j eiv = eu : We can se that becauce u is bounded above, so is g bounded above. Now we now that g is both analytic and bounded, by Liouville’s theorem g is constant. Thus f is constant.
31
IV.5.2 1 2 3 P L K 111 LLL Show that if f (z) is an entire function, and there is a nonempty disk such that f (z) does not attain any values in the disk, then f (z) is constant.
Solution If f (z) does not attain values in the disk jw cj < ", then 1= (f (z) c) is a bounded entire function, hence constant by Liouville’s theorem, and f (z) is constant. Let w0 be the center of the disk that f omits. Then 1= (f (z) w0 )
32
IV.5.2 1 2 3 P
L K LLL Show that if f (z) is an entire function, and there is a nonempty disk such that f (z) does not attain any values in the disk, then f (z) is constant.
Solution (A. Kumjian) Let f be such an entire function. Then there is a disc D = Dr (z0 ) for some z0 2 C and r > 0 such that f (z) 62 D for all z 2 C. That is, jf (z) z0 j r for all z 2 C. We de…ne an entire function g by the formula g (z) =
1 for all z 2 C: f (z) z0
It follows that jg (z)j 1=r for all z 2 C. Hence, g is a bounded entire function an so by Liouville’s Theorem g must be a constant function. The desired result now follows immediately.
33
If f does not attain values in the disk jw cj < ", then 1= (f hence constant by Liouville‘s theorem, and f is constant.
34
c) is bounded,
IV.5.3 1 2 3 P
L K LLL A function f (z) on the complex plane is double periodic if there are two periods w0 and w1 of f (z) that do not lie on the same line thought the origin (that is, w0 and w1 are linearly independent over the reals, and f (z + w0 ) = f (z + w1 ) = f (z) for all complex numbers z). Prove that the only entire functions that are doubly periodic are the constants.
Solution The "lattice" of points mw0 +nw1 , m; n 2 Z are the corners of parallelograms that cover C, each parallelogram being a translate F + mw0 + nw1 of "the fundamental region" F as shown in the …gure: Figure IV.5.3 Thus each 2 C can be written = z + mw0 + nw1 with z 2 F . Since we assume f ( ) = f (z), it follows that f i bounded, thus a constant.
35
If jf (z)j 2 M for jzj 6 R. Suppose jw0 j, jw0 j 6 R. Let P = parallelogram with vertices 0; w0 ; w1 ; w0 + w1 , and suppose for z 2 P . Any z 2 C have form + mw0 + nw1 where 2 P . Thus jf (z)j = jf ( )j 6 M , so f (z) is bounded and entire.
36
IV.5.4 1 2 3 P L K Suppose that f (z) is an entire function such that f (z) =z n is bounded for jzj R. Show that f (z) is a polynomial of degree at most n. What can be said if f (z) =z n is bounded on the entire complex plane?
Solution Apply the Cauchy estimates for f (m+1) (z) to disk jz z0 j < R, and let R ! 1, to obtain f (m+1) (z0 ) = 0. If f (z) is a polynomial of degree 6 m such that f (z) =z m is bounded near 0, then f (z) = cz m .
37
IV.5.5 1 2 3 P L K Show that if V (z) is the velocity vector …eld for a ‡uid ‡ow in the entire complex plane, and if the speed jV (z)j is bounded, then V (z) is a constant ‡ow. Solution Use V (z) = f 0 (z) (section III.6 p92), when f is entire. Get f 0 constant, by Liouville‘s theorem, i.e. V (z) is constant.
38
IV.6.1 1 2 3 P
L K LLL Let L be a line in the complex plane. Suppose f (z) is a continuous complex-valued function on a domain D that is analytic on DnL. Show that f (z) is analytic on D.
Solution Rotate and apply result in text.
39
IV.6.2 1 2 3 P L K Let h (t) be a continuous function on the interval [a; b]. Show that the Fourier transform Z b h (t) e itz dt H (z) = a
is an entire function that satis…es jH (z)j
CeAjyj ;
z = x + iy 2 C;
for some constants A; C > 0. Remark. An entire function satisfying such a growth restriction is called an entire function of …nite type. Solution The analyticity of H (z) follows from the theorem in the text. If jhj 6 M , take C = M (b a) and A = max (jaj ; jbj). We have je itz j = ety . jH (z)j 6 M (b a) max ety 6 M (b a) eAjyj a t b
40
IV.6.3 1 2 3 P L K Let h (t) be a continuous function on a subinterval [a; b] of [0; 1). Show that the Fourier transform H (z), de…ned as above, is bounded in the lower half-plane. Solution
41
IV.6.4 1 2 3 P L K Let be a smooth curve in the plane R2 , let D be a domain in the complex plane, and let P (x; y; ) and Q (x; y; ) be continuous complex-valued functions de…ned for (x; y) on and 2 D. Suppose that the functions depend analytically on for each …xed (x; y) on . Show that Z G ( ) = P (x; y; ) dz + Q (x; y; ) dy is analytic on D. Solution Reduce to Riemann integral and apply theorem in the text.
42
IV.7.1 1 2 3 P L K Find an application for Goursat’s theorem in which it is not patently clear by other means that the function in question is analytic. Solution Answer to this is not known (by me).
43
IV.8.1 1 2 3 P
L K LLL Show from the de…nition that @ @ @ z = 1; z = 0; z = 0; @z @z @z
@ z @z
= 1:
Solution We use the …rst-order di¤erential operators (page 124) @ 1 @ = @z 2 @x
i
@ @y
and
@ 1 @ @ = +i : @z 2 @x @y
We have that @ z @z @ z @z @ z @z @ z @z
1 2 1 = 2 1 = 2 1 = 2 =
@ @ i @x @y @ @ +i @x @y @ @ i @x @y @ @ +i @x @y
1 (1 + 1) = 1 2 1 (x + iy) = (1 1) = 0 2 1 (x iy) = (1 1) = 0 2 1 (x iy) = (1 + 1) = 1 2 (x + iy) =
44
IV.8.2 1 2 3 P
L K LLL Compute @@z (az 2 + bz z + cz 2 ). Use the result to determine where az 2 + bz z + cz 2 is complex-di¤erentiable and where it is analytic. (See Problem II.2.3)
Solution By the Leibnitz rule and the results of Exercise 1, the z –derivative is bz+2cz. The function is complex di¤erentiable at z if and only if bz + 2cz = 0. If b = c = 0, the function is entire. Otherwise this locus is either a point f0g or a straight line through the origin, and there is no open set on which the function is analytic.
45
IV.8.3 1 2 3 P
L K LLL Show that the Jacobian of a smooth function f is given by @f @z
det Jf = Solution Jf = det @f 2 @z
=
1 4
=
1 4
h
@u @x @u @y @f 2 = @z
@u @x
+
@v 2 @u @x @y
@v @y
@v @x @v @y 1 4 2
@u @v @x @y
= @(u+iv) @x
+
@v @x
+
@v @u @v 2 @x + 2 @u @y @x @y
@f @z
2
:
@u @v @y @x
i @(u+iv) @y @u @y
2
2
2
1 4 @u @x
1 4
i @v @u 2 @x = @y
46
@(u+iv) @x @v @y @u @v @x @y
+ i @(u+iv) @y 2
+
@v @x
@v @u @x @y
+
2
@u @y
2
= det Jf
IV.8.4 1 2 3 P
L K LLL
Show that @2 @2 @2 : + = 4 @x2 @y 2 @z@ z Deduce the following, for a smooth complex-valued function h. (a) h is harmonic if and only if @ 2 h=@z@ z = 0. (b) h is harmonic if and only if @h=@z is conjugate-analytic. (c) h is harmonic if and only if @h=@ z = 0. (d) If h is harmonic, then any mth order partial derivative of h is a linear combination of @ m h=@z m and @ m h=@ z m .
Solution @ @ @2 = @x i @y 4 @z@ z (a) clear (b) h harmonic ,
@ @x @2h @ z@z @2h @z@ z
@ + i @y =
=0,
@h @z @ @z
2
@2 @x2
@ i @x@y
2
@ i @y@x +
@2 @y 2
=
@2 @x2
+
@2 @y 2
is analytic
@h =0, = 0. Since @h = @@hz , this ????? @h is (c) h harmonic , @z @z @z @h analytic , @ z is conjugate analytic. @ k+l h (d) If h is analytic, then @z k @ z l = 0 unless l = 0. Only non-vanishing derivak+l @mh m tive is @zm = h (z). If h is conjugate analytic, ????? @@zk @zhl = 0 unless l = 0. If h is harmonic, then h = analytic + conjugate analytic, so all derivatives l k of h vanish except for @@zhk = @@zhk .
47
R d Log zjz=1 = 2 i z1 z=1 = 2 i (f) jz 1 ij=5=4 (zLog1)z2 dz = 21!i dz R ( e z )(z2 4) e z (2z) d e z (g) jzj=1 z2 (z2dz 4)ez = 2 i dz =2 i z2 4 (z 2 4)2 z=0
=
2 i4 16
(h) 2 i
R
=
i 2
dz =2 jz 1j=2 z(z 2 4)ez 1 1 + 8e2 = i 12 4
i (z2 +
1 4)ez
1 4e2
z=0
1 + 2 i (z+2)e z
48
= z=2
z=0
IV.8.5 1 2 3 P
L K LLL With dz = dx idy, show that for a smooth function f (z) that df =
Solution dx + df = @f @x @f dz @z
@f dy @y = 12 @f @x
i @f (dx + idy) + 21 @y h 1 @f 1 @f 1 @f i @f 1 @f dx + dy + dx + dy + dy 2 @x 2 @y 2 @x 2 @y 2 @x +
@f dz @z
@f @f dz + : @z @z
49
@f @x
+ i @f (dx @y
@f dx @y
idy) = i @f @f dy + dx = df @x @y
IV.8.6 1 2 3 P L K Show that if D is a domain with smooth boundary, and if f (z) and g (z) are analytic on D [ @D, then ZZ Z f (z) g (z) dz = 2i f (z) g 0 (z) dx dy: @D
D
Compare this formula with Exercise 1.4.
Solution R RR @ RR By (8.4), @D f (z) g (z) dz = 2i D @E (f g) dz = 2i D f gdz @g @f @g 0 Using the Leibnitz rule, obtain @@z (f g) = R f @z + g @z = RRf @z = f g To get Ex1.4, take f = 1, g (z) = z, get @D z dz = 2i D dxdt = 2iArea
50
IV.8.7 1 2 3 P
L K LLL Show that the Taylor series expansion at z0 = 0 of a smooth function f (z), through the quadratic terms, is given by @f 1 @2f @f @2f 2 (0) z+ (0) z+ (0) jzj2 +o jzj3 : f (z) = f (0)+ (0) z + 2 2 @z @z 2 @z @z@ z Solution Since a smooth function f is a linear combination of the functions 1; z; z; z 2 ; z 2 ; jzj2 and a reminder term O jzj3 , it su¢ ces to check the formula for these six functions. The formula holds fro each of these. For instance, for jzj2 = z z 2 @2f are harmonic. @f (0) = 0, @f (0) = 0, @@ zf2 (0) = 0, f (0) = 0, @z@ (0) = 1. @z @z z 3 3 The term O jzj has Taylor series 0 + O jzj . That ????? it.
51
IV.8.8 1 2 3 P
L K LLL Establish the following version of the chain rule for smooth complexvalued functions w = w (z) and h = h (w). @h @w @h @ w @ (h w) = + ; @z @w @ z @w @z @ @h @w @h @ w (h w) = + : @z @w @z @ w @z Solution @h Suppose w (0) = 0, h (0) = 0, write h (w) = aw + bw + O jwj2 , a = @w (0), 2 @w @w @h b = @ w (0) and w (z) = w + w + O jzj , = @z (0), b = @ z (0) then h (w (z)) = a z + z + O jzj2 + b z + z + O jzj2 + O jwj2 = a z + a z + b z + b z + O jzj2 = a + b z + (a + b ) z + O jzj2 This gives, @h w @h @w @h @w (0) = a + b = @w + @@hw @w = @w + @@hw @@wz @z @z @z @ w=@ z @z @h w @z
(0) = a + b =
@h @w @w @z
+
@h @w
@w = @z |{z}
@ w=@z
52
@h @w @w @z
+
@h @ w @ w @z
IV.8.9 1 2 3 P
L K LLL Show with the aid of the preceding exercise that if both h (w) and w (z) are analytic, then (h w) (z) and (h w)0 (z) = h0 (w (z)) w0 (z).
Solution If the functions h and w are analytic, then @h@ zw = 0 + 0 = 0 so h @h @w @ h w = @w + 0 = h0 (w (z)) w0 (z) analytic. (h w)0 (z) = @z @z
53
w is
IV.8.10 1 2 3 P L K Let g (z) be a continuously di¤erentiable function on the complex plane that is zero outside of som compact set. Show that ZZ 1 @g 1 dx dy; w 2 C: g (w) = w C @z z
Remark. If we integrate this formally by parts, we obtain ZZ 1 @ 1 g (w) = g (z) dx dy: @z z w C
Thus the "distribution derivative" of 1= ( (z w)) with respect to z is the point mass w ("Dirac delta-function"), in the sense that it is equal to 0 away from w, and it is in…nite at w in such a way that its integral (total mass) is equal to 1. Solution Apply Pompeeiu‘s formula to a large disk fjzj < Rg.
54
V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8
1
V.1.1 1 2 3 P L K (Harmonic Series) Show that n X 1 k=1
Deduce that the series estimate
P1
1 k=1 k
1 k
log n:
k
does not converge. Hint. Use the
Z
k+1
k
1 dx: x
Solution Let k be a nonnegative integer. Then since the function x 7! 1=x is decreasing on the positive reals, we have for all x 2 [k; k + 1] that 1=x 1=k . Hence locking at the oversum in the interval [k; k + 1] Z k+1 1 1 1 ((k + 1) k) = dx: k k x k
Therefore, for n n X 1 k=1
k
1
n Z X k=1
k
k+1
1 dx = x
Z
n+1
1
1 dx = log (n + 1) x
log n:
Where the last inequality follows from the fact that x 7 P ! log x is an increasing function on the positive reals. And thus is the sum nk=1 k1 have not limit as n ! 1 and does not converge.
2
V.1.2 1 2 3 P L K Show that if p < 1, then the series Exercise 1 and the comparison test.
P1
k=1
1=k p diverges. Hint. Use
Solution If k > 1 then, k p < k for 0 < p < 1, then 1 1 > : p k k Thus we have 1 1 X X 1 1 > : p k k k=1 k=1
P By the comparison test, we have that since the sum nk=1 1=k diverge as Pthe n ! 1 by Exercise 1, also the sum nk=1 1=k p diverge as n ! 1, which was to be shown.
3
V.1.3 1 2 3 P L K P1
Show that if p > 1, then the series
k=1
n X 1 S < p k (p k=1 Rk Hint. Use the estimate k1p < k 1 dx . xp
1=k p converges to S, where
1 1) np
1
:
Solution Let k be a nonnegative integer. Then since the function x 7! 1=x is decreasing on the positive reals, we have for all x 2 [k 1; k] that 1=k 1=x. Becauce p > 1, it follows that 1=k p 1=xp . Hence locking att the undersum in the interval [k 1; k] Z k 1 1 1 (k (k 1)) = p dx p p k k k 1 x Therefore n X 1 kp k=1
n Z X k=1
k
k 1
Z n dx x p+1 dx = 1+ = 1+ p xp p+1 1 x
n
= 1+ 1
1 p
n 1 p
p+1
n p
p+1
1
Hence the serie converge to S, where S
1+
1 p
1
=
p p
1
;
since its terms are > 0. Now we have that Z N N X 1 dx 6 =1+ p k xp n k=n+1
x p+1 p+1
as N ! 1. Thus we have
4
N
= n
n
p+1
p
N 1
p+1
!
1
1+
1 p
1
:
thus
N X 1 =S kp k=n+1
S
n X 1 n 6 p k p k=1
N X 1 kp k=n+1
(p
5
p+1
1
=
1 1) np
(p
1 1) np
1
:
1
V.1.4 1 2 3 P L K Show that the series 1 X ( 1)k+1 =1 k k=1
1 1 + 2 3
1 + 4
converges. Hint. Show that the partial sums of the series satisfy S2 < S 4 < S 6 < < S5 < S3 < S1 . Solution We de…ne the sum Sn by Sn =
n X ( 1)k+1 =1 k k=1
1 1 + 2 3
1 + :::: 4
This is the “alternating series test”, which is standard signs alternate, and terms ! 0, jtermsj # , so series converge.
6
V.1.5 1 2 3 P L K Show that the series 1 1 1 1 1 1 1 1 + + + + + 3 2 5 7 4 9 11 6 converges to 3S=2, where S is the sum of the series in Exercise 4. (It turns out that S = log 2.) Hint. Organize the terms in the series in Exercise 4 in groups of four, and relate it to the groups of three in the above series. 1+
Solution We have the series S from exercise V.1.5 and arrange the terms in groups of four 1 1 1 1 1 1 1 1 1 1 + + + + + 2 3 4 5 6 7 8 9 10 11 And now we arrange the terms in groups of four S=1
S= 1
1 1 + 2 3
1 1 + 4 5
1 1 + 6 7
1 1 + 8 9
1 1 + 10 11
1 + ::: 12
1 + ::: 12
Now we rearrange the terms in groups of two, and divide the terms by 2 S 1 = 2 2
1 1 + 4 6
1 1 + 8 10
1 + ::: 12
Now add one group of four from the sum S with a group of two from the sum S=2, and so on 3 S = 2 =
1 3 1 1+ 3 1+
1 1 1 1 1 1 1 + + 2 + + 2 + :::+ = 4 5 7 8 9 11 12 1 1 1 1 1 1 1 + + + + + ::: 2 5 7 4 9 11 6
2
Since series S in Exercise 4 converge, thus also 3S=2 converge by ......
7
V.1.6 1 2 3 P L K Show that Z R
1 = [log (log k)]R e = log (log R) e k log k P 1 diverges while 1 k=1 k(log k)2 converges.
log e (log e) ! 1
Solution 1 (Comparison Test) For the …rst sum n+1 2X
1 k log k k=2n +1 there are 2n terms between k = 2n +1 and k = 2n+1 , each of the terms greater 1 1 than 2n+1 log , so these 2n terms have sum greater than 2(n+1) . Thus 2n+1 log 2 series diverges, by comparison with the harmonic series. The other series is treated similarly, using an upper estimate. For the other sum n+1 2X
1 k (log k)2 k=2n +1 there are 2n terms between k = 2n + 1 and k = 2n+1 , each of the terms less 1 1 , so these 2n terms have sum less than 2(n+1) . Thus series than 2n+1 log 2n+1 log 2 converges, by comparison with the harmonic series. Solution 2 (Integral Test) Note that x log x and x (log x)2 are increasing for x > 1, so the terms of both series are positive and decreasing. Vi can use Cauchy’s integralkriterium and look att them corresponding integrals. Becauce of di¢ cult with the integrand we split the integral into two parts Z R 1 = [log (log x)]R log (log e) ! 1; e = log (log R) e x log x 8
as R ! 1.
Z
e
R
1 = x (log x)2
1 log x
R
= e
1 log e
as R ! 1. We have that the series 1 X k=1
diverges, while the series
1 X k=1
konverges.
1 k log k
1 k (log k)2
9
1 ! 1; log R
V.1.7 1 2 3 P L K Show that the series
P
ak converges if and only if
Solution n P If Sn is the partial-sum ak , then Sn diately from
n P
k=1
ak = S n
Sm 1 , when Sn =
1
=
n P
n P
1 k=m k(log k)2
converges.
ak . This follows imme-
k=m
ak is the nth partial-sum of
k=1
k=m
the series. d xk V.2.1 dx = k+x2k
Sm
Pk=n
(k+x2k ) xk (2k)x2k 1 = 0 at k 2 xk 1 + kx3k 1 = 2kx3k 1 , 2 2k (1+x ) p Function " for 0 6 x 6 k 1=2k , # for k 2 xk 1 = k 3k 1 , k = x2k , k = xk . ) p k x k 1 p x > k 1=2k , ! 0 as x ! 1. k+x = worst-case estimation. ) 2k 6 k+k = 2 k Converge to 0 uniformly. kxk
1
10
V.2.1 1 2 3 P L K Show that fk (x) = xk = k + xk converges uniformly to 0 on [0; 1). Hint. Determine the worst-case estimator "k by calculus. Solution Suppose fk (x) converge pointwise to a function f (x) as k ! 1. We have fk (x) =
xk ; k + x2k
thus 0
fk (x) =
kxk
1
k + x2k
xk (2k) x2k
(k + x2k )2
1
;
0
there fk (x) = 0 for x1 = 0 and x2 = k 1=(2k) . We have x 0 k 1=(2k) 0 fk (x) 0 + 0 fk (x) % & and the function is increasing for 0 x k 1=(2k) , and decreasing for x for x k 1=2k , and fk (x) ! 0 as x ! 1. We have that fk (x) attains its maximum for x = k 1(2k) . We determine f (x) applying the works case estimator "k to the function fk (x), p 1 k xk p = lim = 0: lim f (x) = lim k!1 2 k k!1 k + x2k k!1 k + k The series fk (x) = xk = k + xk converge uniformly to 0 on [0; 1).
11
V.2.2 1 2 3 P L K Show that gk = xk = 1 + xk converges pointwise on [0; 1) but not uniformly. What is the limit function? On which subsets of [0; 1) does the sequence converge uniformly? Solution Suppose gk (x) converge pointwise to a function g (x) as k ! 1, we start to show that is is an increasing function 0
gk (x) =
kxk
xk kxk (1 + xk )2 1
1
=
kxk 1 >0 (1 + xk )2
also is the function gk (x) increasing, now we take the limit in three intervals. If 0 x < 1, in this interval we have that xk ! 0 as k ! 1, and we have xk 0 = = 0: k k!1 1 + x 1+0
g (x) = lim If x = 1, then
1 1 1k = = : k k!1 1 + 1 1+1 2 k If x > 1, in this interval we have that x ! 0 as k ! 1, and we have g (1) = lim
xk 1 1 g (x) = lim = lim = = 1: k!1 1 + xk k!1 1=xk + 1 0+1 We also have that gk (x) ! g (x), where 8 < 0 for 0 x < 1; 1 for x = 1; g (x) = : 2 1 for x > 1
Convergence is uniform on [0; 1 "] for any " > 0. Not uniform on [0; 1], because limit function is not continuous at 1. Convergence is uniform on [1 + "; +1] for any " > 0.
12
V.2.3 1 2 3 P L K Show that fk (z) = z k =k converges uniformly for jzj < 1. Show that 0 fk (z) does not converge uniformly for jzj < 1. What can be said 0 about uniform convergence of fk (z)? Solution Suppose fk (z) converge pointwise to a function f (z) as k ! 1 in the interval jzj 1, in this domain we have that z k =k 1=k thus f (z) = lim
k!1
zk k
1 = 0: k!1 k lim
Thus fk (z) converges uniformly for jzj < 1. 0 0 Now suppose fk (z) converge pointwise to a function f (z) as k ! 1 in the interval jzj 1, now we must split the interval when we take the limit 0
f (z) = lim z k k!1
1
=
0
0 for jzj < 1; 1 for jzj = 1:
If jzj = 1, then fk (z) = 1 so the series can not converge uniformly for 0 jzj < 1. But for any " > 0 the series fk (z) = z k 1 converges uniformly for jzj 1 ".
13
V.2.4 1 2 3 P L K Show that 1 X 1 xk k 2 1 + x2k k=1
1 < x < +1.
converges uniformly for
Solution We = k12 , where the sum P1apply the Weierstrass M test (p. 135.)k with Mk 2k 1 + x for all 1 < x < k=0 Mk converges. First we show that x +1. We have xk xk
1 1 + x2k = 1 + x2k x2k < 1 + x2k = 1 + x2k
for jxj 1 for jxj > 1
This establishes the fact that xk 1 + x2k for all 1 < x < +1. Hence, for each k 1 and
1
1 < x < +1, 1 xk k 2 1 + x2k
Thus convergence is uniform on
1 = Mk : k2
1 < x < +1 by the Weierstrass M test.
14
V.2.5 1 2 3 P L K For which real numbers x does
P1
xk k 1+x2k
converge?
Solution Suppose gk (x) converge pointwise to a function g (x) as k ! 1, and take limit in three intervals If x < 1, in this interval we have that xk ! 0 as k ! 1, and we have X X 1 X 1 xk xk < < = k + xk k k 1 + x2k 1=x 1=x k=1 k=1 k=1
1 X 1 k=1
1
1
1
which is a convergent series. If x = 1, then xk = x2k = 1 then
X 1 xk = k 1 + x2k 2k k=1
1 X 1 k=1
1
which is a divergent series. If x > 1, in this interval we have that xk ! 1 as k ! 1, and we have X1 X 1 xk 1 = < k 1 + x2k k 1=xk + xk xk k=1 k=1
1 X 1 k=1
1
1
which is a convergent series.
15
V.2.6 1 2 3 P L K Show that for each " > 0, the series for x 1 + ".
P1
xk k 1+x2k
converges uniformly
Solution We have xk ; 1 + x2k
fk (x) = thus 0
fk (x) =
kxk
1
1 + x2k
xk (2k) x2k
1
(1 + x2k )2
=
xk 1 x3k 1 (1 + x2k )2
0
there fk (x) = 0 for x1 = 0 and x2 = 1. We have x 0 1 0 fk (x) 0 + 0 fk (x) % & and the function is increasing for 0 x 1, and decreasing for x 1, and fk (x) ! 0 as x ! 1. We have that fk (x) attains its maximum for x = 1. If x = 1, then xk = x2k = 1 then X 1 xk = k 1 + x2k 2k k=1
1 X 1 k=1
1
which is a divergent series. If x > 1, in this interval we have that xk ! 1 as k ! 1, and we have 1 X 1 k=1
xk k 1 + x2k
=
1 X 1 k=1 1 X k=1
(1 + ")k 2k
k 1 + (1 + ") 1 1 k (1 + ")k
16
=
1 X 1 k=1
1 k
k 1= (1 + ") + (1 + ")k
V.2.7 1 2 3 P
L K LLL Let an be aPbounded sequence of complex numbers. Show that for each " > 0, z the series 1 converges uniformly for Re z 1 + ". Here we choose n=1 an n the principal branch of n z .
Solution 1 P an n z , jan j 6 C, Re z > 1 + ". Apply Weierstrass M-test, jan n z j 6 n=1 P P C Cn Re z 6 n1+" = Mn , Mn < 1 ) an n z converges uniformly for Re z > 1 + ".
17
V.2.8 1 2 3 P
L K LLL P zk Show that converges uniformly for jzj < 1. k2 Solution Apply the Weierstrass M
test with M = 1=k 2 in jzj
1 1 1 X X zk jzjk X 1 = < ; k2 k2 k2 k=1 k=1 k=1
which is a convergent series by Weierstass M verges uniformly for jzj < 1.
18
1, we …nd that
test. Thus the series con-
V.2.9 1 2 3 P
L K LLL P zk Show that does not converges uniformly for jzj < 1. k Solution Suppose that we have uniform convergence for jzj < 1, then the series must converge for jzj 1 Apply the Weierstrass M test with M = 1=k in jzj 1, we …nd that 1 1 1 X X jzjk X 1 zk = < ; 2 k k k k=1 k=1 k=1
which is a divergent series by Weierstass M converge uniformly for jzj < 1.
19
test. Thus the series do not
V.2.10 1 2 3 P L K Shoe that if a sequence of functions ffk (x)g converges uniformly on Ej for 1 j n, then the sequence converges uniformly on the union E = E1 [ E2 [ [ En . Solution Use Cauchy criterion. Let " > 0. Choose Nj such that jfn (x) fk (x)j < k for m; k > Nj and x 2 Ej . Let N = max fN1 ; : : : ; Nn g. Then jfm (x) fk (x)j < " for m; k > N and x 2 E1 [ : : : [ En . ) ffm g converges uniformly.
20
V.2.11 1 2 3 P L K Suppose that E is a bounded subset of a domain D C at a positive distance from the boundary of D, that is > 0 such that jz wj for all z 2 E and w 2 CnD. Show that E can be covered by a …nite number of closed disks contained in D. Hint. Consider all closed disks with centers at points (m + ni) =10 and radius =10 that meet E. Solution Follow the hint. There are only …nitely many of the disks in ????? (because E is bounded); and they cover E, because the collection of all disks centred at (m + ni) =10 of radius =10 covers the complex plan, and they are contained in D, because dist ("; @D) > .
21
V.2.12 1 2 3 P L K Let f (z) be analytic on a domain D, and suppose jf (z)j M for all z 2 D. Show that for each > 0 and m 1, f (m) (z) m!M= m for all z 2 D whose distance from @D is at least . Use this to show that if ffk (z)g is a sequence of analytic functions on D that converges to f (z) on D, then for (m) each m the derivatives fk (z) converge uniformly to f (m) (z) on each subset of D at a positive distance from D. Solution Use the Cauchy estimates for f (m) (w) on a disk centred at w of radius d (w; @D), get f (m) (w) 6 m!m m .
22
<
V.3.1 1 2 3 P
L K LLL Find the radius of convergence of the following power series: 1 1 1 P P P 3k z k kk (a) 2k z k (d) (g) zk k k 4 +5 1+2k kk (b) (c)
k=0 1 P
k=0 1 P
k k z 6k
2 k
k z
(e)
(f)
k=0 1 P
k=1 1 P k=1
k=1
2k z 2k k2 +k
(h)
z 2k 4k k k
(i)
k!1
k=3 1 P
k=1
Solution (a) We apply the ratio test, thus R = lim
k=1 1 P
(log k)k=2 z k k!z k kk
2k 1 1 ak = lim k+1 = lim = : k!1 2 k!1 2 ak+1 2
(b) We apply the ratio test, thus R = lim
k!1
ak k 6k+1 6k 6 = lim k = lim = lim = 6: k!1 6 k + 1 k!1 k + 1 k!1 1 + 1=k ak+1
(c) We apply the ratio test, thus R = lim
k!1
ak k2 k2 1 = lim = lim = lim = 1: 2 2 k!1 (k + 1) k!1 k + 2k + 1 k!1 1 + 2=k + 1=k 2 ak+1
(d) We apply the ratio test, thus
R = lim
k!1
ak 3k 4k+1 + 5k+1 = lim k = k!1 4 + 5k ak+1 3k+1 3k 5k+1 = lim
k!1
3k+1 5k
4 k+1 5 4 k 5
5
+1
+1 23
= lim
k!1
3
4 k+1 5 4 k 5
+1
+1
5 = : 3
(e) We apply the ratio test, thus
R = lim
k!1
ak 2k z 2k (k + 1)2 + k + 1 = lim 2 = lim k!1 k + k k!1 ak+1 2k+1 z 2(k+1) = lim
k!1
1 + 2=k 1
p The convergence radius R = 1= 2.
1 = 2 jzj2
k+2 k
0; 1;
1 = 2 jzj2
2 jzj2 < 1 : 2 jzj2 > 1
(f) We apply the ratio test, thus ak z 2k 4k+1 (k + 1)k+1 4 (k + 1) (k + 1)k R = lim = lim k k = lim = k!1 ak+1 k!1 4 k k!1 z 2(k+1) z2 kk 4 (k + 1) k!1 jzj2
= lim
1 + 1=k 1
k
= 1:
(g) We apply the ratio test, thus ak kk 1 + 2k+1 (k + 1)k+1 k k 1 + 2k+1 (k + 1)k+1 = lim = R = lim = lim k!1 1 + 2k k k k!1 ak+1 k!1 (k + 1)k (1 + 2k k k ) (k + 1) (k + 1)k+1 ! k 1=k 1 2k+1 (k + 1)k = lim + = k!1 1 + 1=k (1 + 2k k k ) (k + 1) 1 + 2k k k ! k k 1=k 1 k+1 1 +2 = 2: = lim k!1 1 + 1=k (1 + 2k k k ) (k + 1) k 1= (2k k k ) + 1 (h) We apply the root test, thus 24
R=
lim
k!1
1 p k
jak j
= lim
k!1
q k
1 (log k)k=2
=
1 = 0: lim log k
k!1
(i) We apply the ratio test, thus ak k! (k + 1)k+1 R = lim = lim k = lim k!1 ak+1 k!1 k k!1 (k + 1)!
25
k+1 k
k
= lim
k!1
1 1+ k
k
= e:
V.3.2 1 2 3 P
L K LLL Determine for which z the following series converge. 1 1 1 P P P (a) (z 1)k (c) 2m (z 2)m (e) nn (z 3)n (b)
k=1 1 P
k=0
(z i)k k!
(d)
m=0 1 P m=0
(z+1)m m2
(f)
n=3
Solution (a) We apply the ratio test for ak = 1, thus R = lim
k!1
n=1 1 P
2n n2
(z
2
i)n
1 ak = lim = 1: k!1 1 ak+1
hence the radius of convergence is R = 1. Therefore, the series converges for all z satisfying jz 1j < 1 and diverges for all z satisfying jz 1j > 1. Now consider z such that jz 1j = 1 then jz
P1
1j = 1:
Since k=1 1 diverges, the power series diverges for such z also. Hence, the given series converges for all z satisfying jz 1j < 1. (b) We apply the ratio test for ak = 1=k!, thus R = lim
k!1
ak 1 (k + 1)! = lim = lim (k + 1) = 1: k!1 k!1 ak+1 k! 1
Therefore, the series converges for all z in C. (c) We apply the ratio test for am = 2m , thus R = lim
m!1
2m 1 1 am = lim m+1 = lim = : m!1 2 m!1 2 am+1 2
hence the radius of convergence is R = 1=2. Therefore, the series converges for all z satisfying jz 2j < 1=2 and diverges for all z satisfying jz 2j > 1=2. Now consider z such that jz 2j = 1=2 then 26
m
j2 (z
m
m
2) j = 2
1 2
m
= 1:
P Since 1 m=1 1 converges, the power series diverges for such z also. Hence, the given series converges for all z satisfying jz 2j < 1=2. (d) We apply the ratio test for am = 1=m2 , thus
R = lim
m!1
1 (m + 1)2 am m2 + 2m + 1 1 + 2=m + 1=m2 = lim 2 = lim = lim = 1: m!1 m m!1 m!1 am+1 1 m2 1
hence the radius of convergence is R = 1. Therefore, the series converges for all z satisfying jz + ij < 1 and diverges for all z satisfying jz + ij > 1. Now consider z such that jz + ij = 1 then (z + i)m 1m 1 = 2 = 2: 2 m m m
P1 2 Since m=1 1=m converges, the power series converges for such z also. Hence, the given series converges for all z satisfying jz + ij 1. (e) We apply the ratio test for an = nn , thus nn an nn = lim = lim R = lim = lim n n!1 (n + 1)n+1 n!1 n!1 an+1 n!1 (n + 1) (n + 1) Therefore, the series diverges for all z satisfying jz such that jz 3j = 0 then jnn (z
3)n j = nn 0n = 0:
n n+1
n
3j > 0. Now consider z
P Since 1 n=1 0 converges, the power series converges for z = 3. Hence, the given series only converges for z = 3. (f) We apply the ratio test for ak = 2k =k 2 , thus 27
1 = 0: n+1
ak 1 2k (k + 1)2 k 2 + 2k + 1 1 + 2=k + 1=k 2 R = lim = : = lim 2 = lim = lim k+1 2 k!1 ak+1 k!1 k k!1 k!1 2 2k 2 2 hence the radius of convergence is R = 1=2. Therefore, the series converges for all z satisfying jz 2 ij < 1=2 and diverges for all z satisfying jz 2 ij > 1=2. Now consider z such that jz 2 ij = 1=2 then 2n (z n2
2
n
i)
2n = 2 n
1 2
n
=
1 : n2
P 2 Since 1 k=1 1=n converges, the power series converges for such z also. Hence, the given series converges for all z satisfying jz 2 ij 1=2.
28
V.3.3 1 2 3 P
L K LLL Find the radius of convergence of the following series. 1 P n (a) z 3 = z + z 3 + z 9 + z 27 + z 81 + n=0 P p (b) z = z 2 + z 3 + z 5 + z 7 + z 11 + p prime
Solutions (a) (b)
R=
1 lim sup
p k
jak j
=
Neither series converges at z = 1, so R = 1.
29
1 = 1: 1
V.3.4 1 2 3 P L K P Show that the function de…ned by f (z) = z n! is analytic on the open unit disk fjzj < 1g. Show that jf (r )j ! +1 as r ! 1 whenever is a root of unity. Remark. Thus f (z) does not extend analytically to any larger open set than the open unit disk. Solution Cauchy-Hadamard formula gives R = 1 (radius of convergence) ) f (z) is analytic for jzj < 1. Suppose is an N th root of unity, N = 1. Then 1 P n!=N (r )n! = rn! N = rn! for n > N . Now rn! " 1 as r " 1, because it is increasing, and the …nite partial sums ! 1 as r ! 1. ) jf (r )j ! +1 as r ! 1.
30
V.3.5 1 2 3 P
L K LLL What functions are represented by the following power series? 1 1 P P (a) kz k (b) k2z k k=1
k=1
Solution (a) P k Di¤erentiate the geometric series 1 k=0 z , obtain 1 X
1
k
z =
k=0
1
z
)
Multiply by z obtain
1 X
1 X
kz k
1
=
k=1
kz k =
k=1
z z)2
(1
(1
1 z)2
:
:
(b) P k Di¤erentiate the geometric series 1 k=0 z twice, obtain 1 X
1
k
z =
k=0
1
z
)
1 X
kz
k 1
=
k=1
1 (1
2
z)
)
1 X
k (k
1) z k
k=2
2
=
2 (1
Multiply by z 2 obtain 1 X
k2
k z2 =
k=2
We get 1 X k=2
2 k
k z =
1 X k=2
kz k +
2z 2 : (1 z)3
2z 2 z 3 = (1 z) (1 z)2
31
z+
2z 2 (1 z)3
z)3
:
V.3.6 1 2 3 P L K P P Show that the series ak z k , the di¤erentiated series kak z k 1 , and the P ak k+1 integrated series z all have the same radius of convergence. k+1
Solution p p Can use Cauchy-Hadamard formula, and k k = 1. Then lim sup k jak j = k!1 q p ja j k lim sup k k jak j = lim sup k k+1 , so all series have the same radius of converk!1
k!1
gence. Can also use the characterization of R as the radius of the largest disk to which the function extends and ?????. This largest disk is clearly the same for f , f 0 , and ????.
32
V.3.7 1 2 3 P L K Consider the series 1 X
2 + ( 1)k
k
zk :
k=0
Use the Cauchy-Hadamard formula to …nd the radius of convergence of the series. What happens when the ratio test is applied? Evaluate explicitly the sum of the series.
Solution We apply the Cauchy-Hadamard formula where ak = 2 + ( 1)k observe that r p k k k jak j = 2 + ( 1)k = 2 + ( 1)k
k
. First
Note that
3 if k is even, 1 if k is odd.
2 + ( 1)k = Hence p k
jak j =
3 if k is even, 1 if k is odd.
Hence, since the lim sup of this sequence is 3 we have by the Cauchy-Hadamard formula R=
1 lim sup
Applying the ratio test we have ak = ak+1
2 + ( 1)k 2 + ( 1)k+1
1 = : 3 jak j
p k
k k+1
33
=
3k 1 3k+1
if k is even, if k is odd.
Since the sequence does not converge, the ratio test is inconclusive, in that limk!1 jak j = jak+1 j does not exist. However since jak j = jak+1 j 3, it does converge for jzj < 1=3. Then we have with jzj < 1=3 1 X
2 + ( 1)k
k
zk =
k=0
=
1 X
2n 2n 2n
2 + ( 1)
z
k=0
=
1 X n=0
+
1 X
2 + ( 1)2n+1
2n+1 2n+1
1 X
1 X
z
=
k=0
2n 2n
3 z +
1 X
2n+1 2n+1
1
z
=
n=0
n=0
9z
2 n
+z
z2
n
=
n=0
=
z 1 + : 1 9z 2 1 z 2
Note: The series is convergent because j9z 2 j ; jz 2 j < 1 since jzj < 1=3, and that the series have singularities at z = 1=3 and z = 1.
34
V.3.7 p Let L = lim sup k jak j. The de…nition of n ln supnpimplies that given " > 0, p 9 > 0, 9N 3 for k > N we have k jak j 6 L+", but k jak j > L " for in…nitely P many k 0 s. If k > N , then jak j < (L + ")k . If jzj < 1= (L + "), then ak z k converges. (L + ") jzj < 1, k z k 6 (L + ")k jzjk = ((L + ") jzj)k converges by comparison with geometric series. ) R > 1= (L + ") ) jak j > (L ")k . If jzj > L ", then ak z k > (L ")k (L 1")k for in…nitely many k 0 s. ), and P ak z k diverge. ) R 6 L1 " . Let " ! 0, get R 6 L1 .
35
V.3.8 1 2 3 P L K Write out a proof of the Cauchy-Hadamard formula (3.4). Solution
36
V.4.1 1 2 3 P
L K LLL Find the radius of convergence of the power series for the following functions, expanded about the indicated point. (d) Log z about z = 1 + 2i; (a) z 1 1 about z = i; 1 (b) cos z about z = 0; (e) z 3=2 about z = 3; 1 z i (c) cosh about z = 0; (f) about z = 2i: z z3 z Solution (d) Since the values of Log z are unbounded in any neighborhood of the origin it may not be extended to an analytic function on any open disk containing the p origin. But Log z is de…ned and analytic on the disk of radius j1 + p2ij = 5 centered at 1 + 2i. Hence, the desired radius of convergence is 5 by the second Corollary on p. 146.
(a) R =
p
2 (b) R = =2 (c) R = =2 (d) R =
37
p
5 (e) R = 3 (f) R = 2
V.4.2 1 2 3 P
L K LLL Show that the radius of convergence of the power series expansion of (z 2 p about z = 2 is 7.
Solution Rewrite f (z) as
z+1 . Singularities of f (z) are at (z e2 i=3 )(z+e2 i=3 ) p distance from 2 to nearest singularity is 7.
38
e2
i=3
1) = (z 3
, and
1)
V.4.3 1 2 3 P
L K LLL Find the power series expansion of Log z about the point z = i 2. Show p that the radius of convergence of the series is R = 5. Explain why this does not contradict the discontinuity of Log z at z = 2. Solution (From Hints and Solutions) p Log z extends to be analytic for jz (i 2)j < 5, through the extension does not coincide with Log z in the part of the disk in the lower half–plane.
39
V.4.4 1 2 3 P
L K LLL Suppose f (z) is analytic at z = 0 and satis…es f (z) = z + f (z)2 . What is the radius of convergence of the power series expansion of f (z) about z = 0? Solution (From Hints and Solutions) p 1 4z =2. Near 0 the function coincides with one of the branches of 1 The radius of convergence of the power series of either branch is 1=4, which is the distance to the singularity at 1=4.
40
V.4.5 1 2 3 P L K Deduce the identity eiz = cos z + i sin z from the power series expansions. Solution 1 P eiz =
n=0
in z n n!
=
P
n even
+
P
n odd
=
1 P
m=0
( 1)m z 2m (2m)!
+i
1 P
m=0
in the …rst and n = 2m + 1 in the second series.
41
( 1)m z 2m+1 , (2m+1)!
where n = 2m
V.4.6 1 2 3 P
L K LLL Find the power series expansions of cosh z and sinh z about z = 0. What are the radii of convergence of the series? Solution 1 P (a) cosh z = cos iz =
n=0
(b) sinh z = 1 P z 2n+1 = (2n+1)!
z 2n (2n)!
=1+
i sin (iz) = ( i) (iz)
z2 2!
+
z4 4!
(iz)3 3!
n=0
42
+ ::: +
(iz)5 5!
::: = z +
z3 3!
+
z5 5!
+ :::
V.4.7 1 2 3 P L K Find the power series expansion of the principal branch Tan 1 (z) of the inverse tangent function about z = 0. What is the radius of convergence of the series? Hint. Find it by integrating its derivative (a geometric series) term by term.
Solution We have d Tan dz
1
thus Tan
1
X 1 z= = 1 + z2 k=0 1
z=
1 X
( 1)k
k=0
And we have R = 1.
43
z2
z 2k+1 : 2k + 1
k
;
V.4.8 1 2 3 P L K Expand Log (1 + iz) and Log (1 iz) in power series about z = 0. By comparing power series expansions (see the preceding exercise), establish the identity Tan
1
z=
1 Log 2i
1 + iz 1 iz
:
(See Exercise 5 in section I.8)
Solution
Log (1 + iz) =
1 X k=0
( i)
k+1
z k+1 ; k+1
Log (1
iz) =
1 X
ik+1
k=0
Thus X 1 1 + iz 2i X 2j z 2j+1 z 2j+1 Log = = ( 1)j : i 2i 1 iz 2i j=0 2j + 1 2j + 1 j=0 1
1
44
z k+1 : k+1
V.4.9 1 2 3 P L K Let a be real, and consider the branch of z a that is real and positive on (0; 1). Expand z a in a power series about z = 1. What is the radius of convergence of the series? Write down the series explicitly. Solution f (z) = z a = ea Log z , Re z > 0. Assume a 6= 0; 1; 2; : : :. f (m) (z) = a (a 1) : : : (a , f (m) (1) = a (a 1) : : : (a m + 1) 1 P (m) m+1) a am (z 1)m , am = f m!(1) = a(a 1):::(a f (z) = = m , m > 1, and m! m=0
a0 = a1 =
a 0
. f (z) =
1 P
m=0
a m
(z
1)m . R = 1 since
m ! 1. If a is an integer > 0 get R = 1.
45
am+1 am
=
a m m+1
! 1 as
m + 1) z a
m
V.4.10 (From Hints and Solutions) 1 2 3 P L K Recall that for a complex number , the binomial coe¢ cient " choose n" is de…ned by
0
= 1;
and
n
(
=
1)
( n!
n + 1)
;
n
1:
Find the radius of convergence of the binomial series 1 X n=0
n
zn:
Show that the binomial series represents the principal branch of the function (1 + z) . For which does the binomial series reduce to a polynomial? Solution Use f (n) = ( 1) : : : ( n + 1) (1 + z) n and the formula for the coe¢ cient of z n . The series reduces to a polynomial for = 0; 1; 2; : : :. Otherwise radius of convergence is 1, which is distance to the singularity at 1. We can obtain the radius of convergence for the series also from the ratio test, (n+1)! 1):::( n+1) an = ( (1):::( = n+1n ! 1 as n ! 1. an+1 n+1)( n) n!
46
V.4.11 1 2 3 P L K For …xed n
0, de…ne the function Jn (z) by the power series Jn (z) =
1 X k=0
( 1)k z n+2k : k! (n + k)!2n+2k
Show that Jn (z) is an entire function. Show that w = Jn (z) satis…es the di¤erential equation n2 z2
1 w00 + w0 + 1 z
w = 0:
Remark. This is Bessel’s di¤erential equation, and Jn (z) is Bessel’s function of order n. Solution 1 P Jn (z) = Jn0 (z) =
Jn00
(z) =
k=0 1 P
k=0 1 P
k=0
( 1)k z n+2k k!(n+k)!2n+2k
=
zn n!2n
+ O (z n+2 )
(
1)k (n+2k)z n+2k k!(n+k)!2n+2k
(
1)k (n+2k)(n+2k 1)z n+2k k!(n+k)!2n+2k
1
=
zn 1 (n 1)!2n 2
=
+ O (z n+2 ) zn 2 (n 2)!2n
Term multiplying z n+2k in z 2 Jn00 + zJn0 + (z 2
+ O (z n )
n2 ) Jn is
(
1)k (n+2k)(n+2k 1) k!(n+k)!2n+2k
+
( 1)k (n+2k) ( 1)k n2 k!(n+k)!2n+2k k!(n+k)!2n+2k ( 1)k = (k 1)!(n+k 1)!2n+2k 2 ( 1)k k!(n+k)!2n+2k
(n + 2k) (n + 2k
1) + (n + 2k)
n2
4k (n + k)
=0
Should cheek separately to the sum terms of J0 + J1 , when we divide by 0 above, ???? work out. (constant term ????? J0 , z term for J1 ). It works, because the case k = 0 we replace 1= (k 1)! by k=k! = 0. Ratio test gives radius of convergence = 1. (It’s not clear).
1
47
V.4.12 1 2 3 P
L K LLL Suppose that the analytic function f (z) has power series expansion P n an z . Show that if f (z) is an even function, then an = 0 for n odd. Show that if f (z) is an odd function, then an = 0 for n even. Solution If f (z) is analytic f ( z) is also analytic in the same region P1 in Dr n(0) then n and P1 f ( z) = 2n+1n=0 ( 1) an z . If f is even then f (z) f ( z) = 0, and so = 0. A power series is 0 i¤ all of its contents are 0 (since n=0 2a2n+1 z (n) an = f (0) =n! which shows an = 0 for n odd. The result for f (z) odd is analogous.
48
f (z) even, f ( z) = f (z) . Chain rule: f 0 ( z) = f 0 (z), f 00 ( z) = f 00 (z), etc. f (k) ( z) = ( 1)k f (k) (z). f (k) (0) = ( 1)k f (k) (0) ) f (k) (0) = 0 for k odd. ) ak = f (k) (0) =k! for k odd. If f (z) is odd, f ( z) = f (z), same argument gives ak = 0 for k even. Alternatively, apply ever result to zf (z) which is even.
49
V.4.13 1 2 3 P L K Prove the following version of L’Hospitals’s rule. If f (z) and g (z) are analytic, f (z0 ) = g (z0 ) = 0, and g (z) is not identically zero, then f 0 (z) f (z) = lim 0 ; z!z0 g (z) z!z0 g (z) lim
in the sense that either both limits are …nite and equal, or both limits are in…nite. Solution 1 P Assume z0 = 0 , f (z) = ak z k = f 0 (0) z +O (z 2 ), g (z) = g 0 (0) z +O (z 2 ) = k=1 8 < 1 if g (0) = 0 and f (0) 6= 0 or k < N 0 (0) aN (z) if g (0) = 0 , f (0) 6= 0, k = N g 0 (0) z+aN z n +O z N +1 . fg(z) ! fg0 (0) if g 0 (0) 6= 0 ! : bN 0 if g (0) = 0 , f (0) 6= 0, k > N 0 (0)+ka z k 1 +O z k 0 0 0 f ( ) f (z) k f (z) (0) = 0 ! fg0 (0) if g 0 (0) 6= 0 N 1 N , 0 g 0 (z) 8 g (0)+N bN z +O(z ) g (z) g (0) = 0 and either f (0) 6= 0 or k < N < 1 N aN aN = g (0) = 0 , f (0) 6= 0, k = N ! bN : N bN 0 g (0) = 0 , f (0) 6= 0, k > N Then the limits are equal in all cases.
50
V.4.14 1 2 3 P L K Let f be continuous function on the unit circle T = fjzj = 1g. Show that f can be approximated uniformly on T by a sequence of polynomials in z if and only if f has an extension F that is continuous on the closed disk fjzj 1g and analytic on the interior fjzj < 1g. Hint. To approximate such an F , consider dilates Fr (z) = F (rz). Solution If f has an analytic extension F , then Fr (z) = F (rz) has power series 1 P P F (z) = an z k , jzj < 1. Fr (z) = an rn z k , jzj < 1r . Fr converge uniformly n=j
for jzj 6 1, to F (rz) and F (rz) ! f (z) uniformly for jzj = 1 as v " 1. For " > 0, take r < 1 with jFr (z) F (z)j < " for jzj = 1, then take N such that N N P P an rn z n Fv (z) 6 2" for jzj = 1. Then an rn z n F (z) 6 2" + 2" = " n=0
n=0
for jzj = 1. Conversely if pn are polynomials, pn ! f uniformly for jzj = 1, then jpn pm j ! 0 uniformly for jzj = 1 as n; m ! 1, so by the maximum principle, jpn pm j ! 0 uniformly for jzj 6 1. ) fpn g converges uniformly to same function F (z) for jzj 6 1 . Since pn (z) ! f (z) for jzj = 1, F (z) = f (z) for jzj = 1. Since pn is analytic for jzj < 1, also F (z) is analytic for jzj < 1.
51
V.5.1 1 2 3 P
L K LLL Expand the following functions in power series about 1 2 2 (a) z21+1 (b) z3z 1 (c) e1=z (d) z sinh (1=z) Solution (b) Denote the above function by f and de…ne g by (1=w)2 w = g (w) = f (1=w) = 3 1 w3 (1=w) 1
for w 6= 0
and g (0) = 0. Note that g is analytic on the disk D1 (0) with power series representation g (w) =
1 X w = w w3 1 w3 k=0 3
for all w 2 D1 (0) (note that jw j
1 if jwj
f (z) = g (1=z) =
1 X
(1=z)
(a) 1 P
n=0
(b)
1 z 2 +1
=1
z2 + z4
1 P
::: =
n=0
( 1)n z 2n+2 z2 z3 1 1=z 2
(c) e
= =
1 1 z 1 1=z 3 1 P
n=0
(d) z sinh
1 z
=
1 1 n! z 2n
=z
1 z
+
1 z
1 P
n=0
1 3!z 3
+
1 z 3n
=
1 5!z 5
=
1 X
w3n+1
k=0
1). Hence, we have
3n+1
=
1 X k=0
k=0
for all z with jzj > 1.
n
1 ( 1)n z 2n z12 1+1=z 2 =
1 P
n=0
1 z 3n+1
+ ::: =
1 P
n=0
52
1 1 (2n+1)! z 2n
1 z 3n+1
n 1 ( 1) z 2 z 2n
=
V.5.2 1 2 3 P L K Suppose f (z) is analytic at 1, with series expansion (5.1). With the notation f (1) = b0 and f 0 (1) = b1 , show that f 0 (1) = lim z jf (z) z!1
Solution 1 P f (z) =
k=0
bk zk
0
= f (1)+ f (1) + z
f 0 (1) as z ! 1, since
1 P
k=1
1 P
k=2
bk . zk
z [f (z)
f (1)j :
f (1)] = f 0 (1)+
bk+1 wk ! 0 as w ! 0.
53
1 P
k=1
bk+1 zk
!
V.5.3 1 2 3 P L K Suppose f (z) is analytic at 1, with series expansion (5.1). Let 0 be the smallest number such that f (z) extends to be analytic for jzj > . Show that the series (5.1) converges absolutely for jzj > and diverges for jzj < . Solution (A. Kumjian) Recall that f is analytic at 1 i¤ there is a function at 0 such that f (z) = g (w) with w = 1=z wherever f (z) makes sense. We may suppose that f (z) is de…ned for jzj > and that f is analytic at these points. Then setting R = 1= if > 0 and R = 1 if = 0, we have that g is analytic on DR (z0 ) and cannot be extended to an analytic function on any larger open disk centered at z0 . Hence, g has a power series representation g (w) =
1 X
bk w k
k=0
for jwj < R;
with radius of convergence R. Thus, the series converges absolutely for jwj < R and diverges for jwj > R. It follows that by substituting w = 1=z, we have 1 X bk f (z) = zk k=0
for jzj > ;
where the series converges absolutely for jzj >
54
and diverges for jzj < .
V.5.4 (From Hints and Solutions) 1 2 3 P L K Let E be a bounded subset of the complex plane C over which area integrals can be de…ned, and set ZZ dx dy f (w) = ; w 2 CnE; z E w
where z = x + iy. Show that f (w) is analytic at 1, and …nd a formula for the coe¢ cients of the power series of f (w) at 1 in descending powers of w. Hint. Use a geometric series expansion. Solution P n n+1 If jzj 6 M for z 2 E, and R > M , then 1= (w z) = z =w converges uniformly for z 2 E and jwj > R . Integrate term by term, obtain f (w) = 1 RR n P bn , jwj > R, where b = z dxdy. n n+1 w N =0
E
55
V.5.5 (From Hints and Solutions) 1 2 3 P L K Determine explicitly the function f (w) de…ned in Exercise 4, in the case that E = fjwj 1g is the unit disk. Hint. There are two formulae for f (w), one valid for jwj 1 and the other for jwj 1. Be sure thy agree for jwj = 1. Solution To …nd the formula for jwj < 1 , break the integral into two pieces corresponding toZ Zjzj > jwj and Z Z to jzj < jwj, and use geometric series. f (w) = RR n 1 RR dxdy dxdy dxdy z dxdy = + = (1)+(2). If jwj < 1, get = w z w z w z D jzjjwj | {z } | {z } (1) (2) RR n RR 1 i(n 1) r drd e = 0 , f (w) = dxdy = w , jwj > 1. w z RR RR RR P dxdy 1 1 zn If jwj < 1, get (1) w1 = dxdy = dxdy = w1 jwj2 = n 1 z=w w w w w (2)
jzj
RR
1>jzj>jwj
f (w) =
dxdy w z
=
RR
1 dxdy z 1 w=z
=w; jwj > 1 w; jwj < 1
jzj6jwj
=
1 P
wn
n=0
56
RR
jzj6jwj
dxdy z n+1
=0
V.6.1 1 2 3 P
L K LLL Calculate the terms through order seven of the power series expansion about z = 0 of the function 1= cos z.
Solution 1 =1+ cos z 2
z2 2! 4
z4 4!
z6 6!
+ O (z 8 ) + (: : :)2 + (: : :)3 + O (z)
1 + z2! + a4 z + a6 z 6 + O (z 8 ) 2 1 5 z 4 : 4!1 + 2!1 = 14 24 = 24 , z6 : sin z cos z
z3 3!
= z
z:1 z 3 : 21 z 5 : 5!1 1 z 7 : 12
1 3! 1 2
+
z5 5!
z7 7!
+ :::
1+
3 2 + 2!1 = 18 2!4! 5 4 1 6 z2 + 24 z z + 2 12
1 = 31 6 1 5 1 1 5 16 4 + 24 = 120 + 24 = 120 = 30 3! 12 2 5 + 215! 7!1 = 7 6 5 2 57! 7+3 7 1 = 24 3! 420 175+21 1 = 265 = 7 653 7! 7! 432
=
1 2
57
=
2 15
1 24
:::
=
1 12
V.6.2 1 2 3 P
L K LLL Calculate the terms through order …ve of the power series expansion about z = 0 of the function z= sin z.
Solution z =z sin z
z z 3 =3!+z 5 =5! :::
z2 3!
=
1 1 z 2 =3!+z 4 =5! ::: 2
6 z4 + z7! : : : + (: : :) + (: : :)3 + O (z 2 ) = 5! 1 + z 2 =6 + a4 z 4 + a6 z 6 + O (z 7 ) 1 1 7 1 a4 = 5!1 + (3!) = 103603 = 360 2 = 36 120 6(1 14)+4 5 7 2 1 62 a6 = 7!1 = 3!5! + (3!) = 1403!7!78 = 3!7! = 32 3 = 3!7! 2 z 7z 4 = 1 + z6 + 360 + O (z 6 ) sin z
=1+
58
V.6.3 1 2 3 P
L K LLL
Show that ez 1 1 3 3 4 11 5 = 1 + z2 z + z z + 1+z 2 3 8 30 Show that the general term of the power series is given by n
an = ( 1)
1 2!
( 1)n + ; n!
1 + 3!
n
2:
What is the radius of convergence of the series?
Solution We have X 1 ( 1)k z k = 1+z k=0 1
for all jzj < 1
and
1 X zk k=0
k!
for all z 2 C:
Since the given function can be expressed as product of these two we have ez 1 z X cn z n = e = 1+z 1+z n=0 1
where
for all jzj < 1
( 1)n ( 1)n 1 ( 1)n 2 ( 1)0 + + + + an = 0! 1! 2! n! by formula (6.1). Hence, c0 = 1, c1 = 0 and for n 2 we have, since 0! = 1! = 1,
an = ( 1)n
1
( 1 + 1)+
( 1)n 2!
2
+
+
The radius of convergence is 1.
59
( 1)0 1 = ( 1)n n! 2!
1 + 3!
+
( 1)n : n!
Solve ez = (1 + z) = a0 + a1 z + a2 z 2 + : : : by multiplying 9 by (1 + z) and coma0 = 1 > > Recursively = a + a = 1 0 1 z paring with the coe¢ cients of e . ) a0 = 1 a1 + a2 = 1=2 > > an = 1=n! a n ; an 1 + an = 1=n! 1 + (n 1 2)! : : : + ( 1)n 0!1 (the two last terms cancel.) The So an = n!1 (n 1)! radius of convergence is 1 because the function has a pole at 1 . (In fact, ( 1)n an ! e 1 when n ! 1 .)
60
V.6.4 1 2 3 P L K De…ne the Bernoulli numbers Bn by z z2 z4 z6 cot (z=2) = 1 B1 B2 B3 : 2 2! 4! 6! Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the …rst three Bernoulli numbers. Solution 2 4 6 z cot (z=2) = 1 B1 z2! B2 z4! B3 z6! : : :, 2 cot (z=2) = cos(z=2) is odd, so z2 cot (z=2) is even, so only even terms appear in sin(z=2) the series. The function has singularities at z=2 = , i.e., at 2 . Distance from 0 to nearest singularity = 2 = radius of convergence. w w cos sin w
Find the two Bernoullinumbers B1 +B2 . w cot w = = h i 2 2 4 2 w4 w2 1 w2! + w4! + O (w6 ) 1 + w6 + + O (w6 ) 120 6 1
w2 2 1 6 1 2 w 3
+
w4 24
1+
w2 6
+
1 36
1 120
1 1
4 w2 + w4! 2! 4 w2 + w5! 3!
w6 +::: 6! w6 +::: 7!
=
w4 + O (w6 ) =
7 1 1 1 1+ w2 + 360 + 24 w4 + O (w6 ) = 2 12 1 4 6 1 w + O (w ) 45 2 4 4 cot w = 1 B1 w2! B2 2 4!w ::: 22 B 1 1 1 24 1 1 ) 2! = 3 , B1 = 6 , 4! B2 = 45 , B2 = 244!45 , B2 = 30 For B3 , must keep all powers ????? to and ????? w8 . Better to just ????? for B1 and B2 .
61
V.6.5 1 2 3 P L K De…ne the Euler numbers En by X En 1 = zn: cosh z n! n=0 1
What is the radius of convergence of the series? Show that En = 0 for n odd. Find the …rst four nonzero Euler numbers. Solution 1 P 1 = cosh z
n=0
En n z , n!
cosh z has zeros at
2
3 2
i;
i; : : :. Distance from 0 to
nearest singularity is =2 = R . Since cosh z is even, En = 0 for n odd. 1 1 z2 z4 z6 = (1+z2 =2!+z 4 =4!+:::) = 1 + E2 2! + E4 4! + E6 6! = cosh z z2 2!
1 E2 2! E6 6!
+
z4 4!
+ ::: +
= =
1 , E2 = 2! 1 1 + 2 2!4! 6!
E6 =
1+6 5
z2 2!
+
z4 4!
E4 = 4!1 4! 1 , E6 = (2!)3 6! = 29 6 33
1,
+
z6 6!
+ 2!1 61
2
::: 2
=
1 4
5 3 = 29
62
(: : :)3 , E0 = 1 1 24
=
90 =
5 24
, E4 = 5
61
V.6.6 1 2 3 P L K Show that the coe¢ cients of a power series "depend continuously" on the function they represent, in the following sense. If ffm (z)g is a sequence of analytic functions that converges uniformly to f (z) for jzj > , and fm (z) =
1 X
k
ak;m z ;
k=0
then for each k
f (z) =
m X k=0
0, we have ak;m ! ak as m ! 1.
63
ak z k ;
V.7.1 1 2 3 P Find (a) (b) (c)
L K LLL the zeros and orders z 2 +1 (d) cos z 1 z2 1 1 1 + z5 (e) cos zz 1 z z 2 sin z (f) coszz2 1
of zeros of the following functions (g) ez 1 (h) sinh2 z + cosh2 z z (i) Log (principal value) z
Solution (c) Let f (z) = z 2 sin z. Then f has zeros at all integer multiples of i.e. z0 = k for k 2 Z. Since sin z has a zero of order 1 at 0, so sin z = zh (z) for some analytic function h with h (0) 6= 0. Thus, f (z) = z 3 h (z) and so f has a zero of order three at 0. Note that f 0 (z) = 2z sin z + z 2 cos z. Hence, for any other zero z0 = k where k 2 Zn f0g, we have f 0 (k ) = 2k sin k + (k )2 cos k = ( 1)k k 6= 0: Thus, f has a simple zero at z0 = k when k 6= 0. (d) Note that g (z) = cos z 1 = 0 i¤ z = 2k for k 2 Z. We will show that all the zeros are double zeros. First observe that g 0 (z) = sin z and so g 0 (2k ) = sin 2k = 0 for all k 2 Z. But g 00 (z) = cos z and so g 0 (2k ) = cos 2k = 1 for all k 2 Z. Hence, all zeros are double zeros.
64
(a) simple zeros at i, (b) simple zeros at e i=4 , e3 i=4 , (c) triple zero at 0, simple zeros at n , n = 1; 2; : : :, (d) double zeros at n , n = 1; 2 : : :, (h) simple zeros at i=4 + n i=2, n = 1; 2; : : :, (i) no zeros. 1 1 4 (b) z + z5 = (z + 1) z15 = (z z1 ) (z z2 ) (z z3 ) (z z4 ) z15 , where zj = ei( =4+j=4 2 ) , these four points are simple zeros. y y (d) cos z = cos (x + iy) = cos x cos (iy) sin x sin (iy), cos (iy) = e 2+e = cosh y, sin (iy) =
e
y
2i
ey
=i
ey e 2
y
= i sinh y, cos z = 1 ) sin x sinh y =
0 ) y = 0 or x = n . If x = n , then cos (n ) cosh (n ) = ( 1)n cosh (n ). This is = 1 only if n = 0, since cos (n ) > 1 for n 6= 0. Get x = 0, y = 0. If y = 0, then cos x = 1 only at x = 2n , n = 0; 1; 2; : : :. Solutions d (cos z 1) = sin z = 0 at these points, are z = 0; 2 ; 4 ; : : :. Since dz 2 d (cos z 1) = cos z 6= 0 , at these points. The zeros are double zeros. dz 2 (e) cos zz 1 , has only zeros at z = 2n , n = 0; 1; 2; : : : . Simple poles at ) 1 d z = 0 . At the points dz = cos zz 1 = z( sin z) z2(cos(z 1)) = cos(2n for n 6= 0. (2n )2 Other zeros are double. Double zeros at z = 2n , n = 0; 1; 2; : : :. 2 (g) ez 1 = z + z2! + : : :, thus a simple zero at 0. Since ez 1 is periodic, we have simple zeros at i2 k, k 2 Z . (i) Log z = 0 when jzj = 1 and Arg z = 0, thus at z = 1. We write 3 2 Log z = Log (1 (1 z)) = 1 z (1 2z) + (1 3z) + : : :, so we have a simple zero at 1.
65
V.7.2 (From Hints and Solutions) 1 2 3 P L K Determine which of the functions in the preceding exercise are analytic at 1, and determine the orders of any zeros at 1. Solution (a) analytic at 1 , (b) analytic at 1, simple zero, (c) –(i) not analytic at 1.
66
V.7.3 1 2 3 P
L K LLL Show that the zeros of sin z and tan z are all simple. Solution Zeros of sin z are at n , 1 < n < 1, since cos (n ) = 1 6= 0, the zeros are simple.
67
d dz
sin zjz=n =
V.7.4 1 2 3 P
L K LLL Show that cos (z + w) = cos z cos w identity for z and w real.
sin z sin w, assuming the corresponding
Solution cos (z + w) = cos z cos w + sin z sin w = F (z; w) is entire in z for each …xed w and entire in w for each …xed z. F (z; w) = 0 for z; w real. Apply theorem, with D = C, E = R. Get F (z; w) = 0, so cos (z + w) = cos z cos w sin z sin w.
68
V.7.5 1 2 3 P L K Show that Z
1
e
zt2 +2wt
dt =
1
r
z
ew
2 =z
z; w 2 C; Re z > 0;
;
where we take the principal branch of the square root. Compare the result to Exercise IV.3.1. Hint. Show that the integral is analytic in z and w, and evaluate it for z = x >p 0 and w real by making a change of variable and using the known value for z = 1 and w = 0. Solution R1 at2 2bt e dt, note the integral is improper, so a limit process may be used 1
for analytic. Re a < 0. x > 0, R1
e
s2
e s ds =
1
R1
1 R1 1
R1
R1
eat
2
1
=2)
(s
e
e
2
=4
2
ds = e
1
e e
s2
e s ds =
zt2
e
zbt
p
dt =
2bt
=4
p1 x
dt =
Z1
e
(s
1
e p
2
=4
,
R1
e
xt2
e
2bt
dt =
1
2 =z
z
eb
|
69
p1 x
p
{z
e
R1
e
s2
2b p s x
1
=2)
ds
b2 =x
}
, x real
ds. F ( ) =
V.7.6 1 2 3 P
L K LLL Suppose f (z) is analytic on a domain D and z0 2 D. Show that if f (m) (z0 ) = 0 for m 1, then f (z) is constant on D. Solution Suppose that f (m) (z0 ) = 0 for m and de…ne g on D by g (z) = f (z) f (z0 ). Then it su¢ ces to show that g is identically zero. Suppose not, then since g (m) (z0 ) for m 0, the power series expansion for g is trivial and therefore g is identically zero on a disk on nonzero radius centered at z0 but this violates the …rst theorem on page 156 that asserts that an analytic function which is not identically zero only has isolated zeros. This result may also be proved using the Uniqueness Principle (the second theorem on page 156).
70
V.7.7 1 2 3 P
L K LLL Show that if u (x; y) is a harmonic function on a domain D such that all the partial derivatives of u (x; y) vanish at the same point of D, then u (x; y) is constant on D. Solution
71
V.7.8 1 2 3 P L K With the convention that the function that is identically zero has a zero of in…nite order at each point, show that if f (z) and g (z) have zeros of order n and m respectively at z0 , then f (z) + g (z) has a zero of order k min (n; m). Show that strict inequality can occur here, but that equality holds whenever m 6= n. Solution
72
V.7.9 1 2 3 P
L K LLL Show that the analytic function f (z) has a zero of order N at z0 , then f (z) = g (z)N for some function g (z) analytic near z0 and satisfying g 0 (z) 6= 0. Solution (From Hints and Solutions) Write f (z) = (z z0 )N h (z), where h (z) has a convergent power series and h (z0 ) 6= 0. Take g (z) = (z z0 ) e(log(h(z)))=N for an appropriate branch of the logarithm.
73
V.7.10 1 2 3 P L K Show that if f (z) is a continuous function on a domain D such that f (z)N is analytic on D for some integer N , then f (z) is analytic on D. Solution f (z)N analytic. Zeros of f (z) are isolated, for this need maximum principle. f (z) is analytic, except possibly at the isolated points where f (z) = 0. (By the Riemann‘s theorem, f (z) is analytic ????? there but don’t have this yet) Write f (z)N = (z z0 )m h (z), h (z) analytic, h (z) 6= 0. Then h (z) has N th root near z0 , by preceding exercise. f (z)N = g (z)N (z z0 )m , (f (z) =g (z))N = (z z0 )m . (z z0 )m=N is continuous in a neighborhood of z0 . f (z) =g (z) returns to original value, doing circle around z z0 . m is an integral multiple of N .
74
V.7.11 1 2 3 P L K Show that if f (z) is a nonconstant analytic function on a domain D, then the image under f (z) of any open set is open. Remark. This is the open mapping theorem for analytic functions. The proof is easy when f 0 (z) 6= 0, since the Jacobian of f (z) coincides with jf 0 (z)j2 . Use Exercise 9 to deal with the points where f 0 (z) is zero. Solution If f 0 (z0 ) 6= 0, then f maps open disks centred at z0 onto open sets, so f (D) contains a disk centred at f (z0 ). If f 0 (z0 ) = 0, assume f (z0 ) = 0, write f (z) = (z z0 )N h (z), h (z0 ) 6= 0. Then f has a N th root near z0 , f (z) = g (z)N , g 0 (z0 ) 6= 0. g covers a ????? disk, so f cover disk centred at 0.
75
V.7.12 1 2 3 P L K Show that the open mapping theorem for analytic functions implies the maximum principle for analytic functions. Solutions Clearly open mapping theorem ) strict maximum principle for analytic functions, since a non constant analytic function can’t attain maximum at z0 and cover a disk centred at z0 . Then strict maximum principle ) maximum principle.
76
V.7.13 (From Hints and Solutions) 1 2 3 P L K Let fn (z) be a sequence of analytic functions on a domain D such that fn (D) D, and suppose that fn (z) converges normally to f (z) on D. Show that either f (D) D, or else f (D) consists of a single point on @D. Solution f (D) D [ @D. If f (z) is not constant, then f (D) is open, and f (D) cannot contain any point of @D.
77
V.7.14 1 2 3 P L K A set E is discrete if every point of E is isolated. Show that a closed discrete subset of a domain D either is …nite or can be arranged in a sequence fzk g that accumulates only on f1g [ @D. Solution Let Kn = fz 2 D : d (z; @D) > 1=n; jzj 6 ng. En is compact, only …nitely many points of E belongs to Kn ????? points, shading with those in E \ K1 , the E \ (K2 nK1 ) ,?????
78
V8.1 1 2 3 P L K p Suppose that the principal branch of z 2 1 is continued analytically from z = 2 around the …gure eight path indicated above. What is the analytic continuation of the function at the end of the path? Answer the same question 1=3 1=3 form the functions (z 3 1) and (z 6 1) . Solution p z 2 1, phase change i at +1, 1 at 1, i at +1, returns to initial value. p 3 z3 p 1, phase change ei =3 at +1, ei =3 at +1, returns to e2 i=3 times initial value. 3 z 6 1, phase change ei =3 at +1, e 2 i=3 at 1, ei =3 at +1, returns to initial value.
79
V8.2 1 2 3 P L K has an analytic continShow that f (z) = Log z = (z 1) 12 (z 1)2 + it uation around the unit circle (t) = e , 0 t 2 . Determine explicitly the power series ft for each t. How is f2 related to f0 ? Solution P m 1 df = z1 = , f (m) (z) = ( 1) (m 1)! dz zm 1 f (m) eit 1 P P ( ) m it it Series is (z e ) = f (e ) + ( m! m=0
ft (z) = it +
1 P
m=1
m=1
(
itm 1)m 1 e m
(z
it m
1)m
1 e
itm
m
(z
e ) , get f2 (z) = f0 (z) + 2 i .
80
m
eit )
V.8.3 1 2 3 P L K p Show that each branch of z can be continued analytically along any path in Cn f0g, and show that the radius of convergencepof the power series ft (z) representing the continuation is j (t)j. Show that z cannot be analytically along path containing 0. Solution
81
V.8.4 1 2 3 P L K Let f (z) be analytic on a domain D, …x z0 2 D, and let f (z) = be the expansion of f (z) about z0 . Let F (z) =
Z
z
z0
f ( )d =
1 X an (z n + 1 n=0
P
an (z
z0 )n
z0 )n+1
be the inde…nite integral of f (z) for z near z0 . Show that F (z) can be continued analytically along any path in D starting at z0 . What happens in the case D = Cn f0g, z0 = 1, and f (z) = 1=z? What happens in the case that D is star-shaped? Solution
82
V.8.5 1 2 3 P L K Show that the function de…ned by X n f (z) = z2 = z + z2 + z4 + z8 +
is analytic on the open unit disk fjzj < 1g, and that it cannot be extended analytically to any larger open set. Hint. Observe that f (z) = z +f (z 2 ), and that f (r) ! +1 as r ! 1. Solution
83
V.8.6 1 2 3 P L K P Suppose f (z) = an z n , where an = 0 except for n in a sequence nk that satis…es nk+1 =nk 1 + for some > 0. Suppose further that the series has radius of convergence R = 1. Show that f (z) does not extend analytically to any point of the unit circle. Remark. Such a sequence with large gaps between successive nonzero terms is called a lacunary sequence. This result is the Hadamard gap theorem. There is a slick proof. If f (z) extends analytically across z = 1, consider g (w) = f (wm (1 + w) =2), where m is a large integer. Show that the power series for g (w) has radius of convergence r > 1, and that this implies that the power series of f (z) converges for jz 1j < ". Solution
84
V.8.7 1 2 3 P L K P Suppose f (z) = an z n , where the series has radius of convergence R < 1. Show that there is an angle such that f (z) does not have an analytic continuation along the path (t) = tei , 0 t R. Determine the radius of convergence of the power series expansion of f (z) about tei . Solution
85
V.8.8 1 2 3 P L K Let f (z) be analytic at z0 , and let (t), a t b, be a path such that (a) = z0 . If f (z) cannot be continued analytically along , show that there is a parameter value t1 such that there is an analytic continuation ft (z) for a t < t1 , and the radius of convergence of the power series ft (z) tends to 0 as t ! t1 . Solution
86
V.8.9 1 2 3 P L K Let P (z; w) be a polynomial in z and w, of degree n in w. Suppose that f (z) is analytic at z0 and satis…es P (z; f (z)) = 0. Show that if ft (z) is any analytic continuation of f (z) along any path starting at z0 , then P (z; ft (z)) = 0 for all t. Remark. An analytic function f (z) satis…es a polynomial equation P (z; f (z)) = 0 is called an algebraic function. For p n instance, the branches of z are algebraic functions, since the satisfy z wn = 0. Solution
87
V.8.10 1 2 3 P L K Let D be the punctured disk f0 < jzj < "g, suppose f (z) is analytic at z0 2 D, and ew0 = z0 . Show that f (z) has an analytic continuation along any path in D starting at z0 if and only if there is an analytic function g (w) in the half-plane fRe w < log "g such that f (ew ) = g (w) for w near w0 . Remark. If f (z) does not extend analytically to D but has an analytic continuation along any path in D, we say that f (z) has a branch point at z = 0. For the proof, use the fact that any path in D starting at z0 is the composition of a unique path in the half-plane starting at w0 and the exponential function ew . Solution
88
VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8
1
VI.1.1 1 2 3 P L K 111 LLL Find all possible Laurent expansions centered at 0 of the following functions. 1 (a) z21 z (b) zz+11 (c) (z2 1)(z 2 4) Solution (a) In the region 0 < jzj < 1, we have 1 z2
z
=
1X k z = z k=0
1 X
1
1 1 = z1 z
z
k 1
= [k
1 = n] =
1 X
zn:
n= 1
k=0
In the regionjzj > 1, we have 1 z2
1 1 = 2 z z 1
1 z
1 1 X = 2 z k=0
1 z
k
1 1 X = 2 z z k=0
k 2
=[ k
2 = n] =
2 X
zn:
n= 1
(b) In the region jzj < 1 we have 2 =1 z+1
z 1 =1 z+1
X 2 ( 1)n z n = =1 2 ( z) n=0 1
1
1 2
1 X
( 1)n z n :
n=1
In the region jzj > 1 we have z 1 =1 z+1 =1
2
2 =1 z+1
2 z1
1
=1
1 z
1 X ( 1)k = [k + 1 = n] = 1 k+1 z k=0
(c) From computation, we have 2
2
2 X ( 1)k = z k=0 z k 1
1 X ( 1)n zn n=1
1
=1+2
1 X ( 1)n : n z n=1
(z 2
1 1) (z 2
4)
1 1 1 1 + 2 3z 1 3 z2 4
=
In the region jzj < 1, we have
(z 2
1 1) (z 2
4)
=
1 1 1 1 1 1 1 1 + = = 2 2 2 3z 1 3z 4 31 z 12 1 z42 1 1 1 1 X z 2n 1 X 2n 1 X z 4 4 n z 2n : = = 3 n=0 12 n=0 4n 12 n=0
In the region 1 < jzj < 2, we have
(z 2
1 1) (z 2
4) =
=
1 1 1 1 + 2 = 2 3z 1 3z 4
1 1 X 1 3z 2 k=0 z 2n
1 1 2 3z 1
1 X z 2n = 12 n=0 4n 1
1 1 34 1
1 z2
1 1 X 2k z 3 k= 1
z2 4
=
1 X z 2n 12 n=0 4n 1
In the region jzj > 2, we have
(z 2
=
1 1) (z 2
4)
1 1 1 1 + 3 z2 1 3 z2 1 1 + = 3 z 2 1 z12 1 1 X 1 = + 3z 2 n=0 z 2n
=
= 4 1 1 = 3 z 2 1 z42 1 1 X 4n = 3z 2 n=0 z 2n
1X 1 1 X 4n 1X n + = (4 3 n=0 z 2(n+1) 3 n=0 z 2(n+1) 3 n=0 1
1
1
3
1) z
2(n+1)
= [n + 1 =
1 1 X = 4 3 k= 1
1 k
k] =
1 z 2k :
VI.1.2 1 2 3 P L K 111 LLL For each of the functions in Exercise 1, …nd the Laurent expansion centered at z = 1 that converges at z = 12 . Determine the largest open set on which each series converges. Solution (a) Partial fractions give us 1 z2 Laurent series for
1 = z
1=z
1 z+1
1
1 1 1 = z + 1 1 z+1
=
=
1 X k=0
Laurent series for 1=z 1 z
1 = 1 z+1
z
1 1 + z z 1
=
2
=
1 (z + 1)
k+1
1 X 1 = z + 1 k=0 (z + 1)k 1
= [k + 1 =
n] =
1 X
(z + 1)n
n= 1
1 1 X (z + 1)n = 2 n=0 2n 1
1 1 = 2 1 z+1 2
1 1 X
2n+1
(z + 1)n
n=0
Thus we have 1 z2
z
=
1 X
1 1 2n+1
k= 1
The function 1= (z 2 (b) We have
The function (z
an (z + 1)n ; where an =
for n for n
1 0
z) converges on the set 1 < jz + 1j < 2.
z 1 2 =1 z+1 z+1 1) = (z + 1) converges on the set 0 < jz + 1j < 1. 4
(c) Partial fractions give us 1 1 1 1 1 1 1 1 + + = 4) 6 z 1 6 z + 1 12 z 2 12 z + 2 1 1 1 1 1 1 1 1 = + + = 6 z + 1 2 6 z + 1 12 z + 1 3 12 z + 1 + 1 1 1 1 1 1 1 1 1 = z+1 + z+1 1 = 12 1 6 z + 1 36 1 12 (z + 1) 1 + z+1 2 3 1 1 1 1 X (z + 1)n 1 X ( 1)k 1 X (z + 1)n 1 1 + = = 12 n=0 2n 6 z + 1 36 n=0 3n 12 k=0 (z + 1)k+1
(z 2
=
1 1) (z 2
=
1 X (z + 1)n 1 1 + 12 n=0 2n 6z +1 1
1 1 X ( 1)n (z + 1)n + 12 n=0 6 1
The function 1= ((z 2
1) (z 2
1 12
1 n= 1 1 X (z + 1)n 1 X + ( 1)n (z + 1)n = n 36 n=0 3 12 1
(z + 1) 1 +
1 X 12 n=0 1
1
12
1
2n
36 3n
4)) converges on the set 1 < jz + 1j < 2.
5
(z + 1)n ;
VI.1.3 1 2 3 P
L K LLL Recall the powe series for the Bessel function Jn (z), n 0, given in Exercise V.4.11, and de…ne J n (z) = ( 1)n Jn (z). For …xed w 2 C, establish the Laurent series expansion exp
hw 2
(z
1 i X 1=z) = Jn (w) z n ; 1
From the coe¢ cient formula (1.4), deduce that Z 2 1 Jn (z) = ei(z sin n ) ; 2 0
0 < jzj < 1:
z 2 C:
Remark. This Laurent expansion is called Schlömlich formula. Solution (a) f (z) = tan z, 3 < jzj < 4 sin z , singularities at f (z) = tan z = cos z Simple poles at 2 + n sin z lim (z =2) cos = sin 2 lim cos zz z z! =2
sin z lim (z + =2) cos = sin z
z!
=2
z! =2 2
2
+ n ; n = 0; 1; 2; : : :
=2 cos =2
sin =2 sin =2
z+ =2 cos z cos( =2) =2
lim
z! 1 . z =2
=
=
=
1 sin( =2) sin( =2)
=
1
) Principal parts at 2 are ) f0 (z) = f (z) + z 1 =2 + z+1 =2 is analytic for jzj < 32 f1 (z) = z 1=2 z+1 =2 is analytic for jzj > 2 ; ! 0 at ????? ) f (z) = f0 (z) + 6 f1 (z) is the Laurent decomposition for 3 < jzj < 4, also for 2 < jzj < 32 . n P1 2 z+ 2 +z 2 2z 2z 1 (b) f1 (z) = = = = 2 2 2 2 2 2 k=0 z =4 z =4 z 4 z n P1 2 1 2 n=0 4 , converges for jzj > =2. z 2n+1 (c) f (z) is odd, so f0 (z) is odd, and a0 = a2 = 0. H H H H z 1 tan z a1 = 21 i jzj=3 tan dz = + dz 2 z 2 i z2 jzj=" jz =2j=" jz+ =2j=" H tan z sin z 1 At z 0, z2 , jzj=" = 2 i z cos z z H 1 1 1 tan z 4 , , = 24 2 i At z 2 2 2 2 z z z =2 z =2 jzH =2j=" z 1 1 4 1 At z , tan = 24 2 i 2 z+ =2 , 2 z2 z 2 z+ =2 jz+ =2j=" 6
(d) Since f0 (z) = f (z) f1 (z) has poles at jzj < 32 , the series converges for jzj < 32 .
7
3 2
, otherwise is analytic for
VI.1.4 1 2 3 P
L K LLL Suppose that f (z) = f0 (z) + f1 (z) is the Laurent decomposition of an analytic function f (z) on the annulus fA < jzj < Bg. Show that if f (z) is an even function, then f0 (z) and f1 (z) are even functions, and the Laurent series expansion of f (z) has only even powers of z. Show that if f (z) is an odd function, then f0 (z) and f1 (z) are odd functions, ant the Laurent series expansion of f (z) has only odd powers of z. Solution 1 H f (z) P dz, or de…nition of f (z)+f (z), f (z) = an z n , f ( z) = Use an = 0 1 n+1 z 1 jzj=r P ( 1)n an z n . f even ) an = an for n odd ) an = 0 for n odd. ) f0 (z) + f1 (z) have only even ????? ?????.
8
VI.1.5 1 2 3 P
L K LLL Suppose f (z) is analytic on the punctured plane D = Cn f0g. Show that there is a constant c such that f (z) c=z has a primitive in D. Give a formula for c in terms of an integral of f (z).
Solution By the Laurent Expansion Theorem we have for all z 2 D and r > 0 f (z) =
1 X
ak z
k= 1
Setting c = a
1
=
1 2 i
R
k
1 ak = 2 i
where
Z
f( )
Cr (0)
(
z0 )k+1
d :
f ( ) d we have X ak z k : f (z) c=z =
Cr (0)
k6= 1
R Recall that an analytic function g has a primitive on a region D i¤ g (z) dz = 0 for every piecewise smooth closed path in D. Let be such a path in D P and observe the series k6= 1 ak z k converges uniformly on any closed annulus fz : r jzj sg with 0 < r < s and hence on . Hence, by the …rst theorem on page 126, we may integrate the series termwise we obtain: Z X Z X X Z k ak 0 = 0; ak z k dz = ak z = f (z) c=zdz = k6= 1
k6= 1
since z k has a primitive if k 6=
1. Hence, f (z)
9
k6= 1
c=z has a primitive in D.
VI.1.6 1 2 3 P L K Fix an annulus D = fa < jzj < bg, and let f (z) be a continuous function on its boundary @D. Show that f (z) can be approximated uniformly on @D by polynomials in z and 1=z if and only if f (z) has continuous extension to the closed annulus D [ @D that is analytic on D. Solution Use the conformity theorem for polynomial approximation, V5 14. If f is analytic, continuous ????? values, unit f = f0 +f1 , its Laurent decomposing. Then f0 + f1 have continuous boundary values. Can approximate f0 (z) uniformly by polynomials in z for f1 (z) uniformly by polynomials in 1=z. (?????? change of variable w = 1=z ). That does it. If f = lim gn (z) uniformly on @D, then by the maximum principle jgk gj j ! 0 uniformly on D, so gk ! f uniformly on D, and clearly F n@D = f .
10
VI.1.7 1 2 3 P L K Show that a harmonic function u on an annulus fA < jzj < Bg has a unique expansion u rei
=
1 X
an rn cos (n ) +
n= 1
X
bn rn sin (n ) + c log r;
n6=0
which is uniformly convergent on each circle in the annulus. Show that for each r, A < r < B, the coe¢ cients an ; bn and c satisfy n
an r + a n
bn r + b
nr
n
nr
n
= =
1 1
1 a0 + c log r = 2
Z Z
Z
u rei
cos (n ) d ;
n 6= 0;
u rei
sin (n ) d ;
n 6= 0;
u rei
d :
Hint. Use a decomposition of the form u = Re f +c log jzj, where f is analytic on the annulus. (See Exercise III.3.4.) Solution By III.3.4, we have u = Re f +c log jzj for some e: Write itsPLaurent series k f (z) = 1 k= a ck z , A < jzj < B. Then for Pck = k k , wek have k Re f P = 1 k= 1 k Re z + k Im z k k u= 1 k= 1 k r cos (k ) + k r sin (k ) + c log The series converging uniformly on ????? The term w= 0 ????? Multiply by cos n , ????? use get for R R orthogonality, R n 6= 0, n 2 n u (r; ) cos n d = n r cos n d + n r cos2 n d = n n ( nr + nr ) R u (r; ) cos n d , n 6= 0. ) n rn + n r n = 1 R 1 n n The formula for n r u (r; ) sin n d , n 6= 0, is similar. nr R = R R 1 If we just integrate, we get u (r; ) = 21 c log rd + 2 0 d = 0 + c log r Note: 11
VI.1.8 (ej) H Multiply f by a unimodular constant, assume jf (z)j 6 M , f (z) dz = R2
f ei
d = 2 M . Take real parts, have
0
R2
jzj=1
Re f ei
iei
d = 2 M,
0
Re f ei iei d 6 f ei 6 M . If have strict inequality somewhere, R2 M . Since f ei iei then < 2 M . We conclude that Re f ei iei d 0
M , f ei 0. ) f ei iei M , we have Im f ei iei d f (z) iM z. ) f (z) = cz for a constant, jcj = M .
ie
i
M,
VI.1.9 (ej) For the analyticity, di¤erentiate by hand. (See i Exercise III.1.6). The Derivh R h(z) h(z) ative is H 0 (w) = lim 1w dz z (w+ w) z w w!1 R R h(z) w dz = (zh(z)dz = lim 1w (z (w+ w))(z w) w)2 w!1
12
6
VI.2.1 1 2 3 P
L K LLL Find the isolated singularities of the following functions, and determine where they are removable, essential, or poles. Determine the order of any pole, and …nd the principal part at each pole. 2 sin z (a) z= (z 2 1) (d) tan z = cos (g) Log 1 z1 z z (e) z 2 sin z1 (h) (zLog1)z3 (b) zze 2 1 2 2z cos z (c) e 1 (f) (i) e1=(z +1) z
z2
2 =4
Solution (a) (J.R.J Groves) with g (z) = We have that z = 1 is a pole of order 2 since (z2 z 1)2 = (zg(z) 1)2 z and g (z) is analytic near z = 1. Similarly we have, z = 1 is a pole (z+1)2 h(z) z of order 2 since (z z1)2 = (z+1) and h (z) is analytic near 2 with h (z) = (z 1)2 z = 1. (b) (J.R.J Groves) z g(z) zez We have that z = 1 is a pole of order 1 since zze 2 1 = z 1 with g (z) = z+1 , and g (z) is analytic near z = 1. Similarly, z = 1 is a pole of order 1 since z zez = h(z) with h (z) = zze 1 , and h (z) is analytic near z = 1. z2 1 z+1 (c) (A. Kumjian) Note that the given function is analytic on the punctured plane D = Cn f0g. It has an isolated singularity at z = 0. For z 6= 0 we have ! 1 1 1 X 1 1 X 2j+1 j e2z 1 (2z)k 1 X 2k z k = = z : 1+ = z z k! z k! z (j + 1)! j=0 k=0 k=1
Since the function has a power series expansion for all z 6= 0, it extends to an analytic function at z = 0. Hence, the function has a removable singularity at 0. (d) (J.R.J Groves) We have that cos z has simple zeros at z = =2 + k for k 2 Z. So cos1 z has simple poles at z = =2 + k for k 2 Z. Since sin z is analytic and di¤erent sin z has simple poles at z = =2 + k for k 2 Z. to 0 at these points, tan z = cos z (e) (A. Kumjian)
13
The function is analytic on the punctured plane D = Cn f0g. It has an isolated singularity at z = 0. For z 6= 0 we have 2
z sin
1 z
1 X ( 1)k =z (2k + 1)! k=0 2
1 z
2k+1
=
1 X ( 1)k 1 = 2k (2k + 1)! z k=0
1
1 X 1 ( 1)k =z+ : 2k (2k + 1)! z 1 k=1
Hence z = 0 is an essential singularity for the function since there are an in…nite number of nonzero terms in the Laurent series with negative exponents. (e) (J.R.J Groves) The function z 2 sin z1 has an isolated singularity at z = 0. Since z 2 sin
1 z
= z2
1 1 X X 1 ( 1)n 1 ( 1)n = ; 2n+1 2n 1 (2n + 1)! z (2n + 1)! z n=0 n=0
z = 0 is an essential singularity. (f) (J.R.J Groves) The function z 2 ( =2)2 = (z =2) (z + =2) has simple zeros at z = =2 and z = =2. The function cos z has a simple zero at z = =2. So we suspect a removable singularity. Set w = z =2. Then
z2
cos (z) = (z ( =2)2
cos (z) cos (w + =2) = = =2) (z + =2) w (w + )
1 w+
sin w : w
Since we know that sin (w) =w = 1 w2 =3! + : : :, we deduce that our function has a removable singularity at w = 0 and so at z = =2. In a similar fashion, it also has a removable singularity at z = =2. (g) (J.R.J Groves) The function Log (z) is analytic on the region Cn ( 1; 0], it follows that Log 1 z1 is analytic on Cn [0; 1]. There are no isolated singularities. (h) (A. Kumjian) First observe that for z 2 C with jz 1j < 1 we have 1 1 = z 1 + (z
1)
=
1 X k=0
14
( 1)k (z
1)k :
Since Log z is a primitive of 1=z, its power series expansion centered at z0 = 1 my be obtained from that of 1=z by integrating termwise: we obtain (note Log 1 = 0) 1 X ( 1)k+1 (z Log z = k k=1
for jz
1j < 1. Hence, we have for 0 < jz
1)k ;
1j < 1
1 1 X X Log z 1 ( 1)k+1 ( 1)k+1 k = (z 1) = (z k k (z 1)3 (z 1)3 k=1 k=1 1 X 1 1 ( 1)j = + (z 1)j 2 2 (z 1) k + 3 (z 1) j=0
1)k
3
=
Hence, z = 0 is a pole of order 2. The principal part is given by (z 11)2 2(z1 1) . (h) (J.R.J Groves) We have that z = 1 is a pole of order 2. This follows from the fact that 3 2 for jz 1j < 1 so that Log z = (z 1) (z 21) + (z 31) 1 Log z 3 = (z 1) (z 1)2
1 1 1 + 2 z 1 3
(i) (J.R.J Groves) There are isolated singularities at z = z2
:
i and z = i. Moreover,
i 1 1 = +1 2 z+i
1 z
i
so that 1
i
e z2 +1 = e 2 i
1 z+i
e
i 1 2 z i
i
= e2
1 z+i
X1 ( i)n 1 : n n=0 2 n! (z i)n 1
1
since e 2 z+i is analytic and non-zero at z = i, e z2 +1 has an essential singularity at z = i. Similarly, it has an essential singularity at z = i.
15
(a) Double poles at 1 , principal parts (1=4) (z 1)2 . z 1 ez ez + z+1 . Principal part (b) Singularity at z = 1, simple poles zze 2 1 = 2 z 1 1=2 1=2 at z = 1, z+1 at z = 1. z 1 (d) tan z = sin z= cos z, poles at z = =2 + n , n = 0; 1; 2; : : :. cos z = sin ( =2 + n ) (z =2 n ) + O (z =2 n )3 sin( =2+n )+O((z =2+n )2 ) tan z = sin( =2+n )(z =2 n )+O (:::)3 = z =21 n + O (z =2 n ) Poles ( ) are simple, principal part 1= (z =2 n ) . 2 2 (f) cos z= (z =4), singularities at =2 are both removable. (g) De…ned and analytic for 1 1=z 2 Cn [ 1; 0), 1=z 1 2 Cn [0; 1), 1=z 2 Cn [1; 1), z 2 Cn [0; 1]. No isolated singularities. Rz R2 dw RP (h) Isolated singularity at z = 1, Log z = dw = = ( 1)n (w 1)n dw = w 1+w 1 1 P
1
n
( 1) (z 1) n+1
n=0 Log z (z 1)3
=
n+1
1 (z 1)2
Log z = (z 1 z(z 1)
1)
(z 1) 2
2
+
(z 1) 3
3
:::
+ analytic
2 Double pole, principal part (z 11)2 z(z1 2) at z = 1. (i) e1=(z +1) , singularities at z = i. Values of 1= (z 2 + 1) approach 1 from all directions as z ! i, 2 so values of e1=(z +1) do not converge, and singularities are essential.????? 2 2 check e1(1 t ) = f (it) ! +1 as t " 1 or t # 1, e1(1 t ) = f (it) ! 0 as t # 1 or t " 1, so no limit at i.
16
VI.2.2 1 2 3 P
L K LLL Find the radius of convergence of the power series for the following functions, expanded about the indicated point. (a) zz4 11 ; about z = 3 + i; (c) sinz z ; about z = i; cos z z2 (b) z2 2 =4 ; about z = 0; (d) sin3 z ; about z = i: Solution (a) Removable singularity at z = 1, poles at i; 1. R = 3 = ?????? to i (nearest singularity). (b) singularities at =2 are removable. p 2. (c) removable at z = 0, poles at ; 2 . R=j i j= 2 2 (d) z = sin z, simple poles at z = 0, R = .
17
VI.2.3 1 2 3 P L K Consider the function f (z) = tan z in the annulus f3 < jzj < 4g. Let f (z) = f0 (z) + f1 (z) be the Laurent decomposition of f (z), so that f0 (z) is analytic for jzj < 4, and f1 (z) is analytic for jzj > 3 and vanishes at 1. (a) Obtain an explicit expression for f1 (z). (b) Write down the series expansion for f1 (z), and determine the largest domain on which it converges. (c) Obtain the coe¢ cients a0 ; a1 and a2 of the power series expansion of f0 (z). (d) What is the radius of convergence of the power series expansion for f0 (z)? Solution (a) tan z has two poles in the disk jzj < 4, simple poles at =2, principal parts 1= (z =2). If f1 (z) = 1= (z =2) 1 (z + =2), then f0 (z) = f (z) f1 (z) is analytic for jzj < 4, and f1 (z) is analytic for jzj > 3 and ! 0 as z ! 1. By uniqueness, f (z) = f1 (z) + f2 (z) is the Laurent decomposition. z+ =2+z =2 = (b) Use geometric series. Converges for jzj > =2. f1 (z) = 2 =4 z2 1 n P 2 1 2z 2z = , converges for jzj > =2. 2 =4 = z2 z2 4 z2 2
1 P
n=0
k=0
2
4
n
1
z 2n+1
(c) a0 = a2 = 0; a1 = 1 + 8= 2 . (d) Since f0 (z) = f (z) f1 (z) has poles at 3 =2, otherwise is analytic for jzj < 3, the series converges for jzj < 3 =2, and R = 3 =2.
18
VI.2.4 1 2 3 P L K Suppose f (z) is meromorphic on the disk fjzj < sg, with only a …nite number of poles in the disk. Show that the Laurent decomposition of f (z) with respect to the annulus fs " < jzj < sg has the form f (z) = f0 (z) + f1 (z), where f1 (z) is the sum of the principal parts of f (z) at its poles Solution Since f1 (z) is a sum of principal parts, with poles inside fjzj 6 "g, f1 (z) is analytic for jzj > ", and f1 (z) ! 0 as z ! 1. Further f0 (z) = f (z) f1 (z) is analytic at ????? ??????, hence for jzj < 5. By the uniqueness of the Laurent decomposition, f is f0 (z) + f1 (z) = f (z). Use the uniqueness of the Laurent decomposition.
19
VI.2.5 1 2 3 P L K By estimating the coe¢ cients of the Laurent series, prove that if z0 is an isolated singularity of f , and if (z z0 ) f (z) ! 0 as z ! z0 , then z0 is removable. Give a second proof based on Morea’s theorem. Solution Suppose that z0 is an isolated singularity of f and that (z z0 )N f (z) is bounded near z0 . Then by Riemann’s Theorem on removable singularities (see p. 172), (z z0 )N f (z) has a removable singularity at z0 , the Laurent expansion is of the following form: there is a r > 0 so that (z
N
z0 ) f (z) =
1 X
ak (z
z0 )k
k=0
for all 0 < jz
z0 j < r. And hence, we have f (z) =
1 X
ak (z
z0 )k
N
=
1 X
aj+N (z
z0 )j :
j= N
k=0
It follows that z0 is a pole of order at most N unless ak = 0 for all k = 0; 1; : : : ; N 1, in which case z0 is a removable singularity.
20
VI.2.6 1 2 3 P
L K LLL Show that if f (z) is continuous on a domain D, and if f (z)8 is analytic on D, then f (z) is analytic on D. Solution
21
VI.2.7 1 2 3 P
L K LLL Show that if z0 is an isolated singularity of f (z), and if (z z0 )N f (z) is bounded near z0 , then z0 is either removable or a pole of order at most N .
Solution Suppose that z0 is an isolated singularity of f and that (z z0 )N f (z) is bounded near z0 . Then by Riemann’s Theorem on removable singularities (see p. 172), (z z0 )N f (z) has a removable singularity at z0 , the Laurent expansion is of the following form: there is r > 0 so that (z
N
z0 ) f (z) =
1 X
ak (z
z0 )k
k=0
for all 0 < jz
z0 j < r. And hence, we have f (z) =
1 X
ak (z
k N
z0 )
=
1 X
aj+N (z
z0 )j :
j= N
k=0
It follows that z0 is a pole of order at most N unless ak = 0 for all k = 0; 1; : : : ; N 1, in which case z0 is a removable singularity.
22
VI.2.8 1 2 3 P L K A meromorphic function f at z0 is said to have order N at z0 if f (z) = (z z0 )N g (z) for some analytic function g at z0 such that g (z0 ) 6= 0. The order of the function 0 is de…ned to be +1. Show that (a) order (f g; z0 ) = order (f; z0 ) + order (g; z0 ), (b) order (1=f; z0 ) = order (f; z0 ), (c) order (f + g; z0 ) min forder (f; z0 ) ; order (g; z0 )g. Show that the inequality can occur in (c), but that equality hold in (c) whenever f and g have di¤erent orders at z0 . Solution
23
VI.2.9 1 2 3 P L K Recall that "f (z) = O (h (z)) as z ! z0 " means that there is a constant C such that jf (z)j C jh (z)j for z near z0 . Show that if z0 is an isolated singularity of an analytic function f (z), and if f (z) = O ((z z0 )m ) as z ! z0 , then the Laurent coe¢ cients of f (z) are 0 for k < m, that is the Laurent series of f (z) has the form f (z) = am (z
z0 ) + am+1 (z
z0 )m+1 +
:
Remark. This shows that the use of notation O (z m ) in section V.6 is consistent. Solution
24
VI.2.10 1 2 3 P
L K LLL Show that if f (z) and g (z) are analytic functions that both have the same order N 0 at z0 , then f (z) f (N ) (z0 ) = (N ) : z!z0 g (z) g (z0 ) lim
Solution
25
VI.2.11 1 2 3 P L K P Suppose f (z) = ak z k is analytic for jzj < R, and suppose that f (z) extends to be meromorphic for jzj < R + ", with only one pole z0 on the circle jzj = R. Show that ak =ak+1 ! z0 as k ! 1. Solution P If f (z) = an z n is analytic for jzj < R, and is meromorphic for jzj < R + ", with poles N on the circle jzj = R, then an = O nN 1 . If P of ?????6 f (z) = an z n as only one pole on the circle jzj = R, of ????? N , with A( 1)N +1 kN (1 + O (1=k)), so f (z) = A= (z z0 )N + lower order, then ak = (N 1)!z N +k+1 0
N P P Al + bk z k that ak =ak+1 ! z0 as n ! 1. Proof: Write f (z) = (z z0 )l l=0 P analytic, h (z) = bk z k analytic. We have jbk j 6 1=sk for some s > 1 j P 1 1 z R, bk = O 1= jz0 jk . z 1z0 = z10 1 z=z = , ( 1)d (z(l z1)!)l = z0 0 zj 1 z0
1 P
j=l
1 (z z0 )l 1 (z z0 )l
j=0
j(j 1):::(j l+1)z j z0l
=
( 1)l+1 (l 1)!
=
( 1)l+1 (l 1)!
1 P
k=0 1 P k=0
l
1 z0k+l+1
(k + l) : : : (k + 1) z k
1 z0k+l+1
(k + l) : : : (k + 1) z k
26
0
0
VI.2.12 1 2 3 P
L K LLL Show that if z0 is an isolated singularity of f (z) that is not removable, then z0 is an essential singularity for ef (z) .
Solution Apply Arzela‘-Weierstrass theorem. Suppose z0 = 0, an isolated singularity of f (z), not removable. 0 essential ) values of f (z) ???? on C as z ! 0, ! values of ef (z) ????? on C as z ! 0. 0 a pole ) f (z) ! 1, as z ! 0, monotone, values of f (z) ?????? of 0 cover the extremum of some ?????? fjwj > Rg. In any neighbourhood of 0 there are values of ef (z) = ew that are near 0 (e m ) and that are near to (e+m ), so ef (z) does not have a limit as z ! 0. 0 is essential singularity of ef (z) .
27
VI.2.13 (From Hints and Solutions) 1 2 3 P L K Let S be a sequence converging to a point z0 2 C, and let f (z) be analytic on some disk centered at z0 except possibly at the points of S and z0 . Show that either f (z) extends to be meromorphic on some neighborhood of z0 , or else for any complex number L there is a sequence fwj g such that wj ! z0 and f (wj ) ! L. Solution Suppose values of f (z) do not cluster at L as z ! z0 . Then g (z) = 1= (f (z) L) is bounded for jz z0 j < "; z 6= zj . Apply Riemann‘s theorem …rst to the zj `s for j large, then to z0 , to see that g (z) extends to be analytic for jz z0 j < ", and f (z) is meromorphic there
28
VI.2.14 1 2 3 P L K Suppose u rei is harmonic on a punctured disk f0 < r < 1g, with Laurent series as in Exercise 7 of Section 1. Suppose > 0 is such that r u rei ! 0 as r ! 0. Show that an = 0 = bn for n . Solution 1 1 P P u rei harmonic, 0 < r < 1. u = an rn cos n + bn rn sin n + c log r. n=0
n6=0
Let m > , m > 0. Have am rm + a m r m = O (r ), am r2m + a m = O (r +m ) as r ! 0. Let r ! 0, get a m = 0 for m > . Similarly b m = 0 for m > .
29
VI.2.15 1 2 3 P L K Suppose u (z) is harmonic on the punctured disk f0 < jzj < g. Show that if u (z) !0 log (1= jzj)
as z ! 0, then u (z) extends to be harmonic at 0. What can you say if you know only that ju (z)j C log (1= jzj) for some …xed constant C and 0 < jzj < ? Solution Assume u (z) = O (log (1= jzj)) as z ! 0. Consider Laurent series as in R 14. Have. For m > 1, get u rei d = O (log (1=r)) am rm + a m r m = O (log (1=r)), am r2m + a m = O (rm log (1=r)), a m = 0 for all m > 0, and b m = 0 for m > 0. u rei = Re f rei + c log r, when f is analytic, c = Re f (rei ) u(r) c log r = log(1=r) + log(1=r) , c = 0, and u = Re f is analytic at constant. log(1=r) z = 0.
30
VI.3.1 1 2 3 P L K Consider the functions in Exercise 1 of Section 2 above. Determine which have isolated singularities at 1, and classify them. Solution
31
VI.3.2 1 2 3 P L K Suppose that f (z) is an entire function that is not a polynomial. What kind of singularity can f (z) have at 1? Solution
32
VI.3.3 1 2 3 P L K Show that if f (z) is nonconstant entire function, then ef (z) has an essential singularity at z = 1. Solution
33
VI.3.4 1 2 3 P L K Show that each branch of the following functions is meromorphic at 1, and obtain the series expansion for each branch at p p 1. 5=2 3 2 3 (a) (z 1) (b) (z 1) (c) z 2 1=z Solution
34
VI.4.1 1 2 3 P
L K LLL Find the partial fractions decompositions of the following functions. z 1 (a) z21 z (c) (z+1)(z12 +2z+2) (e) z+1 2 z 2 +1 1 (b) z(z (d) (z2 +1) (f) zz2 4z+3 2 2 1) z 6 (a) 1 z2
z
=
A B + z z 1
A = lim z f (z) = z!0
B = lim (z z!1
1 z2
z
1 z
1
1) f (z) =
=
=
1
z=0
1 z
=1 z=1
1 1 + z z 1
(b) z2 + 1 A B C = + + 2 z (z 1) z z 1 z+1 A = lim z f (z) = z!0
z2 + 1 z2 1
=
1
z=0 2
z +1 =1 z!1 z (z + 1) z=1 z2 + 1 C = lim (z + 1) f (z) = =1 z! 1 z (z 1) z= 1
B = lim (z
1) f (z) =
z2 + 1 = z (z 2 1)
1 1 1 + + z z 1 z+1
(c) 1 (z +
1) (z 2
+ 2z + 2)
=
A B C + + z+1 z+1+i z+1 35
i
A = B = C =
lim z f (z) =
z! 1
(z 2
1 + 2z + 2)
=1 z= 1
1 (z + 1) (z + 1 i) 1 i) f (z) = (z + 1) (z + 1 + i)
=
lim (z + 1 + i) f (z) =
z! 1 i
lim (z + 1
z! 1+i
1 (z +
1) (z 2
+ 2z + 2)
=
1 z+1
1=2 z+1+i
z= 1 i
= z= 1+i
1=2 z+1 i
(d) (/snider 106) man så? A B C D 1 = + + + 2 2 2 (z i) (z + i) z i (z 2 + 1) (z i) 1 1 = 2 z!i 4 (z + i) z=i d 1 i B = lim (z i)2 f (z) = = 2 z!i dz (z + i) z=i 4 1 1 C = lim (z + i)2 f (z) = = 2 z! i 4 (z i) z= i d i 1 D = lim (z + i)2 f (z) = = 2 z! i dz (z i) z= i 4 i)2 f (z) =
A = lim (z
1 = (z 2 + 1)2
1=4 (z i)2
i=4 (z i)
1=4 i=4 2 + z i (z + i)
(e) z 1 A = +B z+1 z+1 A = B =
lim (z + 1) f (z) = z
z! 1
lim f (z) = 1
z!1
36
1jz=
1
=
2
1 2 1 2
z 1 =1 z+1
2 z+1
(f) z2 z2
A =
lim (z + 2) f (z) =
z! 2
B = lim (z z!3
C =
A B 4z + 3 = + +C z 6 z+2 z 3
3) f (z) =
z2
4z + 3 z 3
z 2 4z + 3 z (z + 1)
lim f (z) = 1
z!1
z2 z2
4z + 3 3 = +1 z 6 z+2
37
= z= 2
=0 z=3
3
VI.4.2 1 2 3 P
L K LLL Use the division algorithm to obtain the partial fractions decomposition of the following functions 3 9 z6 (a) zz2 +1 (b) zz6 +11 (c) (z2 +1)(z +1 1)2 (a) z3 + 1 z+1 A B = z + = z + + z2 + 1 z2 + 1 z+i z i
A =
1 1 z+1 = + i z i z= i 2 2 1 1 z+1 = i) f (z) = i z + i z=i 2 2
lim (z + i) f (z) =
z! i
B = lim (z z!i
z+1 ( 1 + i) =2 z3 + 1 =z+ 2 =z+ 2 z +1 z +1 z+i
(1 + i) =2 z i
(b) 3 z9 + 1 1 3 z +1 = z + = z3+ 3 = z3+ 6 6 z 1 z 1 z 1 (z
1 A B C = z3+ + + 2 1) (z + z + 1) z 1 z w z w
Because z 3 1 = 0 have tree roots z1 = 1, z2 = e2 z2 = w, and we see that z3 = w.
A = lim (z z!1
B = lim (z z!w
C = lim (z z!w
1) f (z) =
z2
1 +z+1
= z=1
i=3
, and z3 = e4
1 3
w 1 = w) f (z) = (z 1) (z e 2 i=3 ) z=e2 i=3 3 1 w w)2 f (z) = = 2 i=3 (z 1) (z e ) z=e 2 i=3 3
z9 + 1 1=3 w=3 w=3 = z3 + + + ; 6 z 1 z 1 z w z w 38
where w = e2
i=3
i=3
. Set
Konstigt svar i boken? 1=3 + zw=3w + zw=3w z 1 (c)
1=3 z 1
z6 (z 2 + 1) (z 2 z4
z6 2z 3 + 2z 2
+
w=3 z w
2z + 1) 2z + 1
w=3 z+w
=
z4
=
z6 2z 3 + 2z 2
= z 2 + 2z + 2 +
2z + 1
2z 3 z 2 + 2z 2 (z 2 + 1) (z 1)2
2z 3 z 2 + 2z 2 A B C D = + + + 2 2 z + i z i (z 1) z 1 (z 2 + 1) (z 1)
A =
2z 3 z 2 + 2z 2 = (z i) (z 1)2 z= i 2z 3 z 2 + 2z 2 1 i) f (z) = = 2 4 (z + i) (z 1) z=i 2z 3 z 2 + 2z 2 1 1)2 f (z) = = 2 (z + 1) 2 z=1 3 2 d 2z z + 2z 2 1)2 f (z) = = dz (z 2 + 1) z=1
lim (z + i)2 f (z) =
z! i
B = lim (z z!i
C = lim (z z!1
D = lim (z z!1
39
1 4
5 2
VI.4.3 1 2 3 P
L K LLL Let V be the complex vector space of functions that are analytic on the extended complex plane except possibly at the points 0 and i, where they have poles of order at most two. What is the dimension of V ? Write down explicitly a vector space basis for V . Solution
40
VI.5.1 1 2 3 P L K Show that if f (z) and g (z) have period w, then so do f (z) +g (z) and f (z) g (z). Solution
41
VI.5.2 1 2 3 P L K Expand 1= cos (2 z) in a series of powers of e2 iz that converges in the upper half-plane. Determine where the series converge absolutely and where it converges uniformly. Solution
42
VI.5.3 1 2 3 P L K Expand tan z in a series of powers of exponentials eikz , 1 < k < 1, that converges in the upper half-plane. Also …nd an expansion of tan z as in an exponential series that converges in the lower half-plane. Solution
43
VI.5.4 1 2 3 P L K Let f (z) be an analytic function in the upper half-plane that is periodic, with real period 2 > 0. Suppose that there are A; C > 0 such that jf (x + iy)j CeAy for y > 0. Show that X f (z) = an einz= ; n
A
where the series converges uniformly in each half-plane fy " > 0. Solution
44
"g, for …xed
VI.5.5 1 2 3 P L K Suppose that 1 are periods of a nonzero doubly periodic function f (z), and suppose that there are no periods w of f (z) satisfying 0 < jwj < 1. How many periods of f (z) lie on the unit circle? Describe the possibilities, and sketch the set of periods for each possibility. Solution
45
VI.5.6 1 2 3 P L K We say that w1 and w2 generate the periods of a double periodic function if the periods of the function are precisely the complex numbers of the form mw1 + nw2 where m and n are integers. Show that if w1 and w2 generate the periods of a doubly periodic function f (z), and if 1 and 2 are complex numbers, then 1 and 2 generate the periods of f (z) if and only if there is a 2 2 matrix A with integer entries and with determinant 1 such that A (w1 ; w2 ) = ( 1 ; 2 ). Solution
46
VI.5.7 1 2 3 P L K Let w1 and w2 be two complex numbers that do lie on the same line through 0. Let k 3. Show that the series 1 X
m;n= 1
1 (z
(mw1 + nw2 ))k
converges uniformly on any bounded subset of the complex plane to double periodic meromorphic function f (z), whose periods are generated by w1 and w2 . Strategy. Show that the number of periods in any annulus fN jzj N + 1g is bounded by CN for some constant C. Solution
47
VI.6.1 1 2 3 P L K . Find the Consider the continuous function f ei = j j, i complex Fourier series f e and show that it can be expressed as a cosine series. Sketch the graphs of the …rst three partial sums of the cosine series. Discuss the convergence of the series. Does it converge uniformly? Partial answer. The cosine series is j j=
4 2
cos +
1 1 cos 3 + 2 cos 5 + 2 3 5
Solution
48
:
VI.6.2 1 2 3 P L K < (the principal value of the argument). Find the Let f ei , complex Fourier series of f ei and the sine series of f ei . Show that the complex Fourier series diverges at = , while the sine series converges at . Di¤erentiate the complex Fourier series term by term and determine where the di¤erentiated series converges. Solution
49
VI.6.3 1 2 3 P L K Consider the continuous function f ei = 2 , . Find the i complex Fourier series of f e and show that it can be expressed as a cosine series. Discuss the convergence of the series. Does it converge uniformly? By substituting = 0, show that 2
12
=1
1 1 + 22 32
Solution
50
1 + 42
:
VI.6.4 1 2 3 P L K Consider the continuous function f ei = 4 2 2 2 , . Find i the complex Fourier series of f e and show that it can be expressed as a cosine series. Relate the Fourier series to the series of the function in Exercise 3. Solution
51
VI.6.5 1 2 3 P L K P P ik Show that if ck converges absolutely, then ck e converges absolutely for each , and the series converges uniformly for . Solution
52
VI.6.6 1 2 3 P L K Show that any function f ei on the unit circle with absolutely convergent Fourier series has the form f ei = g ei + h (ei ), where g (z) and h (z) are continuous functions on the unit circle that extend continuously to be analytic on the open unit disk. Solution
53
VI.6.7 1 2 3 P L K If f ei
P
ck eik , and the series converges uniformly to f ei , then Z
i
f e
2
1 X d = jck j2 : 2 k= 1
Remark. This is called Parseval’s identity. Formula (6.6) show Parseval’s identity holds for a function f ei if and only if the partial sums of the Fourier series of f ei converge to f ei in the sense of "mean-square" or "L2 approximation". Solution
54
VI.6.8 1 2 3 P L K By applying Parseval’s identity to the picewise constant function with series (6.5), show that 2
1 1 1 + 2+ 2+ : 2 8 3 5 7 Use this identity and som algebraic manipulation to show that =1+
2
6
=1+
1 1 1 + 2+ 2+ 2 2 3 4
Solution
55
:
VI.6.9 1 2 3 P L K By applying Parseval’s identity to the function of Exercise 1, show that 4
1 1 1 + 4+ 4+ : 4 96 3 5 7 Use this identity and som algebraic manipulation to show that =1+
4
90
=1+
1 1 1 + 4+ 4+ 4 2 3 4
Solution
56
:
VI.6.10 1 2 3 P L K If f (z) is analytic P in some annulus containing the unit circle jzj = 1, with Laurent expansion ak z k , then 1 2
I
jzj=1
jf (z)j2 jdzj =
Solution
57
1 X
k= 1
jak j2 :
VI.6.11 1 2 3 P L K Let f ei be a continuous function on the unit circle, with Fourier series P ck eik . Show that f ei extends to be analytic on some annulus containing the unit circle if and only if there exist r < 1 and C > 0 such that jck j Crjkj for 1 < k < 1. Solution
58
VI.6.12 1 2 3 P L K Using the convergence theorem for Fourier series, prove that every contintinously function on the unit circle in the complex plane can be approximated uniformly there by trigonometric polynomials, that is, by …nite linear combinations of exponentials eik , 1 < k < 1. Strategy. First approximate f eik by a smooth function. Solution
59
VI.6.13 1 2 3 P L K Let D be a domain bounded by a smooth boundary curve of length 2 . We parametrize the boundary of D by arc length s, sot the is given by P boundary iks a smooth periodic function (s), 0 s 2 . Let ck e be the Fourier series of (s). (a) P 2 Show that k jck j2 = 1. Hint. Apply Parseval’s identity to 0 (s) and use j 0 (s)j = 1 for a curve parameterized by arc length. (b) P Show that the area of D is k jck j2 . Hint. Use Exercise IV.1.4. (c) Show that the area of D is , with equality if and only if D is a disk. Remark. This proves the isoperimetric theorem. Among all smooth closed curves of a given length, the curve that surrounds the largest area is a circle. Solution
60
VI.1.14 1 2 3 P L K Show that Z
i
f e
n X
k= m
2 ik
bk e
d = 2
Z
i
f e
n X
k= m
2 ik
ck e
n X d + jbk 2 k= m
ck j2 ;
for any choice of complex numbers bk , m k n. Remark. This i by exponential sums shows that the best mean-square approximate to f e Pn ik , for …xed m and n, is the corresponding partial sum of the Fourier m bk e series. Solution
61
VI.1.15 1 2 3 P L K Show that a continously di¤erentiable function on the unit circle has an absolutely convergent Fourier series. Strategy. Write the Fourier coe¢ cients ck of f ei as ak bk , where ak = 1=ik and bk is the Fourier coe¢ cient of the derivative. qPUse Bessel’ qP s inequality and the Cauchy-Schwarz inequality P 2 j kj j k j2 . k k Solution
62
VI.6.16 1 2 3 P L K Let f ei be a continuous function on the unit circle. Suppose that f ei is piecewise continuously di¤erentiable, in the sense that it has a continuous derivative except at a …nite number of points, at each of which the derivative has limits from the left and from the right. Show that the Fourier series f ei is absolutely convergent. Strategy. Cancel the discontinuities of the derivative using translates of the function in Exercise 3, whose Fourier series is absolutely convergent. Solution
63
VI.6.17 1 2 3 P L K Let f ei be a piecewise continuously di¤erentiable, in the sense that it is continuously di¤erentiable except at a …nite number of points, at each of which both the function and its derivative have limits from the left and from the right. Show that the Fourier series of f ei converges at each point, to f ei if the function is continuous at ei , and otherwise to the average of right. Strategy. Show that the limits of f ei from P the ileft and from the i i i satis…es the hypotheses of bj hj e , where f1 e = f1 e + f e Exercise 15, and each hj ei is obtained from the function of Exercise 2 by change of variable 7 ! j. Solution
64
VII 1 2 3 4 5 6 7 8
1 X X X X X X
2 X X X X X X X
3 X X X X X X X
4
5
X X X X X
X X X X X X
6 X X X X X X
7
8
9
X X X X X X X X
X
1
10 11 12 13 14 15 16 17 18 19 X X
X
X
VII.1.1 Evaluate the following residues. z (d) Res sin (a) Res z21+4 ; 2i ;0 z2 (b) Res (c) Res
1 ; z 2 +4 1 ;1 z5 1
(g) Res
cos z ;0 z2
(e) Res (h) Res (f) Res [cot z; 0] (i) Res
2i
h
i
z ;1 Log z z e ;0 z 5n z +1 ; wk zn 1
Solution a) By rule 4, Res
z2
1 1 ; 2i = +4 2z
1 = 4i
= z=2i
i : 4
b) By rule 4, Res
z2
1 1 ; 2i = +4 2z
= z= 2i
1 i = : 4i 4
c) By rule 4, Res
1 z5
1
;1 =
1 5z 4
z=1
1 = : 5
d) By rule 1, Res
sin z sin z ; 0 = lim = 1: 2 z!0 z z
e) By rule 2, Res f) By rule 3,
h cos z z2
i d ; 0 = lim cos z = z!0 dz
Res [cot z; 0] = Res
sin zjz=0 = 0:
i cos z ;0 = sin z cos z
h cos z 2
= 1: z=0
g) By rule 3, Res
z z ;1 = Log z 1=z
= 1: z=1
h) By Laurent expansion 1+ ez = z5
z 1!
+
z2 2!
+
z3 3! z5
+
z4 4!
+ o (z 5 )
=
1 1 1 1 1 + o (1) ; + + + + z 5 1!z 4 2!z 3 3!z 2 4!z
hence Res
ez 1 1 ;0 = = : 5 z 4! 24
i) We have poles of the form wk = e2 that wkn = 1, by rule 3,
Res
zn + 1 zn + 1 ; w = k zn 1 nz n 1
ik=n
= z=wk
, where k = 0; 1; 2; : : : n
wkn + 1 2 n 1 = nwk nwkn
3
1
=
1, remark
2e2 ik=n 2wk = : n n
VII.1.2 Calculate the residue at each singularity in the complex plane of the following functions. z (d) z21+z (a) e1=z (b) tan z (c) (z2 +1) 2 Solution (a) The function e1=z has a singularity at z = 0, by Laurent expansion e1=z = 1 +
1 1 + + :::: z 2!z 2
Thus by de…nition Res e1=z ; 0 = 1: (b) sin z has isolated singularities at z = The function tan z = cos z n < 1, they are simple poles. By rule 3,
Res [tan z; =2 + n ] = Res (c) The function poles. By rule 2,
Res
z (z 2 +1)2
sin z ; =2 + n cos z
=
sin z sin z
has isolated singularities at z =
z d z ;i = 2 dz (z + i)2 (z 2 + 1)
z d z Res 2; i = 2 dz (z i)2 (z + 1)
= z=i
= z= i
(d)
4
2
+ n,
=
i)2 (z
1:
z= =2+n
i, they are double
(z + i)2 2z (z + i) (z + i)4
(z
1<
2z (z i)4
= 0; z=i
i)
= 0: z= i
The function simple poles. By rule 1,
1 z(z+1)
has isolated singularities at z = 0 and z =
Res
1 1 1 ; 0 = lim = z!0 z + 1 z (z + 1) z+1
Res
1 1 1 ; 1 = lim = z! 1 z z (z + 1) z
5
= 1; z=0
= z= 1
1:
1 they are
VII.1.3 Evaluate the followingR integrals using Rresidue theorem. R z (a) jzj=1 sin dz (c) jzj=2 cosz z dz (e) jz 1j=1 z81 1 dz z2 R R R z z4 dz (f) jz 1=2j=3=2 tanz z dz (b) jzj=2 z2e 1 dz (d) jzj=1 sin z
Solution (a) Using the residue theorem and rule 1, Z
jzj=1
sin z sin z sin z dz = 2 i Res ; 0 = 2 i lim 2 2 z!0 z z z
= 2 i (1) = 2 i:
(b) Using the residue theorem and rule 3, Z
jzj=2
ez z2
1
dz = = 2 i Res
=2 i
ez z2
1 z e ez + 2z z=1 2z
; 1 + Res
ez z2
=2 i z= 1
1 e1
; 1 e
= 1
2
= = 2 i sinh 1:
(c) Using the residue theorem and rule 3, Z
jzj=2
h z i h z i z dz = 2 i Res ; + Res ; = cos z cos z 2 cos z 2 ! z z =2 i + =2 i sin z z= =2 sin z z= =2 2
2
= =
(d) Using the residue theorem and rule 3, 6
2 2 i:
Z
jzj=1
z4 z4 dz = 2 i Res ;0 = 2 i sin z sin z
z4 cos z
= 0: z=0
(e)
VII.1.3e
|z - 1| = 1
Using the residue theorem with the contour in Figure VII.1.3e and rule 4, Z
1
dz = 2 i Res
8 1 jz 1j=1 z 1 1 + 7 =2 i 7 8z z=1 8z
1 z8
z=ei
=4
1
; ei
=4
+ Res
1
z8 1 z8 i = 1 + e i=4 + e i=4 = 4 i =4 z=e p 1 i 1+i i 1+ 2 : 1+ p + p = 4 2 2
1
1 8z 7 i = 4
+
; 1 + Res
(f) z=z The function tanz z = sin has removable singularity at z = 0, and a isolated cos z singularity at z = 2 inside the domain jz 1=2j = 3=2. Using the residue theorem and rule 3,
7
1
;e
i =4
=
Z
jz 1=2j=3=2
tan z dz = z
sin z=z = 2 i Res ; =2 i cos z 2
sin z=z lim z! =2 sin z
=2 i
sin z=z sin z
z= =2
!
=
8
= (2 i) 4i:
2
=
VII.1.4 1 2 3 P L K Suppose P (z) and Q (z) are polynomials such that the zeros of Q (z) are simple zeros at the points z1 ; : : : ; zm ; and deg P (z) < deg Q (z). Show that the partial fractions decomposition P (z) =Q (z) is given by P (z) X P (zj ) 1 = : 0 (z ) z Q (z) Q z j j j=1 m
Solution Since the zeros of Q (z) are simple, the polynomial quotient P (z) =Q (z) has at most simple poles. By Rule 3, the residue at zj is P (z) =Q0 (z) : Hence the principal part of P (z) =Q (z) at zj is P (zj ) 1 : Q0 (zj ) z zj Since deg P (z) < deg Q (z), then P (z) !0 Q (z) as z ! 1, and P (z) =Q (z) has no principal part at 1. By partial fraction decomposition our polynomial quotient P (z) =Q (z) can be written as the sum of its principal parts at its pole, P (z) X P (zj ) 1 = : 0 (z ) z Q (z) Q z j j j=1 m
9
or
VII.1.5 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane that is doubly periodic, and suppose that none of the poles of f (z) lie on the boundary of the period parallelogram P constructed in Section VI.5. By integrating f (z) around of P , show that the sum of the residues at the poles of f (z) in P is zero. Conclude that there is no doubly periodic meromorphic function with only one pole, a simple pole, in the period parallelogram. Solution R Follow directions. @P = 0, the curve integrals over opposite sides cancel. Thus the sum of the residues is zero. In particular, there can not be a single nonzero residue.
10
VII.1.6 Consider the integral Z
@DR
i(z 1=2)2
e 1
e
2 iz
dz;
where DR is the parallelogram with vertices 12 (1 + i) R. (a) p Use the residue theorem to show that the integral is (1 + i) = 2. (b) By parameterizing the sides of the parallelogram, show that the integral tends to Z 1 2 e 2 t dt (1 + i) 1
as R ! 1. (c) Use (a) and (b) to show that Z 1
e
s2
ds =
p
:
1
Solution (a)
VII.1.6 γ2
γ3
z1
γ1
γ4
Set 11
I=
Z
@DR
i.e, we integrate
e 1
i(z 1=2)2
e
2 iz
dz;
2
e i(z 1=2) f (z) = 1 e 2 iz along the parallelogram contour in Figure VII.1.6. The parallelogram has vertices at 1=2 (1 + i) R. Residue at a simple pole at z1 = 0, where by Rule 3, " # 2 2 e i(z 1=2) e i(z 1=2) e i=4 Res ; 0 = = : 1 e 2 iz 2 ie 2 iz 2 i z=0
Using the Residue Theorem, we obtain that Z
@DR
i.e.,
Z
i(z 1=2)2
e 1
e
@DR
e 1
2
dz = 2 i iz
i(z 1=2)2
e
2 iz
(b) Integrate along 1 parametrized by z = let R ! 1. This gives Z
=
Z
R R
R
1+i dz = p : 2 1 2
+ (1 + i) t with
R
t
R, and
i(z 1=2)2
z = 12 + (1 + i) t = dz = (1 + i) dt e 2 R Z R 2 2 e i((1+i)t) (1 + i) e 2 t dt = (1 + i) dt ! 1 2 i(1+i)t R 1+e 1 e 2 i( 2 +(1+i)t) Z 1 2 e 2 t ! (1 + i) dt: 2 i(1+i)t 1 1+e
f (z) dz = 1
Z
e i=4 ; 2 i
e 1
dz = iz
Now integrate along 2 parameterized by z = t 1, and let R ! 1. 12
t+
1 2
+ (1 + i) R with 0
Z
f (z) dz = 2
Z
= =
Z
0
1
e 1 2 e e
1
i(z 1=2)2
e
2
dz = iz
Z
i( t+(1+i)R)2 2 i( t+1=2+(1+i)R)
2 2 e (t 2t)i e 2 (R Rt) sup e (2t 1 2R)i e2 R 0 t 1 1
t + 12 + (1 + i) R dt
z = dz = dt =
1
0
1
Integrate along 3 , parameterized by z = and let R ! 1. This gives Z
sup
=
Z
R R
R
Rt)
2
(R2
Rt)
(2t 1 2R)i e2 R jj
= sup
Rj
1 2
e je
0 t 1
e
2
e2
(R2
Rt)
R
1
+ (1 + i) t with
R
= ! 0: t
R,
i(z 1=2)2
1 z = + (1 + i) t 2 = 2 dz = (1 + i) dt e R Z R 2 i(1+i)t 2 t2 2 e i( 1+(1+i)t) (1 + i) dt e e dt ! = (1 + i) 1 2 i(1+i)t 1+e R 1 e 2 i( 2 +(1+i)t) Z 1 2 i(1+i)t 2 t2 e e dt ! (1 + i) dt: 2 i(1+i)t 1+e 1
f (z) dz = 3
Z
2 2 e (t 2t)i e 2 (R Rt) dt 1 e (2t 1 2R)i e2 R
0 t 1 j1
2 e 2 (R = sup e2 0 t 1 j1
=
Now integrate along and let R ! 1.
e 1
4
dzdz = iz
parameterized by z = t
13
1 2
(1 + i) R with 0
t
1,
Z
f (z) dz = 4
= Z
=
1
0
Z
e i(t 1 e 2
sup
1 i(t
e (2R
dz = iz
2Rt+t2 2t+1)i
1
0 t 1
i(z 1=2)2
z = t 12 (1 + i) R = dz = dt e 2 Z 1 (2R 2Rt+t2 2t+1)i 2 (R2 +R (1+i)R)2 e e dz = 1=2 (1+i)R) (2t 1 2R)i 1 e e 2 R 0
e 1 4
e
e
2
(R2 +R
Rt)
1
(2t 1 2R)i e 2 R
e je
sup
0 t 1 j1
sup 0 t 1
2
(R2 +R
dz = Rt)
(2t 1 2R)i e 2 R jj
2
(R2 +R
j1
e
e
Rt)
Rt)
2 Rj
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that (1 + i)
Z
1 1
e 1+e
2 t2 2 i(1+i)t
dt + 0 + (1 + i)
Z
1 1
e
2 i(1+i)t
1+e
e
2 t2
dt
2 i(1+i)t
dt + 0 = = 2 i
and therefore (1 + i) i.e.,
Z
1
e
2 t2
1
(1 + i)
1 + e 2 i(1+i)t 1+i dt = p ; 2 i(1+i)t 1+e 2
Z
1
e
2 t2
1
(c) Comparing real parts thus yields, Z 1 e
1+i dt = p : 2
1 dt = p ; 2 1 and changing variables as follows gives the desired result 2 t2
14
e i=4 ; 2 i
1 p = 2 i.e.,
Z
1
e
2 t2
dt =
1
Z
p s = p2 t ds = 2 dt
1
e
s2
dt =
1
15
p
:
1 =p 2
Z
1 1
e
s2
ds;
VII.2.1 1 2 3 P L K Show using residue theory that Z 1 dx = ; 2 2 a 1 x +a
a > 0:
Remark. Check the result by evaluating the integral directly, using the arctangent function. Solution
VII.2.1 γ2 z1
γ1
-R
R z2
Set I= and integrate f (z) =
z2
Z
1 1
dx ; x 2 + a2
1 = + a2 (z
1 ia) (z + ia)
along the contour in Figure VII.2.1. Residue at a simple pole at z1 = ia, where by Rule 2, Res
z2
1 1 ; ia = 2 +a 2z
16
= z=ia
i : 2a
Integrate along 1 , and let R ! 1. This gives Z 1 Z Z R dx dx ! = I: f (z) dz = 2 2+a 2 + a2 x x 1 R 1 Integrate along
2,
and let R ! 1. This gives Z 1 f (z) dz R 2 R a2 2
R
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that I +0=2 i
i 2a
;
and therefore Z
1 1
x2
dx = ; + a2 a
a > 0:
Remark. Check of the result by evaluating the integral directly gives, because the integrand is positive the integral can be computed taking a single limit, namely, Z
1 1
dx = lim 2 x + a2 b!1
Z
b b
dx 1 x = lim arctan 2 2 b!1 a x +a a
17
b
= b
a
;
a > 0:
VII.2.2 1 2 3 P L K Show using residue theory that Z 1 dx = 3: 2 2 2 2a 1 (x + a ) Remark. Check the result by di¤erentiating the formula in the preceding exercise with respect to the parameter. Solution
VII.2.2 γ2 z1
γ1
-R
R z2
Set I=
Z
1 1
(x2
dx ; + a2 )2
and integrate f (z) =
(z 2
1 = + a2 )2 (z
1 ia) (z + ia)2 2
along the contour in Figure VII.2.2. Residue at a double pole at z1 = ia, where by Rule 2, Res
1 d 1 2 ; ia = lim = 2 2 z!ia dz (z + ia) (z 2 + a2 ) (z + ia)3 18
= z=ia
i : 4a3
Integrate along Z
1,
and let R ! 1. This gives Z 1 Z R dx dx ! = I: f (z) dz = 2 2 + a2 ) 2 +a2 )2 (x (x 1 R 1
Integrate along
2,
Z
and let R ! 1. This gives 1
f (z) dz
a2 )2
(R2
2
R
R3
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that that i 4a3
I +0=2 i
;
and therefore Z
1 1
(x2
dx = : + a2 )2 2a3
Remark. From Exercise VII.2.1 we have the formula Z 1 1 dx = ; (1) 2 2 a 1 x +a
a > 0:
It is allowed to di¤erentiate both sides in the formula, because for every compact interval with a 6= 0, both the integrand and its derivative are continuous, and the primitive of the derivative is uniformly bounded by a a R independent function h (x) such that h (x) < 1 on every such like interval. We start by di¤erentiate the left hand side in the formula with respect to the parameter a
(2)
d da
Z
1 1
1 dx = 2 x + a2
Z
1 1
2a dx = 2 (x + a2 )2
2a
Z
1 1
(x2
dx ; + a2 )2
and di¤erentiate the right hand side in the formula in the same way (3)
d = da a
By (1) - (3), we have that 19
a2
:
Z
1 1
dx = : (x2 + a2 )2 2a3
20
VII.2.3 Show using residue theory that Z 1 x2 dx : 2 = 2 2 1 (x + 1) Remark. Check the result by combining the preceding two exercises. Solution
VII.2.3 γ2
z1
-R
γ1
R
z2
Set I=
Z
1 1
x2 dx ; (x2 + 1)2
and integrate f (z) =
z2 = (z 2 + 1)2 (z
z2 i)2 (z + i)2
along the contour in Figure VII.2.3. Residue at a double pole at z1 = i, where by Rule 2,
Res
z2 d z2 2z (z + i)2 2z 2 (z + i) ; i = lim = z!i dz (z 2 + i)2 (z 2 + 1)2 (z + i)4
Integrate along
1,
and let R ! 1. This gives 21
= z=i
i : 4
Z
f (z) dz = 1
Integrate along
2,
Z
Z
R R
x2 dx ! (x2 + 1)2
Z
1 1
x2 dx = I: (x2 + 1)2
and let R ! 1. This gives R2
f (z) dz
R
1)2
(R2
2
R
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that I +0=2 i
i 4
;
and therefore Z
1 1
x2 dx = : (x2 + 1)2 2
Remark. We split the integrand using partial fraction, and put a = 1 in the formulas from the preceding two exercises Z 1 Z 1 Z 1 x2 dx dx dx = = : 2 2 = 2 2 2 2 2 1 (x + 1) 1 x +1 1 (x + 1)
22
VII.2.4 Using residue theory, show that Z 1 dx =p : 4 2 1 x +1 Solution
VII.2.4 γ2
γ1
-R
z2
z1
z3
z4
R
Set I= and integrate
f (z) =
z4
1 = +1 z
Z
1 1
dx ; +1
x4
1 1+i p 2
1+i p 2
z
z
1 i p 2
along the contour in Figure VII.2.4. p , where by Rule 2, Residue at a simple pole at z1 = 1+i 2 p 1 1+i 1 2 Res 4 ; p = 3 = ( 1 z +1 4z z=(1+i)=p2 8 2 Residue at a simple pole at z2 =
1+i p , 2
where by Rule 2,
23
z
i)
1p i 2
1 ; Res 4 z +1 Integrate along
Integrate along
1 1+i p = 3 4z 2
p
=
z=( 1+i)= 2
p
2 (1 8
i)
1,
and let R ! 1. This gives Z Z R Z 1 dx dx f (z) dz = ! = I: 4 4 R x +1 1 x +1 1 2,
and let R ! 1. This gives Z 1 R f (z) dz 4 R 1 2
R3
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that ! p p 2 2 I +0=2 i ( 1 i) + (1 i) ; 8 8 and therefore Z
1 1
dx =p : +1 2
x4
24
VII.2.5 Using residue theory, show that Z 1 x2 dx = p : x4 + 1 2 2 0 Solution
VII.2.5 γ2
γ1
-R
z2
z1
z3
z4
R
Set I=
Z
0
and integrate z2 f (z) = 4 = z +1 z
1
x2 dx ) 2I = x4 + 1
Z
1 1
x2 dx x4 + 1
z2 1+i p 2
1+i p 2
z
z
1 i p 2
along the contour in Figure VII.2.5. p , where by Rule 3, Residue at a simple pole at z1 = 1+i 2 p z2 1+i 1 2 Res 4 ; p = (1 = p z +1 4z z=(1+i)= 2 8 2 Residue at a simple pole at z2 =
1+i p , 2
where by Rule 3,
25
z
i) :
1p i 2
z2 ; Res 4 z +1 Integrate along Z
1,
Integrate along
2,
1 1+i p = 4z 2
p
=
z=( 1+i)= 2
p
2 ( 1 8
i) :
and let R ! 1. This gives Z R Z 1 x2 x2 f (z) dz = dx ! dx = 2I: 4 4 R x +1 1 x +1 1 and let R ! 1. This gives Z R2 R f (z) dz R4 1 2
R
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that ! p p 2 2 2I + 0 = 2 i (1 i) + ( 1 i) ; 8 8 and therefore Z
0
1
x2 dx = p : x4 + 1 2 2
26
VII.2.6 Show that Z
1 1
(x2
x dx = + 2x + 2) (x2 + 4)
10
:
Solution
VII.2.6 γ2 z1 z2
γ1
-R
R
z3 z4
Set I= and integrate f (z) =
(z 2
Z
1 1
(x2
x dx + 2x + 2) (x2 + 4)
z = + 2z + 2) (z 2 + 4) (z
( 1 + i)) (z
z 2i) (z
( 1
i)) (z + 2i)
along the contour in Figure VII.2.6. Residue at a simple pole at z1 = 2i, where by Rule 3,
Res
(z 2
z z ; 2i = 2 2 + 2z + 2) (z + 4) (z + 2z + 2) 2z
Residue at a simple pole at z2 =
= z=2i
1 + i, where by Rule 3,
27
1 20
1 i: 10
z z ; 1 + i = (z 2 + 2z + 2) (z 2 + 4) (2z + 2) (z 2 + 4)
Res
Integrate along Z
f (z) dz = 1
1,
Z
= z= 1+i
1 3 + i: 20 20
and let R ! 1. This gives
R R
(x2
x dx ! + 2x + 2) (x2 + 4) Z 1 x ! dx = I: 2 2 1 (x + 2x + 2) (x + 4)
Integrate along 2 , and let R ! 1. This gives Z R f (z) dz 2 (R 2R 2) (R2 4) 2
R
R2
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that I +0=2 i
1 20
1 1 3 i+ + i 10 20 20
and therefore Z
1 1
(x2
x dx = + 2x + 2) (x2 + 4)
28
10
:
VII.2.7 1 2 3 P Show that Z 1
1
L K LLL
cos (ax) dx = p e x4 + 1 2
p a= 2
a a cos p + sin p 2 2
;
a > 0:
Solution
VII.2.7 γ2
γ1
-R
z2
z1
z3
z4
R
Set I= and integrate
f (z) =
eiaz = z4 + 1 z
Z
1 1
cos (ax) dx x4 + 1
a > 0;
eiaz 1+i p 2
1+i p 2
z
z
along the contour in Figure VII.2.7. p , where by Rule 3, Residue at a simple pole at z1 = 1+i 2
29
1 i p 2
z
1p i 2
eiaz eiaz 1 + i p = ; = z4 + 1 4z 3 z=(1+i)=p2 2 p a=p2 p a=p2 p p p 2e 2e 2e cos a= 2 sin a= 2 i 8 8
Res =
Residue at a simple pole at z2 = 1+i eiaz p = 4z 3 2 p p p a= 2 p 2e cos a= 2 2e = 8 Res
eiaz ; z4 + 1
Integrate along Z
f (z) dz = 1
Z
1,
R R
1+i p , 2
p a= 2
p p 2e cos a= 2 i+ 8
p z=( 1+i)= 2 p a= 2
=
p p sin a= 2 2e i 8
p a= 2
p p cos a= 2 2e i 8
p a= 2
I+i
1
p sin a= 2 8
and let R ! 1. This gives
eax dx ! x4 + 1
Z
1 1
cos (ax) dx+i x4 + 1
Z
1 1
sin (ax) dx = I+i x4 + 1
Z
1 1
sin (ax) dx x4 + 1
Using the Residue Theorem and letting R ! 1, we obtain that 1
p sin a= 2 8
where by Rule 3,
Integrate along 2 , and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 1 f (z) dz R ! 0: 4 R 1 R3 2
Z
p a= 2
sin (ax) dx = 2 i x4 + 1
p
and therefore Z 1 cos (ax) dx = p e 4 2 1 x +1
2e
p a= 2
p a= 2
p cos a= 2 i 4
a a cos p + sin p 2 2
30
p
2e
;
p a= 2
p ! sin a= 2 i ; 4
a > 0:
VII.2.8 Show that Z
1 1
cos x dx = : e (x2 + 1)2
Solution
VII.2.8 γ2
z1
γ1
-R
R
z2
Set I=
Z
1 1
cos x dx (x2 + 1)2
and integrate f (z) =
eiz = (z 2 + 1)2 (z
eiz i)2 (z + i)2
along the contour in Figure VII.2.8. Residue at a double pole at z1 = i, where by Rule 2,
Res
eiz d eiz ieiz (z + i)2 2eiz (z + i) ; i = lim == z!i dz (z + i)2 (z 2 + 1)2 (z + i)4
Integrate along
1,
and let R ! 1. This gives 31
= z=i
1
e 2
i
Z
f (z) dz = 1
=
Z
1 1
eix dx ! (x2 + 1)2
Integrate along plane, this gives
2,
Z
Z
1 1
cos x dx + i (x2 + 1)2
Z
1
sin x dx = + 1)2 1 Z 1 sin x =I +i 2 dx: 2 1 (x + 1)
and let R ! 1. Because jeiz j 1
f (z) dz
1)2
(R2
2
R
R3
(x2
1 in the upper half
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that Z 1 sin x e 1 i ; I +i dx + 0 = 2 i 2 2 2 1 (x + 1) and therefore Z
1 1
cos x dx = : e (x2 + 1)2
32
VII.2.9 Show that Z
1 1
sin2 x dx = 1 x2 + 1 2
1 : e2
Solution
VII.2.9 γ2
z1
γ1
-R
R
z2
Set I= and integrate
Z
1 1
f (z) =
sin2 x dx = x2 + 1
Z
1 1
1 cos 2x dx 2 (x2 + 1)
1 ei2z 1 ei2z = 2 (z 2 + 1) 2 (z i) (z + i)
along the contour in Figure VII.2.9. Residue at a simple pole at z1 = i, where by Rule 3, 1 ei2z 1 ei2z Res ;i = 2 (z 2 + 1) 4z Integrate along
z=i
1 , and let R ! 1. This gives
33
1 = ie 4
2
1 i 4
Z
f (z) dz = 1
=
Z
R R
1 ei2x dx ! 2 (x2 + 1)
Integrate along plane, this gives
2,
Z
Z
1 1
1 cos 2x dx + i 2 (x2 + 1)
Z
1
sin 2x dx = 2 1 2 (x + 1) Z 1 sin 2x dx =I +i 2 1 2 (x + 1)
and let R ! 1. Because jeiz j f (z) dz 2
2 2(R2
1)
R
R
Using the Residue Theorem and letting R ! 1, Z 1 sin 2x dx + 0 = 2 i I +i 2 1 2 (x + 1) 1 1
sin2 x dx = 1 x2 + 1 2
34
! 0:
we obtain that 1 ie 4
and therefore
Z
1 in the upper half
1 : e2
2
1 i ; 4
VII.2.10 Show that Z
1 1
cos (ax) e jajb ; dx = x 2 + b2 b
1 < a < 1; b > 0:
For which complex values of the parameters a and b does the integral exist? Where does the integral depend analytically on the parameters? Solution The integral exists only for a real, and Re b 6= 0 or cos (iab) = 0. It depends analytically on b for Re b 6= 0. It depends analytically on a in the intervals 1 < a < 0 and 0 < a < 1. Case 1: a < 0
VII.2.10a z1
γ1
-R
R z2
γ2
Set I= and integrate
Z
1 1
f (z) =
cos ax dx; x 2 + b2
a < 0; b > 0
eiaz = z 2 + b2 (z
eiaz ib) (z + ib)
along the contour in Figure 7.2.10a. (The lower half-plane) Residue at a simple pole at z2 = ib, where by Rule 3, 35
Res Integrate along Z
1
eiaz eiaz ; ib = z 2 + b2 2z
= z= ib
eab i 2b
and let R ! 1. This gives
f (z) dz = 1
=
Z
R R
eiax dx ! x 2 + b2
Z
1 1
cos (ax) dx + i x 2 + b2
Z
1
sin (ax) dx = 2 2 1 x +b Z 1 sin (ax) dx =I +i 2 2 1 x +b
Integrate along 2 and let R ! 1. Because jeiaz j 1 in the lower half plane if a < 0, this gives Z 1 R ! 0: f (z) dz 2 R R b2 2 Using the Residue Theorem and letting R ! 1, we obtain that I +0=
2 i
eab i ; 2b
and hence (1)
I=
eab ; a < 0: b
Case 2: a > 0
36
VII.2.10b γ4
z1
γ3
-R
R
z2
Set I= and integrate
Z
1 1
f (z) =
cos ax dx; x2 + b2
eiaz = z 2 + b2 (z
a > 0; b > 0;
eiaz ib) (z + ib)
along the contour in Figure VII.2.10b. (The upper half-plane) Residue at a simple pole at z1 = ib, where by Rule 3, Res Integrate along Z
3
eiaz eiaz ; ib = z 2 + b2 2z
e ab i 2b
= z=ib
and let R ! 1. This gives
f (z) dz = 3
=
Z
R R
eiax dx ! x 2 + b2
Z
1 1
cos (ax) dx + i x 2 + b2
37
Z
1
sin (ax) dx = 2 2 1 x +b Z 1 sin (ax) =I +i dx 2 2 1 x +b
Integrate along 4 and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 1 f (z) dz ! 0: R 2 R R b2 4 Using the Residue Theorem and letting R ! 1, we obtain that e ab i ; 2b
I +0=2 i and hence (2)
e ab ; a > 0: b
I=
By (1) and (2) we can concluede that Z 1 e jajb cos (ax) dx = ; 2 2 b 1 x +b
38
1 < a < 1; b > 0:
VII.2.11 1 2 3 P L K Evaluate Z
1
cos x dx: + a2 ) (x2 + b2 ) 1 Indicate the range of the parameters a and b. (x2
Solution The integral converges absolutely if a2 ; b2 62 ( 1; 0], a; b 62 iR. It also converges absolutely for certain points on the imaginary axis, so that ia are zeros of cos x, ib are zeros of cos x, but they are not the same zeros. The integrand, is an even function of a and b, so it su¢ ces to evaluate it for a; b in the right half-plane. For this, we can assume that a; b > 0, and extend the formula by analyticity. We can also assume that a 6= b, and get other case in this limit. So we may as well assume that 0 < a < b.
VII.2.11 γ2 z2 z1
-R
γ1
z3
R
z4
R1 x 2 2 = ( 1; 0] i.e. The integral 1 (x2 +acos 2 )(x2 +b2 ) dx converges absolutely if a ; b 2 a; b 2 = iR. It also converges absolutely for certain points on imaginary axis, so that ia are zeros of cos x, ib are zeros of cos x, but they are not same zeros. (Problem should may be simpli…ed to avoid thus points.) The integrand, is an even function of a and b, so it su¢ ces to evaluate it for a; b in the right half-plane. For this, we can assume that a; b > 0, and extend the formula by analyticity. We can assume that a 6= b, and get other case in this limit. So we may as well assume 0 < a < b. 39
Case 1: (a; b > 0; a 6= b) Set I= and integrate
Z
1 1
(x2
cos x dx + a2 ) (x2 + b2 )
eiz = (z 2 + a2 ) (z 2 + b2 ) (z
f (z) =
ia) (z
eiz ib) (z + ia) (z + ib)
along the contour in Figure VII.2.11. Residue at a simple pole at z1 = ai, where by Rule 3, Res
eiz eiz ; ia = (z 2 + a2 ) (z 2 + b2 ) 2z (z 2 + b2 )
= z=ai
e
a
2a (b2
a2 ) i
Residue at a simple pole at z1 = ib, where by Rule 3, eiz eiz Res ; ib = (z 2 + a2 ) (z 2 + b2 ) 2z (z 2 + a2 ) Integrate along Z
1
f (z) dz = 1 Z 1 ! 1
Integrate along this gives Z
Z
e
=
2b (b2
z=bi
b
a2 ) i
and let R ! 1. This gives R
eix dx ! 2 2 2 2 R (x + a ) (x + b ) Z 1 cos x sin x dx + i dx = 2 2 2 2 2 2 2 2 (x + a ) (x + b ) 1 (x + a ) (x + b ) Z 1 sin x =I +i dx 2 2 2 2 1 (x + a ) (x + b ) 2
and let R ! 1. Because jeiz j
f (z) dz 2
(R2
1 a2 ) (R2
b2 )
1 in the upper half plane,
R
R3
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that I +0=2 i
e 2a (b2
a
e a2 ) i 2b (b2
40
b
a2 ) i
;
and therefore Z
1 1
be a cos x dx = (x2 + a2 ) (x2 + b2 ) ab (b2
ae b : a2 )
Case 2: (a; b > 0; a = b) From Case 1 we have Z
1 1
be a cos x dx = (x2 + a2 ) (x2 + b2 ) ab (b2
If we let b ! a, we get Z
1 1
ae b (b = 2 a) ab (b + a)
cos x dx = 3 (1 + a) : 2a (x2 + a2 )2
For a = 1, we have Z
1 1
cos x dx = ; e (x2 + 1)2
which thus yields the answer from Exercise 8.
41
a) e
a
+a e (b a)
a
e
b
:
VII.2.12 Let Q (z) be a polynomial of degree m with no zeros line, and let f (z) be a function that is analytic in the plane and across the real line. Suppose there is b < m jf (z)j jzjb for z in the upper half-plane, jzj > 1. Show Z 1 X f (z) f (x) dx = 2 i Res ; zj ; Q (z) 1 Q (x)
on the real upper half1 such that that
summed over the zeros zj of Q (z) in the upper half-plane. Solution
VII.2.12 γ2 zj - 1
z3
zj z2
-R
γ1
z1
R
Set I= and integrate
Z
1 1
f (x) dx Q (x)
f (z) =
f (z) Q (z)
along the contour in Figure VII.2.12. The sum of the residues at the poles z1 ; z2 ; : : : ; zj in the upper half plane, X
Res
f (z) ; zj : Q (z) 42
Integrate along Z Integrate along Z
1,
and let R ! 1. This gives Z 1 Z R f (x) f (x) dx ! dx = I: f (z) dz = Q (x) Q (x) 1 R 1 2,
and let R ! 1. This gives z = Rei 0
f (z) dz = 2
Z
f Rei d jQ (Rei )j
0
Rb Rm (C + O (1=R))
f Rei jQ (Rei )j
R (C + O (1=R))
Using the Residue Theorem and letting R ! 1, we obtain that I +0=2 i and therefore
Q.E.D.
Z
1 1
X
Res
f (z) ; zj Q (z)
X f (z) f (x) dx = 2 i Res ; zj Q (x) Q (z)
43
R
CR
! 0:
VII.3.1 Show using residue theory that Z 2 cos d =2 2 + cos 0
1
2 p 3
:
Solution
VII.3.1
|z| = 1 z3
z2 z1
Z
2
cos d : 2 + cos 0 A change of variables as in book page 203-204 gives I=
I=
Z
2
0
cos d = 2 + cos
I
jzj=1
1 2
z + z1 2 + 12 z +
1 z
dz 1 = iz i
I
z2 + 1 dz: 3 2 jzj=1 z + 4z + z
Integrate f (z) =
z2 + 1 = z 3 + 4z 2 + z z z
2
z2 + 1 p 3 z
2+
along the unit circle contour in Figure VII.3.1. Residue at a simple pole at z1 = 0, where by Rule 3, Res
z2 + 1 z2 + 1 ; 0 = z 3 + 4z 2 + z 3z 2 + 8z + 1 44
= 1: z=0
p
3
Residue at a simple pole at z2 =
2+
p
3, where by Rule 3,
p z2 + 1 z2 + 1 ; 2+ 3 = 2 Res 3 z + 4z 2 + z 3z + 8z + 1
p z= 2+ 3
Using the Residue Theorem, we thus obtain p ! 2 3 1 ; 3
1 I=2 i i and therefore Z
0
2
cos d =2 2 + cos
45
1
2 p 3
:
=
p 2 3 : 3
VII.3.2 Show using residue theory that Z 2 d 2 =p 2 a + b sin a 0
b2
;
a > b > 0:
Solution
VII.3.2 (a = 7, b = 6)
|z | = 1
z1 z2
Set I=
Z
0
2
1 d : a + b sin
A change of variables as in book page 203-204 gives
I=
Z
2
0
1 d = a + b sin
I
dz = iz
1
jzj=1
a+b
1 2i
z
a
p
1 z
I
jzj=1
2 bz 2 + 2azi
Integrate
f (z) =
bz 2
2 + 2azi
b
2
= b z
a2
b
b2
i
z
p a+ a2 b2 i b
along the unit circle contour in Figure p VII.3.2. a+ a2 b2 Residue at a simple pole at z1 = i, where by Rule 3, b 46
b
dz
2 Res 2 bz + 2azi
b
;
a+
p
a2 b
b2
i
=
2 2bz + 2ai
z=
p
a+ a2 b
Using the Residue Theorem, we thus obtain I=2 i
p
1 a2
b2
i ;
and therefore Z
0
2
1 2 d =p a + b sin a2
47
b2
;
a > b > 0:
b2
= i
p
i a2
b2
:
VII.3.3 Show using residue theory that Z h sin2 d = a 0 a + cos
p
a2
i 1 ;
a > 1:
Solution
VII.3.3 (a = 2)
|z| = 1 z3
z2 z1
Set
I=
Z
0
Z sin2 (2 t) =2 t dt = = d = dt t) 2 a + cos(2 Z 2 Z 2 sin2 t sin2 (2 t) dt = dt: = a + cos(2 t) a + cos t
sin2 d = a + cos
We have 2I =
Z
2
sin2 d : a + cos
0
A change of variables as in book page 203-204 gives
2I =
Z
0
2
sin2 d = a + cos
I
jzj=1
1 2i
z
a+
1 2
2
1 z
z+ 48
1 z
dz = iz
1 2i
I
jzj=1
2
(z 2 1) dz: z 2 (z 2 + 2az + 1)
Integrate 2
(z 2 p a + a2 1
(z 2 1) = f (z) = 2 2 z (z + 2az + 1) z2 z
2
1) z
a
p
a2
1
along the unit circle contour in Figure VII.3.3. Residue at a double pole at z1 = 0 , where by Rule 2, # " 2 2 d (z 2 1) (z 2 1) ; 0 = lim = 2a: Res 2 2 z!0 dz (z 2 + 2az + 1) z (z + 2az + 1) p Residue at a simple pole at z2 = a + a2 1, where by Rule 3, "
2
2
p (z 2 1) =z Res 2 ; a + a2 z + 2az + 1
#
2
(z 2 1) =z 1 = 2z + 2a
2
p = 2 a2 p z= a+ a2 1
Using the Residue Theorem, we thus obtain 2I = 2 i
p 2a + 2 a2
1 2i
1 ;
and therefore Z
0
sin2 d = a + cos
a
49
p
a2
1 ;
a > 1:
1:
VII.3.4 Show using residue theory that Z d 1 + sin2
=
p
2:
Solution
VII.3.4
|z | = 1 z4
z3
z1
z2
Set
I=
Z
1 d = 1 + sin2
= d =
t = dt Z 0
=
1 2 1+ sin (
2
t)
Z
dt =
2
0
1 dt: 1+ sin2 t
We have Z
2
1 d : 1 + sin2 0 A change of variables as in book page 203-204 gives I=
I=
Z
0
2
1 d = 1 + sin2
I
jzj=1
1 1+
1 2i
z 50
1 z
2
dz = iz
4 i
I
jzj=1
z4
z dz: 6z 2 + 1
Integrate
f (z) = =
z4
z
z = 6z 2 + 1 p 2 1 z
p
z z
2+1
p
2+1
p
z
2
1
along the unit circle contour in Figure VII.3.4 p 2 1 , where by Rule 3, Residues at a simple poles at z1;2 = Res
z4
p
z ; 6z 2 + 1
2
1
=
z 4z
3
12z
z=
Using the Residue Theorem, we thus obtain I=2 i
4 i
1 1 p p 8 2 8 2
and therefore Z
p 1 d = 2: 1 + sin2
51
;
p
(
= 2 1)
1 p : 8 2
VII.3.5 Show using residue theory that Z d 1 r2 = 1; 2 1 2r cos + r 2
0
r < 1:
Solution
VII.3.5 (r = 1/2)
|z | = 1 z2
z1
Set
I=
Z
1
1 r2 d = t = = 2 d = dt 2r cos + r 2 Z 0 1 r2 dt = = 2 2r cos ( t) + r 2 2 1 Z 2 dt 1 r2 = 2 1 + 2r cos t + r 2 0
A change of variables as in book page 203-204 gives
52
I=
Z
2
0
1 r2 dt = 1 + 2r cos t + r2 2 I 1 1 r2 = 2 jzj=1 1 + 2r 21 z + z1
r2
dz = iz
+ I 1 1 r2 = dz: 2 i jzj=1 (r + z) (rz + 1)
Integrate f (z) =
1 r2 1 r2 = (r + z) (rz + 1) r (z + r) (z + 1=r)
along the unit circle contour in Figure VII.3.4 Residue at a simple pole at z1 = r, where by Rule 3, 1 r2 1 r2 Res ; r = r (z + r) (z + 1=r) 2rz + r2 + 1
= 1: z= r
Using the Residue Theorem, we thus obtain I=2 i
1 (1) ; 2 i
and therefore Z
1
1 r2 d = 1; 2 2r cos + r 2
53
0
r < 1:
VII.3.6 By expanding both sides of the identity Z 2 1 1 d =p 2 2 0 w + cos w
in a power series at 1, show that Z 2 1 (2k)! cos2k d = ; 2 0 22k (k!)2
1
k
0:
Solution Rewrite both sides in the given identity 1 I= 2
(1)
Z
0
2
d = 1 + cos w
1 w2
1
1=2
;
where the integrand in the left side in the identity can be rewritten as k X 1 k cos ( 1) = : wk 1 + cos w k=0 1
Now, we work on the left side of the identity
Z 2 X Z k 1 1 1 X1 ( 1)k 2 k cos I= ( 1) d = cosk d : k=0 k=0 w k 2 0 wk 2 0 R2 Because 0 cosk d = 0 if k odd, we have 1 X 1 I =1+ 2 k=1
(2)
Z
2
cos2k d
0
1 : w2k
Now, we use the power series expansion (MH p. 192.) for real x < 1,
(1 + x) = 1+ x+
(
1) 2!
x2 +
(
1) ( 3!
we have 54
2)
x3 +
+
n
and
xn +
1<
;
1
=1+ +
1 2
1=2
1 w2
= 1 2
1
1!
w2
3 2
5 2
1
+
7 2
3 2
1 2
1 w2
2! 1
4
+
+
1 2
2
+ 3 2
1 2
3 2
5 2
1 w2
3! 5 2
7 2
1 2k 2
3
+ 1
k
w2 k! w2k X1 1 3 (2k 1) 1 =1+ = k=1 2k k! w2k X1 1 2 3 4 5 : : : (2k 1) 2k 1 = =1+ k=1 2k k! 2 4 6 : : : 2k w2k X1 1 2 3 4 5 : : : (2k 1) 2k 1 = 1+ k=1 2k k! 2k (1 2 3 : : : k) w2k X1 X1 (2k)! 1 (2k)! 1 =1+ = 1 + ; k k 2k k=1 2 k!2 k! w k=1 22k (k!)2 w 2k
4!
i.e., (2)
1
1 w2
1=2
=1+
X1
k=1
(2k)! 1 : 22k (k!)2 w2k
Comparing the two power series expressions (2) and (3) of the both sides in the identity (1) gives us together with the uniqueness theorem for the power series Z 2 1 (2k)! cos2k d = ; k 0: 2 0 22k (k!)2
55
+
=
VII.3.7 Show using residue theory that Z 2 d 2 w ; 2 = (w + cos ) (w2 1)3=2 0
w 2 C n [ 1; 1] :
Specify carefully the branch of the power function. Check you answer by di¤erentiating the integral of 1= (w + cos ) with respect to the parameter w. Solution
VII.3.7 (w = 2)
|z| = 1 z2
z1
Set I=
Z
2
0
d (w + cos )2
A change of variables as in book page 203-204 gives
I=
Z
0
2
d = (w + cos )2
Z
2
0
4z 2 (z 2 +2wz+1) p 2 branch of w
Integrate f (z) =
1 w+
1 2
z+
1 z
dz 4 = 2 iz i
I
z 2 dz: 2 jzj=1 (z + 2wz + 1)
along the unit circle contour in Figure VII.3.7.
We choose 1 on Cn [ 1; 1] that is positive for w 2 (1; 1). It su¢ ces to check the identity for w > p 1. Residue at a double pole at z1 = w + w2 1, where by Rule 2, 56
p z ; w + w2 1 = 2 (z 2 + 2wz + 1) z d = lim p p 2 = z! w+ w2 1 dz z w w2 1 p p 2 z w w2 1 2z z w w2 1 = = p 4 p w2 1 z w 2 z= w+ w 1 p 2 z w w 1 2z = = p 3 p z w w2 1 z= w+ w2 1 1 w = : 2 4 (w 1)3=2
Res
Using the Residue Theorem, we thus obtain 4 I=2 i i and therefore Z
0
2
1 w 4 (w2 1)3=2
2 w d ; 2 = (w + cos ) (w2 1)3=2
!
;
w 2 C n [ 1; 1] :
Check. From Exercise VII.3.6 we have the formula Z 2 1 d 1 (1) =p 2 2 0 w + cos w 1 It is allowed to di¤erentiate both sides in the formula, because for every compact interval with w 2 C n [ 1; 1], both the integrand and its derivative are continuous, and that the primitive of the derivative is uniformly bounded R by an w independent funtion h ( ) such that h ( ) d < 1 on every such like interval on w. We start by di¤erentiating the left hand side in the formula with respect to the parameter w
57
(2)
d dw
1 2
Z
2
0
d w + cos
=
1 2
Z
2
0
d ; (w + cos )2
and di¤erentiate the right hand side in the formula in the same way d 1 p 2 dw w
(3)
1
By (1) - (3), we thus have that Z 2 0
=
w (w2
1)3=2
d 2 w : 2 = 2 (w + cos ) (w 1)3=2
58
:
VII.4.1 By integrating around the keyhole contour, show that Z 1 x a dx = ; 0 < a < 1: 1+x sin ( a) 0
Solution
VII.4.1 γ2
γ1
z1
γ3
γ4
-R
R
Set I=
Z
0
and integrate
1
x a dx 1+x
z a jzj a e ia arg z f (z) = = 1+z 1+z along the keyhole contour in Figure VII.4.1. We make a branchcut for z along the positive real axis, where 0 < arg z < 2 . Residue at a simple pole at z1 = 1, where by Rule 3, Res
jzj
a
e ia arg z ; 1 = jzj 1+z 59
a
e
ia arg z z= 1
=e
ia
:
a
Integrate along Z
f (z) dz = 1
Z
a
e ia arg z dz = 1+z
jzj 1
Integrate along
2
Integrate along
3,
Z
and let R ! 1 and " ! 0+ . This gives
1,
z = xe0i dz = dx
=
Z
R
x a dx ! " 1+x Z 1 x a dx = I: ! 1+x 0
and let R ! 1. This gives then 0 < a < 1 Z R a 2 f (z) dz 2 R ! 0: R 1 Ra 2 and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
=
Z
jzj 3
Integrate along
a
e ia arg z z = xe2 i dz = dz = dx 1+z Z 0 a 2 ia x e ! dx = e 2 1 + x 1 4,
Z
= ia
Z
Z
"
R 1
0
x a e 2 ia dx ! 1+x
x a dx = 1+x
e
2 ia
I:
and let " ! 0+ . This gives then 0 < a < 1 " a 2 " 1 "
f (z) dz 4
2 "1
a
! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0 Multiplying with e
ia
e
2 ia
I +0=2 i
e
, this yields e
ia
e
ia
and hence solving for I,
60
I = 2 i;
ia
:
I=
e
ia
2 i e
ia
=
sin ( a)
;
i.e., Z
0
1
x a dx = ; 1+x sin ( a)
61
0 < a < 1:
VII.4.2 By integrating around the boundary of a pie-slice domain of aperture 2 =b, show that Z 1 dx ; b > 1: = 1 + xb b sin ( =b) 0 Remark. Check the result by changing variable and comparing with Exercise 1. Solution
VII.4.2 (b = 12)
z1
ε
γ3
γ2
γ1
R
Set I=
Z
0
and integrate f (z) =
1
dx ; 1 + xb
1 1 = b 1+z 1 + jzjb eib arg z
along the pie-slice of aperture 2 =b contour in Figure VII.4.2. We make a branch cut for z b along the negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = e i=b , where by Rule 3, Res
1 ;e 1 + zb
i=b
=
1 bz b 62
1
= z=e
i=b
1 e b
i=b
:
Integrate along Z
1,
f (z) dz = 1
=
Z
1 1
jzjb eib arg z
1+
Integrate along
Integrate along
z = xe0i dz = dx
dz =
=
Z
R
1 dx ! b " 1+x Z 1 1 ! dx = I: 1 + xb 0
and let R ! 1. This gives then b > 1
2
Z
Z
and let R ! 1 and " ! 0+ . This gives
1
f (z) dz
Rb
2
3,
2 R 1 b
2 Rb
1
! 0:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
=
Z
1 3
1+
jzjb eib arg z !
Integrate along
Z
0
1
dz =
z = xe2 i=b dz = e2 i=b dx
1 e2 1 + xb
i=b
dx =
=
2 i=b
e
Z
Z
"
R
1
0
1 e2 1 + xb
1 dx = 1 + xb
and let " ! 0+ . This gives then b > 1 Z 1 2 " 2 " f (z) dz ! 0: b b b 1 " 4
i=b
dx !
e2
i=b
I:
4,
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0 Multiplying with e
i=b
e2
i=b
I +0=2 i
1 e b
, this yields e
i=b
e
i=b
and hence solving for I, 63
I=
2 i ; b
i=b
:
I=
(e
i=b
2 i e
i=b )
=
b sin ( =b)
i.e., Z
1
0
dx ; = 1 + xb b sin ( =b)
b > 1:
Remark. We begin with the …rst integral I, we make the substitution xb = t and use the integral from Exercise VII.4.1 where we choose a = 1 1=b. (Because b > 1, then 0 < 1 1=b < 1 and the restriction that 0 < a < 1 for the integral in Exercise VII.4.1 is satis…ed.)
I=
Z
0
1
2
3 xb = t dx 5= dx = 4 x = t1=b 1 + xb 1 1=b 1 dx = b t dt Z Z 1 1=b 1 1 1 t (1 1=b) 1t dt = dt = = b 1+t b 0 1+t 0 1 = = : b sin ( (1=b 1)) b sin ( =b)
64
VII.4.3 By integrating around the keyhole contour, show that Z 1 2 cos ( a) log x dx = ; 0 < a < 1: a x (x + 1) sin2 ( a) 0
Remark. Check the result by di¤erentiating the identity in Exercise 1. Solution
VII.4.3 γ2
γ1
z1
γ3
γ4
-R
R
Set I=
Z
0
and integrate f (z) =
xa
1
xa
log x ; (x + 1)
log x log jzj + i arg z = a ia arg z (x + 1) jzj e (z + 1)
along the keyhole contour in Figure VII.4.3, where the cimicircle 4 have radius ". We make a branch cut for log z along the positive real axis, so that 0 < arg z < 2 . Residue at a simple pole at z1 = 1, where by Rule 3, 65
Res
log jzj + i arg z log jzj + i arg z ; 1 = a ia arg z jzj e (z + 1) jzja eia arg z
Integrate along Z
1,
= ie
ia
:
z= 1
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
=
Z
1
log jzj + i arg z dz = jzja eia arg z (z + 1)
Integrate along
2
Integrate along
3,
z = xe0i dz = dx
Z
R log x = dx ! a " x (x + 1) Z 1 log x dx = I: ! a x (x + 1) 0
and let R ! 1. This gives then 0 < a < 1 q Z log2 R + (2 )2 2 log R f (z) dz 2 R ! 0: a (R R 1) R 2
Z
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
=
Z
3
Z " log x + 2 i log jzj + i arg z z = xe2 i dz = = ! a ia arg z a 2 ia (x + 1) dz = dx jzj e (z + 1) R x e Z 0 log x + 2 i ! dx = a 2 ia (x + 1) 1 x e Z 1 Z 1 log x dx 2 ia 2 ia = e dx 2 ie = a a x (x + 1) x (x + 1) 0 0 = e 2 ia I 2 ie 2 ia J and let " ! 0+ . This gives then 0 < a < 1 p Z log2 " + 2 f (z) dz 2 " 2 "1 a jlog "j ! 0: a " (1 ") 4
Integrate along
4,
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 66
I +0
e
Multiplying with e
ia
2 ia
I
2 ie
2 ia
J +0=2 i
ie
ia
:
, this yields
e
ia
e
ia
I
ia
2 ie
2 2;
J=
and thus 2i sin ( a) I
2 ie
ia
2 2:
J=
Separating real and imaginary parts, we get simultaneous equations 2 sin ( a) J = 2 2 cos ( a) J = 0;
2 sin ( a) I
2
and hence the solutions
I=
Z
0
1
2 log x cos ( a) dx = ; a x (x + 1) sin2 ( a)
Z
J=
0
1
xa
dx = ; (x + 1) sin ( a)
i.e. Z
1
0
2 cos ( a) log x dx = ; a x (x + 1) sin2 ( a)
0 < a < 1:
Remark. From the Exercise VII.4.1 we have the formula Z 1 x a dx = ; (1) 1+x sin ( a) 0
0 < a < 1:
It is allowed to di¤erentiate both sides in the formula, because for every compact interval with 0 < a < 1, both the integrand and its derivative is continuous. And that the primitive of the derivative is uniformly limited by a constant on every compact interval on a, there 0 < a < 1. We begin to di¤erentiate the left side in the formula with respect to the parameter a
(2)
d da
Z
0
1
x a dx = 1+x
Z
1 1
log x x 1+x 67
a
dx =
Z
1 1
xa
log x dx; (x + 1)
and di¤erentiate the right hand side in the formula in the same way (3) By (1) - (3), we have that Z 1 0
d = da sin ( a)
cos ( a) : sin2 ( a)
2 log x cos ( a) dx = : xa (x + 1) sin2 ( a)
68
VII.4.4 For …xed m 2, show that by integrating around the keyhole contour that Z 1 a (a + 1) (a + m 2) x a ; 1 m < a < 1: m dx = (1 + x) (m 1)! sin ( a) 0 Remark. The result can be obtained also by integrating the formula in Exercise 1 by parts. Solution
VII.4.4 γ2
γ1
z1
γ4
-R
γ3
R
Set I=
Z
0
and integrate
1
x a dx (1 + x)m
z a jzj a e ia arg z f (z) = = (1 + z)m (1 + z)m along the keyhole contour in Figure VII.4.4. We make a branchcut for z along the positive real axis, where 0 < arg z < 2 . 69
a
From Sa¤/Snider page 310 we have Res [f (z); z0 ] = lim
1
z!z0 (m
dm 1)! dz m
Residue at a pole of order m at z1 =
1 1
z0 )m f (z)] :
[(z
1
1 dm 1 z a z a = m ; 1 = lim z! 1 (m (1 + z) 1)! dz m 1 1 = lim ( a) ( a 1) ( a m + 2) z ( a (m 1)) = z! 1 (m 1)! 1 = lim ( 1)m 1 a (a + 1) (a + m 2) jzj( a (m 1)) ei( a (m 1)) arg z = z! 1 (m 1)! 1 ( 1)m 1 a (a + 1) (a + m 2) j 1j( a (m 1)) ei( a (m 1)) = = (m 1)! 1 = ( 1)m 1 a (a + 1) (a + m 2) ( 1)m 1 e ia = (m 1)! a (a + 1) (a + m 2) = e ia : (m 1)! Res
Integrate along Z
f (z) dz = 1
and let R ! 1 and " ! 0+ . This gives
1,
Z
1
jzj a e ia arg z dz = (1 + z)m
z = xe0i dz = dx !
Integrate along m 2 Z Integrate along
= Z
f (z) dz 2
3,
R a 2 R (R 1)m
0
2 Rm+a 1
x a dx ! (1 + x)m
x a dx = I: (1 + x)m
m < a < 1, where
! 0:
and let R ! 1 and " ! 0+ . This gives
70
R
"
1
and let R ! 1. This gives then 1
2
Z
Z
f (z) dz = 3
Z
3
Z " a 2 ia x e jzj a e ia arg z z = xe2 i dz = = m m dx ! dz = dx (1 + z) R (1 + x) Z 0 a 2 ia Z 1 x e x a 2 ia ! e e 2 m dx = m dx = (1 + x) 1 (1 + x) 0
Integrate along m 2
4,
Z
and let " ! 0+ . This gives then 1 f (z) dz
4
" (1
ia
I:
m < a < 1, where
a
")m
2 "1
2 "
a
! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0
e
2 ia
Multiplying with e e
ia
ia
I +0=2 i
a (a + 1) (a + m (m 1)!
2)
a (a + 1) (a + m (m 1)!
2)
e
ia
:
, this yields e
ia
I=2 i
;
and hence solving for I,
I=
(e
ia
2i e
a (a + 1) (a + m ia ) (m 1)!
2)
=
a (a + 1) (a + m (m 1)! sin ( a)
i.e. Z
0
1
x a a (a + 1) (a + m m dx = (1 + x) (m 1)! sin ( a)
2)
;
1
m < a < 1:
Remark. Set (this is our induction hypothesis) Z 1 x a a (a + 1) (a + m Im (a) = m dx = (1 + x) (m 1)! sin ( a) 0 71
2)
2)
;
From Exercise VII.4.1 we have Z 1 x b dx = ; 1+x sin ( b) 0
0 < b < 1:
Use Exercise VII.4.1 and integrate by part
sin ( b)
=
Z
0
Z 1 1 x b dx = x b dx = 1+x 1+x 0 Z 1 b+1 1 x x b+1 1 1 = dx = 1 b 1+x 0 1 b (1 + x)2 0 Z 1 1 x b+1 = dx: 1 b 0 (1 + x)2
1
Set b = a + 1 and solve for the integral in the left side in the inequality above Z 1 a x a = ; a 2 dx = sin ( (a + 1)) sin ( a) (1 + x) 0 that is the same as I2 (a), thus our induction hypothesis is valid for m = 2. Now suppose that our formula is valid for m = p, and integrate by parts, b (b + 1) (b + p 2) = (p 1)! sin ( b) Z 1 Z 1 1 x b x b = dx = p dx = (1 + x) (1 + x)p 0 0 Z 1 b+1 1 1 x p x b+1 = dx = p 1 b (1 + x) 0 1 b (1 + x)p+1 0 Z 1 p x b+1 = dx: 1 b 0 (1 + x)p+1 Set b = a + 1 and solve for the integral in the left side in the inequality above
72
Z
0
1
x a dx = (1 + x)p+1 ((a + 1) + p 2) a (a + 1) (a + 2) = = p (p 1)! sin ( (a + 1)) a (a + 1) (a + 2) (a + (p + 1) = p! sin ( a)
2)
;
that is the same as Ip+1 (a), thus our induction hypothesis is valid for m = p + 1. We have showed that the formula Im (a) is valid for all m > 2, that was to be shown.
73
VII.4.5 By integrating a branch of show that Z
1
0
(log z)2 (z+a)(z+b)
around the keyhole contour,
log x (log a)2 (log b)2 dx = ; (x + a) (x + b) 2 (a b)
a; b > 0; a 6= b:
Solution
VII.4.5 (a = 1, b = 2) γ2
z2
γ1
z1
γ4
-R
γ3
R
Set I=
Z
0
and integrate
1
log x dx (x + a) (x + b)
(log z)2 (log jzj + i arg z)2 = f (z) = (z + a) (z + b) (z + a) (z + b) around the keyhole contour in Figure VII.4.5. We make a branchcut for log z along the positive real axis, where 0 < arg z < 2 . Residue at a simple pole at z1 = a, where by Rule 3,
74
"
# (log jzj + i arg z)2 (log jzj + i arg z)2 Res ; a = (z + a) (z + b) (z + b) Residue at a simple pole at z1 =
z= a
b, where by Rule 3,
"
# (log jzj + i arg z)2 (log jzj + i arg z)2 Res ; b = (z + a) (z + b) (z + a) Integrate along Z =
1,
z= b
(log b + i )2 : = a b
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
Z
1
(log jzj + i arg z)2 dz = (z + a) (z + b)
z = xe0i dz = dx !
Integrate along Z
2
3,
= Z
0
Z
R
"
1
(log x)2 dx ! (x + a) (x + b)
(log x)2 dx = J: (x + a) (x + b)
and let R ! 1. This gives
f (z) dz 2
Integrate along Z
(log a + i )2 : = b a
log2 R + (2 )2 2 R (R a) (R b)
4 log2 R ! 0: R
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
Z
Z " (log jzj + i arg z)2 (log x + 2 i)2 z = xe2 i = dz = = dx ! dz = dx (z + a) (z + b) R (x + a) (x + b) 3 Z 0 Z 1 (log x + 2 i)2 (log x + 2 i)2 ! dx = dx = (x + a) (x + b) 1 (x + a) (x + b) 0 Z 1 Z 1 Z 1 (log x)2 log x 4 2 = dx 4 i + dx = (x + a) (x + b) (x + a) (x + b) (x + a) (x + b) 0 0 0 = J 4 iI+K 75
Integrate along Z
4,
and let " ! 0+ . This gives
f (z) dz 4
log2 " + (2 )2 2 " (a ") (b ")
2 " log2 " ! 0: ab
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that J +0
J 4 iI+K + 0 = 2 i
(log jbj + i )2 (log jaj + i )2 + b a a b
!
and hence, setting imaginary parts equal, (log a)2 (log b)2 I= ; 2 (a b) i.e. Z
0
1
log x (log a)2 (log b)2 dx = ; (x + a) (x + b) 2 (a b)
76
a; b > 0; a 6= b:
;
VII.4.6 Using residue theory, show that Z 1 a sin ( a) a 2 cos ( a) x log x dx = ; sin2 ( a) (1 + x)2 0
1 < a < 1:
Solution
VII.4.6 γ2
γ1
z1
γ3
γ4
-R
R
Set I=
Z
0
1
xa log x dx (1 + x)2
and integrate f (z) =
z a log z jzja eia arg z (log jzj + i arg z) = (1 + z)2 (1 + z)2
along the keyhole contour in Figure VII.4.6. We make a branchcut for z a and log z along the real axis, where 0 < arg z < 2 . Residue at a double pole at z1 = 1, where by Rule 3,
77
Res
jzja eia arg z (log jzj + i arg z) ; 1 = (1 + z)2 d a az a log z z a = (z log z) = + dz z z z= 1 a z (a log z + 1) = = z z= 1 jzja eia arg z = (a (log jzj + i arg z) + 1) z
= z= 1
= z= 1
=e Integrate along Z =
Z
1,
ia
( 1
ai) :
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
1
jzja eia arg z (log jzj + i arg z) dz = (1 + z)2
z = xe0i dz = dx !
Integrate along
and let R ! 1. This gives q Z Ra log2 R + (2 )2 f (z) dz 2 R (R 1)2 2
Integrate along
= Z
0
1
Z
R
xa log x 2 dx ! " (1 + x) xa log x dx = I: (1 + x)2
2
3,
2 log R ! 0: R1 a
and let R ! 1 and " ! 0+ . This gives
78
Z =
f (z) dz =
Z
3
Z " a 2 ia x e (log jxj + 2 i) jzja eia arg z (log jzj + i arg z) z = xe2 i = dz = dx ! 2 dz = dx (1 + z) (1 + x)2 R 3 Z 0 a 2 ia Z 1 a x e (log jxj + 2 i) x (log jxj + 2 i) 2 ia ! dx = e dx = 2 (1 + x) (1 + x)2 1 0 Z 1 a Z 1 x log x xa dx 2 ia 2 ia = e dx 2 ie = (1 + x)2 (1 + x)2 0 0 = e2 ia I 2 ie2 ia J and let " ! 0+ . This gives q Z a " log2 " + (2 )2 f (z) dz 2 " (1 ")2 4
Integrate along
4,
2 "1+a jlog "j ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that ia
e2
I +0 multiplying with e e
ia
I
2 ie2
ia
J +0=2 i e
ia
( 1
ai) ;
, this yields
ia
e
ia
I
2 ie
ia
J = 2 i( 1
ai) ;
and thus 2i sin ( a) I
2 ie
ia
J = 2 2a
2 i:
Separating real and imaginary parts, we get simultaneous equations 2 sin ( a) I
2 sin ( a) J = 2 2 a 2 cos ( a) J = 2 ;
and hence the solutions
I=
Z
0
1
xa log x = (1 + x)2
sin ( a) a 2 cos ( a) ; sin2 ( a)
79
J=
Z
0
1
xa dx a ; 2 = sin ( a) (1 + x)
i.e. Z
0
1
xa log x dx = (1 + x)2
sin ( a) a 2 cos ( a) ; sin2 ( a)
80
1 < a < 1:
VII.4.7 Show that Z
0
1
xa 1 ; dx = 1 + xb b sin ( a=b)
0 < a < b:
Determine for which complex values of the parameter a the integral exists (in the sense that the integral of the absolute value is …nite), and evaluate it. Where does the integral depend analytically on the parameter a? Solution
VII.4.7 (b = 12)
ε
z1
γ3
γ2
γ1
R
Set I=
Z
0
and integrate
1
xa 1 ; 1 + xb
za 1 jzja 1 ei(a 1) arg z f (z) = = 1 + zb 1 + jzjb eib arg z
along the pie-slice, of aperture 2 =b, contour in Figure VII.4.7. We make a branch cut for z a 1 along the negative imaginary axis , where =2 < arg z < 3 =2. Residue at a simple pole at z1 = e i=b , where by Rule 3,
81
za 1 Res ;e 1 + zb Integrate along Z
i=b
za = b bz
1
1 e b
=
1
z=e
i=b
and let R ! 1 and " ! 0+ . This gives
1,
1
Z
jzja
1 i(a 1) arg z
e
dz =
1 + jzjb eib arg z
1
z = xe0i dz = dx
=
Z
R
Integrate along
2
Z Integrate along
=
xa 1 dx ! 1 + xb
"
Z
!
Z
:
f (z) dz = =
Z
ia=b
1
0
xa 1 dx = I: 1 + xb
and let R ! 1. This gives then 0 < a < b Ra 1 2 R Rb 1 b
f (z) dz 2
3,
2 bRb
a
! 0:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
jzja 3
1 i(a 1) arg z
e
=
jzjb eib arg z
1+ Z !
0
1
xa 1 e2 i(a 1 + xb
Integrate along
4,
Z
z = xe2 i=b dz = e2 i=b dx
=
1)=b 2 i=b
e
dx =
2 ia=b
e
Z
Z
"
R 1
0
xa 1 e2 i(a 1 + xb
1)=b
xa 1 dx = 1 + xb
e2 e2
i=b
dx !
ia=b
I:
and let " ! 0+ . This gives then 0 < a < b f (z) dz 4
"a 1 2 " 1 "b b
2 "a ! 0: b
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0
e2
and multiplying with e
ia=b
ia=b
I +0=2 i
, this yields 82
1 e b
ia=b
;
e
ia=b
e
ia=b
I=
2 i b
and hence solving for I, I=
b (e
2 i e
ia=b
ia=b )
=
b sin ( a=b)
;
i.e., Z
0
1
xa 1 ; dx = 1 + xb b sin ( a=b)
0 < a < b:
Now we suppose a and b complex and rewrite the nominator in the integrand xa
1
= e(a
1) log x
= e(Re a
1) log x i Im a log x
e
= xRe a 1 ei Im a log x :
The absolute value of the integrand is xa 1 xRe a 1 ei Im a log x xRe a 1 = = ; 1 + xb 1 + xb 1 + xb thus Re a 1 > 1 and Re a 1 b < 1. Thus the integrand depends analytically on the parameter a when 0 < Re a < b.
83
VII.4.8 By integrating a branch of (log z) = (z 3 + 1) around the boundary of an indented sector of aperture 2 =3, show that Z 1 Z 1 log x 2 2 1 2 p : dx = ; dx = x3 + 1 27 x3 + 1 3 3 0 0
Remark. Compare the results with those of Exercise 3 (after changing variables) and Exercise 2. Solution
VII.4.8
γ2
γ3
z1
γ4
z2
ε
z3
γ1
R
Set I=
Z
0
and integrate f (z) =
1
log x dx; x3 + 1
log z = z3 + 1 z
J=
Z
0
1
x3
1 dx +1
log jzj + i arg z e (z e i ) z e5 i=3
i=3
along the indented sector, of aperture 2 =3, contour in Figure VII.4.8. We make a branchcut for log z along the negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = e i=3 , where by Rule 3,
84
Res
log jzj + i arg z ;e z3 + 1
Integrate along Z
1,
i=3
=
log jzj + i arg z 3z 2
z=e
i=3
i e 9
2 i=3
:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
=
Z
1
log jzj + i arg z dz = z3 + 1
Integrate along
z = xe0i dz = dx
and let R ! 1. This gives q 2 Z log2 R + 23 2 R f (z) dz 3 R 1 3 2
Integrate along Z
=
Z
R
log x dx ! 3 " x +1 Z 1 log x ! dx = I: x3 + 1 0
=
2,
3,
2 log R ! 0: 3R2
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
Z " log x + 2 i=3 2 i=3 log jzj + i arg z z = xe2 i=3 = = dz = e dx ! 2 i=3 3 dz = e dx z +1 x3 + 1 R 3 Z 0 Z 1 log x + 2 i=3 2 i=3 log x + 2 i=3 2 i=3 e dx = e dx = ! x3 + 1 x3 + 1 0 1 Z 1 Z log x 2 i 2 i=3 1 1 2 i 2 i=3 2 i=3 = e dx e dx = e2 i=3 I e J: 3 3 x +1 3 x +1 3 0 0 Z
and let " ! 0+ . This gives q 2 Z log2 " + 23 2 " f (z) dz 3 3 1 " 4
Integrate along
4,
2 " jlog "j ! 0: 3
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 85
I + 0 e2 Multiplying with e
i=3
i=3
I
2 i 2 e 3
i=3
i e 9
J +0=2 i
2 i=3
:
, this yields 2 ie 3
2i sin ( =3) I
i=3
J=
2
2 9
;
and thus p
p
i 2 2 J= : 3 3 9 Separating real and imaginary parts, we get the simultaneous equations 3Ii +
3
J
p
p and hence the solutions Z 1 log x dx = I= x3 + 1 0
3
3I
3
J = 29 J = 0; 3
2 2 ; 27
J=
Z
0
2
1
x3
1 2 dx = p : +1 3 3
Remark. We begin with the …rst integral I, we make the substitution x3 = t and use the integral from Exercise VII.4.3, where we choose a = 2=3.
I=
Z
0
1
3 x3 = t log x 5= dx = 4 x = t1=3 x3 + 1 1 2=3 dx = 3 t dt Z 1 1 log t 1 2 cos 23 = dt = 9 0 t2=3 (t + 1) 9 sin2 23 2
=
2 2 : 27
For the second integral J, we use the integral from Exercise VII.4.2, where we choose b = 3. Z 1 2 dx = J= = p : b 1+x 3 sin ( =3) 3 3 0
86
VII.4.9 By integrating around an appropriate contour and using the results of Exercise 8, show that Z
1
0
10 3 (log x)2 p : dx = x3 + 1 81 3
Solution
VII.4.9
γ2
γ3
z1
γ4
z2
ε
z3
γ1
R
Set I=
Z
0
Integrate f (z) =
(log z)2 = z3 + 1 z
1
(log x)2 dx x3 + 1
(logjzj+i arg z)2
e
i=3
(z
e i) z
e5
i=3
along the pie-slice contour with 0 < arg z < 2 =3 in Figure VII.4.9. We make a branchcut for log z along the negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = e i=3 , where by Rule 3, "
(log jzj + i arg z)2 Res ;e z3 + 1
i=3
#
=
(log jzj + i arg z)2 3z 2 87
2
= z=e
i=3
27
e
2 i=3
:
Integrate along Z
1,
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
Z
1
(log jzj + i arg z)2 dz = z3 + 1
z = xe0i dz = dx
=
Z
"
! Integrate along Z
2,
Z =
=
Z
log2 R + 23 R3 1
2
3,
Z
0
(log x)2 dx ! x3 + 1 1
(log x)2 dx = I: x3 + 1
and let R ! 1. This gives
f (z) dz
Integrate along
R
2
2 log2 R ! 0: 3R2
2 R 3
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
Z " (log x + 2 i=3)2 2 i=3 (log jzj + i arg z)2 z = xe2 i=3 dz = e dx ! = dz = e2 i=3 dx z3 + 1 x3 + 1 R 3 Z 0 Z 1 (log x + 2 i=3)2 2 i=3 (log x + 2 i=3)2 2 i=3 ! e dx = e dx = x3 + 1 x3 + 1 1 0 Z 1 Z Z (log x)2 4 i 2 i=3 1 log x 4 2 2 i=3 1 1 2 i=3 e dx e dx+ e dx = x3 + 1 3 x3 + 1 9 x3 + 1 0 0 0 Z Z 4 i 2 i=3 1 log x 4 2 2 i=3 1 1 2 i=3 = e I e dx+ e : 3 x3 + 1 9 x3 + 1 0 0
Integrate along Z
4,
and let " ! 0+ . This gives log2 " + 23 1 "3
f (z) dz 4
2
2 " 3
2 " log2 " ! 0: 3
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 2 i=3
I+0 e
4 i 2 I e 3
i=3
Z
0
1
log x 4 2 2 dx+ e x3 + 1 9 88
i=3
Z
0
1
1 +0 = 2 i 3 x +1
2
27
e
2 i=3
:
multiplying with e
e
i=3
e
i=3
i=3
, this yields
4 i I 3
Z
1
log x e x3 + 1
0
i=3
dx+
2
4 9
Z
1
0
1 e 3 x +1
i=3
dx =
and thus, substituting the values for the integrals from Exercise 9, 2i sin ( =3) I
4 i e 3
i=3
2 2 27
+
2
4 9
e
i=3
And hence, setting imaginary parts equal, p
3I+
2 3 4 3 12 3 + = ; 81 81 27
and therefore Z
0
1
10 2 (log x)2 p : dx = x3 + 1 81 3
89
2 p 3 3
=
2 3 i: 27
2 3 i; 27
VII.4.10 By integrating a branch of (log z) = (z 3 1) around the boundary of an indented half-disk and using the result of Exercise 8, show that Z 1 log x 4 2 dx = : x3 1 27 0 Solution
VII.4.10 γ4 z2
γ6 -R
γ5
γ2 γ1 z 1
γ3 R
z3
Set I=
Z
0
Integrate f (z) =
log z = z3 1 (z
1
log x dx x3 1
log jzj + i arg z 1) z e2 i=3 z e4
i=3
along the indented half-disk contour in Figure VII.4.10. We make a branchcut for log z along the negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1, where by Rule 3, Res
log jzj + i arg z log jzj + i arg z ;1 = 3 z 1 3z 2
Residue at a simple pole at z2 = e2
i=3
= 0: z=1
, where by Rule 3,
90
Res
log jzj + i arg z 2 ;e z3 1
Integrate along Z
f (z) dz + 1
=
and
1
Z
Z
3,
i=3
=
log jzj + i arg z 3z 2
= z=e2
i=3
2 i e 9
4 i=3
:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
1
log jzj + i arg z dz + z3 1
Z
3
log jzj + i arg z dz = z3 1
z = xe0i = dz = dx Z R log x log x dx + dx ! 3 3 x 1 1 1+" x Z !
= =
Z
1 "
"
1
0
Integrate along
2,
log x dx = I: x3 1
and let " ! 0+ . This gives Z f (z) dz ! i ( 0) 0 = 0: 2
Integrate along
and let R ! 1. This gives p Z log2 R + 2 f (z) dz R 3 R 1 4
Integrate along
4,
5,
log R ! 0: R2
and let R ! 1 and " ! 0+ . This gives
91
Z
f (z) dz = 5
=
Z
5
log jzj + i arg z z = dz = 3 dz = z 1 Z 0 log x + i ! e i dx = 3 x 1 1 Z 1 log x = dx i x3 + 1 0 =
xe i e i dx Z 1
=
"
R
log x + i e id ! 3 x 1
log x + i dx = x3 + 1 0 Z 1 1 dx = 3 x +1 0 Z 1 Z 1 log x 1 dx i dx: 3 3 x +1 x +1 0 0
and let " ! 0+ . This gives p Z log2 " + f (z) dz " 1 "3 6
Integrate along
Z
6,
" jlog "j ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0+0
Z
0
1
log x dx x3 + 1
i
Z
0
1
x3
1 dx + 0 = 2 i +1
2 i e 27
4 i=3
and thus, substituting the values for the integrals from Exercise 9, 2 2 27
2 p I i = 3 3 And hence, setting imaginary parts equal, 2 2 4 2 I+ = ; 27 18 and therefore Z
0
1
log x 4 2 dx = : x3 1 27
92
2
4 9
e
4 i=3
:
;
VII.5.1 Use the keyhole contour indented on the lower edge of the axis at x = 1 to show that Z 1 log x 2 2 dx = ; 0 < a < 1: xa (x 1) 1 cos (2 a) 0 Solution
VII.5.1 γ2
γ1
-R
γ5
γ6
z γ4
γ3
R
Set I= and integrate f (z) =
za
Z
1 1
xa
log x dx (x 1)
log z (log jzj + i arg z) = a(logjzj+i arg z) (z 1) e (z 1)
along the contour in Figure VII.5.1. We make a branchcut for log z along the positive imaginary axis, where 0 < arg z < 2 . Thus f (z) is analytic on the keyhole domain. It extends analytically to (0; 1) from above, and the 93
apparent singularity at z = 1 is removable. However, the extension to (0; 1) from below has a simple pole at z = 1. Residue at a simple pole at z = 1, where by Rule 3, Res
(log jzj + i arg z) (log jzj + i arg z) ; 1 = ea(logjzj+i arg z) (z 1) ea(logjzj+i arg z)
Integrate along Z
1,
= 2 ie
2 ia
:
z=1
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 1
=
Z
1
(log jzj + i arg z) dz = a(logjzj+i arg z) (z e 1)
z = xe0i dz = dx
Z
R log x = dx ! a 1) " x (x Z 1 log x ! dx = I: xa (x 1) 0
and let " ! 0+ . This gives when 0 < a < 1 q Z log2 R + (2 )2 2 log R f (z) dz 2 R ! 0: a R (R 1) Ra 2
Integrate along
2,
Integrate along
3
Z
=
=
and
5,
and let R ! 1 and " ! 0+ . This gives
Z
f (z) dz+ f (z) dz = 5 Z Z (log jzj + i arg z) (log jzj + i arg z) = dz+ dz = a(logjzj+i arg z) a(logjzj+i arg z) (z e (z 1) e 1) 3 5 Z 1+" Z " (log x + 2 i) (log x + 2 i) z = xe2 i 2 ia 2 ia =e dx+e dx ! a dz = dx xa (x 1) 1) R 1 " x (x Z 0 Z 1 (log x + 2 i) (log x + 2 i) 2 ia 2 ia !e dx = e dx = a 1) xa (x 1) 1 x (x 0 Z 1 Z 1 log x 1 2 ia 2 ia e dx 2 ie dx = e 2 ia I 2 ie 2 ia J: a a x (x 1) x (x 1) 0 0
3
Integrate along
4,
use fractional residue formula and let " ! 0+ . This gives 94
Z
4
f (z) dz !
i(
0) 2 ie
2 ia
= 2 2e
2 ia
:
and let " ! 0+ . This gives 0 < a < 1 q Z log2 " + (2 )2 2 " 2 "1 a jlog "j ! 0: f (z) dz a (1 " ") 6
Integrate along
6,
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I
2 ia
e
Multiplying with e2
ia
I
2 ie
2 ia
2 ia
J + 2 2e
+ 0 = 0:
, this yields Ie2
ia
I
2 iJ + 2
2
= 0:
We equate real parts (cos (2 a) and therefore Z
0
1
log x dx = a x (x 1) 1
1) I + 2
2
= 0;
2 2 ; cos (2 a)
95
0 < a < 1:
VII.5.2 Show using residue theory that Z 1 sin (ax) dx = 2 1 x (x + 1)
1
e
a
;
a > 0:
Hint. Replace sin (az) by eiaz , and integrate around the boundary of a halfdisk indented at z = 0. Solution
VII.5.2 γ4
-R
z1 γ2 z2
γ1
γ3
R
z3
Set I= and integrate f (z) =
Z
1 1
sin ax dx x (x2 + 1)
eiaz eiaz = z (z 2 + 1) z (z i) (z + i)
along the contour in Figure VII.5.2. Residue at a simple pole at z1 = 0, where by Rule 3, eiaz eiaz Res ;0 = 2 z (z 2 + 1) 3z + 1
= 1: z=0
Residue at a simple pole at z2 = i, where by Rule 3, 96
Res Integrate along Z
f (z) dz + 1
1
Z
eiaz eiaz ; i = z (z 2 + 1) 3z 2 + 1
and
3,
= z=i
a
e 2
:
and let R ! 1 and " ! 0+ . This gives
f (z) dz =
3 Z eiaz eiaz z = x dz + dz = = = 2 2 dz = dx z (z + 1) z (z + 1) 3 1 Z " Z R eiax eiax = dx + dx ! 2 2 R x (x + 1) " x (x + 1) Z 1 Z 1 Z 1 eiax cos ax sin ax ! dx = dx + i dx = 2 2 2 1 x (x + 1) 1 x (x + 1) 1 x (x + 1) Z 1 cos ax = dx + iI: 2 1 x (x + 1)
Z
Integrate along
2,
use fractional residue formula and let " ! 0+ . This gives Z f (z) dz ! i ( 0) 1 = i: 2
Integrate along 4 , and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 1 f (z) dz R ! 0: R (R2 1) R2 4 Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that Z 1 cos ax e a dx + iI i + 0 = 2 i ; 2 2 1 x (x + 1)
and hence, setting imaginary parts equal,
e a;
I= i.e.,
97
Z
1 1
sin ax dx = x (x2 + 1)
1
98
e
a
;
a > 0:
VII.5.3 Show using residue theory that Z 1 2 sin (ax) dx = ; 2 2 2 ax) 1 x(
a > 0:
Solution
VII.5.3 γ8
γ2 -R
z3
γ1
γ4 z1
γ3
γ6 γ5
z2
γ7 R
Set I= and integrate f (z) =
Z
1 1
sin (ax) dx x ( 2 a2 x 2 )
eiaz = z ( 2 a2 z 2 )
a2 z (z
eiaz =a) (z + =a)
along the contour in Figure VII.5.3. Residue at a simple pole at z1 = 0, where by Rule 3, Res
eiaz ;0 = z ( 2 a2 z 2 )
2
eiaz 3a2 z 2
= z=0
1 2
:
Residue at a simple pole at z2 = =a, where by Rule 3, Res
eiaz ; = z ( 2 a2 z 2 ) a 99
2
eiaz 3a2 z 2
= z= =a
1 2
2
:
Residue at a simple pole at z3 = eiaz ; z ( 2 a2 z 2 )
Res Integrate along Z
1;
f (z) dz + 1
Z =
+
3;
5
and
f (z) dz + 3
Z
Z
1
=a, where by Rule 3,
a 7,
Z
=
f (z) dz + 5
eiaz dz + z ( 2 a2 z 2 ) e z(
2
2
= z=
=a
1 2
2
:
and let R ! 1 and " ! 0+ . This gives
iaz
5
eiaz 3a2 z 2
a2 z 2 )
dz +
Z
Z
3
7
Z
f (z) dz = 7
eiaz dz+ z ( 2 a2 z 2 )
eiaz dz = z ( 2 a2 z 2 )
z = x = dz = dx Z =a " Z " eiax eiax = dx + dx+ 2 x ( 2 a2 x 2 ) a2 x 2 ) R =a+" x ( Z =a " Z R eiax eiax + dx + dx ! 2 x ( 2 a2 x 2 ) a2 x2 ) " =a+" x ( Z 1 eiax dx = ! 2 a2 x 2 ) 1 x( Z 1 Z 1 cos (ax) sin (ax) = dx + i dx = 2 2 2 2 ax) a2 x 2 ) 1 x( 1 x( Z 1 cos (ax) = dx + iI: 2 a2 x 2 ) 1 x( =
use fractional residue formula and let " ! 0+ . This gives Z 1 1 f (z) dz ! i ( 0) = i: 2 2 2 2
Integrate along
2,
Integrate along
4,
use fractional residue formula and let " ! 0+ . This gives Z 1 1 f (z) dz ! i ( 0) 2 = i: 4
100
Integrate along
use fractional residue formula and let " ! 0+ . This gives Z 1 1 f (z) dz ! i ( 0) i: = 2 2 2 6
6,
Integrate along 8 , and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 1 R f (z) dz ! 0: 2 2) R (a R2 a2 R 2 2 Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that Z 1 cos ax 1 1 1 dx + iI i i i = 0; 2 2 2 1 x (x + 1) and hence, setting imaginary parts equal I
1 2
1
1 = 0; 2
i.e., Z
1 1
2 sin (ax) dx = ; 2 2 2 x( ax)
101
a > 0:
VII.5.4 Show using residue theory that Z 1 1 cos x dx = : x2 2 0 Solution
VII.5.4 γ4
γ2
-R
z1
γ1
γ3
R
Set I=
Z
1
1
0
and integrate
cos x dx ) 2I = x2
f (z) =
1
Z
1 1
1
cos x dx x2
eiz z2
along the contour in Figure VII.5.4. Residue at a double pole at z1 = 0, where by Rule 2, Res Integrate along
1
eiz z2
1
and
d 1 z!0 dz
; 0 = lim 3,
eiz =
ieiz
z=0
=
i
and let R ! 1 and " ! 0+ . This gives
102
Z
f (z) dz + 1
Z
f (z) dz = 3
=
Z
" R
Integrate along
Z
eiz
1
z2
1
dz +
Z
1 3
eiz z2
dz =
z = x = = dz = dx Z R Z 1 1 eix 1 eix 1 eix dx + dx ! dx = x2 x2 x2 " 1 Z 1 Z 1 Z 1 1 cos x sin x sin x dx i dx = 2I i dx: 2 2 2 x 1 1 x 1 x 2,
use fractional residue formula and let " ! 0+ . This gives Z f (z) dz ! i ( 0) ( i) = : 2
Integrate along 4 , and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 2 2 R ! 0: f (z) dz 2 R R 4 Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that Z 1 sin x 2I i dx + 0 = 0; 2 1 x
and hence, setting imaginary parts equal 2I
= 0;
1
cos x dx = : x2 2
i.e., Z
0
1
103
VII.5.5 By integrating (e 2iz 1) =z 2 over appropriate indented contours and using Cauchy´s theorem, show that Z 1 sin2 x dx = : 2 1 x Solution
VII.5.5 γ2
γ3
-R
γ4 z1
γ1
R
Set I= and integrate
Z
1 1
sin2 x dx = x2
Z
1 1
1
cos 2x dx 2x2
ei2z 1 f (z) = z2 along the contour in Figure VII.5.5. Residue at a double pole at z1 = 0, where by Rule 2, Res Integrate along
ei2z 1 d i2z ; 0 = lim e z!0 dz z2 1
and
3,
1 = 2iei2z
z=0
= 2i
and let R ! 1 and " ! 0+ . This gives
104
Z
f (z) dz + 1
Z
Z
f (z) dz = 3
ei2z 1 dz + z2
Z
ei2z 1 dz = z2
z = x = dz = dx 1 3 Z " i2x Z R i2x Z 1 i2x e 1 e 1 e 1 = dx + dx ! dx = x2 x2 x2 R " 1 Z 1 Z 1 Z 1 sin 2x sin 2x cos 2x 1 dx + i dx = 2I + i dx: = x2 x2 x2 1 1 1 =
Integrate along
2,
use fractional residue formula and let " ! 0+ . This gives Z f (z) dz ! i ( 0) 2i = 2 : 2
Integrate along 4 , and let R ! 1. Because jeiaz j 1 in the upper half plane if a > 0, this gives Z 2 2 R ! 0: f (z) dz 2 R R 4 Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that Z 1 sin 2x 2I i dx + 2 + 0 = 0; 2 1 2x
and hence, setting real parts equal,
2I + 2 = 0; i.e., Z
1 1
sin2 x dx = : x2
105
VII.5.6 By integrating a branch of (log z) = (z 3 1) around the boundary of an indented sector of aperture 2 =3, show that Z 1 log x 4 2 dx = : x3 1 27 0
Remark. See also Exercise 4.10. Solution
VII.5.6
γ5 z2
γ4
γ6 γ8
γ7
γ2 γ1
z1
γ3 R
z3
Set I=
Z
0
and integrate f (z) =
log z = z3 1 (z
1
log x dx x3 1
e0i )
log jzj + i arg z z e2 i=3 z
e4
i=3
along the contour in Figure VII.5.6. We make a branchcut for log z along the negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1, where by Rule 3, Res
log jzj + i arg z log jzj + i arg z ;1 = 3 z 1 3z 2
Residue at a simple pole at z2 = e2
i=3
= 0: z=1
, where by Rule 3,
106
Res
log jzj + i arg z 2 ;e z3 1
Integrate along Z
f (z) dz+ 1
Z
1
and
1
Z
3,
i=3
=
log jzj + i arg z 3z 2
= z=e2 +
i=3
2 i 2 e 9
i=3
:
and let R ! 1 and " ! 0 . This gives
f (z) dz = 3
Z log jzj + i arg z log jzj + i arg z z = xe0i dz+ dz = = 3 dz = dx z3 1 z 1 3 Z 1 " Z R Z 1 log x log x log x = dx+ dx ! dx = I: 3 3 x 1 1 x3 1 " 1+" x 0
Integrate along
use fractional residue formula and let " ! 0+ . This gives Z f (z) dz ! i ( 0) 0 = 0:
2,
2
Integrate along
and let R ! 1. This gives q 2 Z log2 R + 23 2 R f (z) dz 3 R 1 3 4
Integrate along Z
f (z) dz + 5
Z
4,
and
5
Z
7,
2 log R ! 0: R2
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 7
Z log jzj + i arg z log jzj + i arg z z = xe2 i=3 = = dz+ dz = dz = e2 i=3 dx z3 1 z3 1 5 7 Z 1+" Z " log x + 2 i=3 2 i=3 log x + 2 i=3 2 i=3 = e dx + e dx ! x3 1 x3 1 R 1 " Z 0 Z 1 Z log x + 2 i=3 log x 2 i 2 i=3 1 1 2 i=3 2 i=3 !e dx = e dx e dx = x3 1 x3 1 3 x3 1 1 0 0 Z 2 i 2 i=3 1 1 2 i=3 = e I e dx: 3 x3 1 0 Integrate along
6,
use fractional residue formula and let " ! 0+ . This gives 107
Z
6
f (z) dz !
i(
0)
2 i 2 e 9
i=3
and let " ! 0+ . This gives q 2 Z log2 " + 23 2 " f (z) dz 3 3 1 " 8
Integrate along
=
2
2 9
e2
i=3
:
8,
2 " jlog "j ! 0: 3
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that Z 2 i 2 i=3 1 1 2 2 2 i=3 2 i=3 dx + I +0 e I e e = 0: 3 x3 1 9 0 Multiplying with e I
2 i=3
1 2
, this yields p ! Z 3 2 i 1 1 2 2 i I dx + = 0: 2 3 0 x3 1 9
We equate real parts 3 2 2 I+ = 0; 2 9 and hence Z
0
1
log x 4 2 dx = : x3 1 27
108
VII.6.1 Integrate 1= (1
x2 ) directly, using partial fractions, and show that Z 1 dx PV = 0: 1 x2 0
Show that Z
1
0
Z
dx = +1; 1 x2
1
1
dx = 1 x2
1:
Solution Partial fractions gives 1 x2
1
PV
Z
0
= lim
"!0
1
dx = lim 1 x2 "!0
1 [log (1 + x) 2 = lim
"!0
Z
=
1 "
0
log (1
1 2
dx + 1 x2 x)]10
1+x 1 log 2 1 x 1 "!0 2
= lim
log
1 1 + 1+x 1 x
1 " 0
2
"
Z
+
1
1+"
dx 1 x2
1 [log j1 + xj 2
1 1+x + log 2 1 x
" "
log
2+" "
= log j1 !
1
xj]1 1+"
=
=
1+"
=
1 = lim log "!0 2
2 " 2+"
= 0:
Inspection gives Z
0
1
dx = lim 1 x2 "!0+
Z
1 "
dx 1 = lim+ [log (1 + x) log (1 x)]10 " = 2 "!0 2 1 x 0 1 " 1 1+x 1 2 " = lim+ log = lim+ log = +1 : "!0 2 "!0 2 1 x 0 " 109
and the second integral Z
1
1
dx = lim 1 x2 "!0+ = lim+ "!0
Z
1
1 dx = lim+ [log (1 + x) log (1 x)]1 1+" = 2 "!0 2 x 1+" 1 1 1+x 1 2+" 1 log log ! 1: = lim+ "!0 2 2 1 x 1+" "
110
VII.6.2 Obtain the principal value in Exercise 1 by taking imaginary parts of the identity (5.4) p.211 in the preceding section and making a change of variable Solution Identity (5.4) Z
1
0
Z log x dx+ x2 1
1 " 1
Z log jxj + i dx+ x2 1
Taking imaginary part of (5.4), get Z 1 " Z 0 dx + 2 x2 1 1 1+" x
0 1+"
1
Z log jxj + i log z dx+ dz = 0: 2 x2 1 1 C" z
dx + Im
C"
Note
( )
Im
Z
C"
1 " 1
x2
1
dx +
Z
x, and by using ( ) we get
0 1+"
x2 Z
1
dx =
1
1+"
x2
log z dz = 0: z2 1
log z dz ! 0; z2 1
as " ! 0, by estimation on p. 211. Making the change of variables, x ! Z
Z
1
x = dx = Z 1 dx + 0
x dx
=
"
x2
1
dx ! 0 as " ! 0;
which gives that PV
Z
0
1
dx = 0: 1 x2
Note. For a linear change of variables, we can take the limit either before or after change of variable. Otherwise we can not do this.
111
VII.6.3 By integrating around the boundary of an indented half-disk in the upper half-plane, show that Z 1 1 a dx = ; 1 < a < 1: PV 2 2 a) a +1 1 (x + 1) (x Solution
VII.6.3 (a = 2) γ4
z2
-R
γ1
γ2 z1
z3
γ3
R
Set I = PV and integrate f (z) =
Z
1 1
1 (z 2 + 1) (z
(x2
a)
1 + 1) (x
=
(z
a)
dx
1 i) (z + i) (z
a)
along the contour in Figure VII.6.3. Residue at a simple pole at z1 = a, where by Rule 1, Res
(z 2
1 + 1) (z
a)
;a =
z2
1 +1
= z=a
Residue at a simple pole at z1 = i, where by Rule 3,
112
a2
1 : +1
Res
1 (z 2 + 1) (z
a)
Integrate along Z
f (z) dz + 1
=
and
1
Z
Za
;i = 3,
3z
1 2az + 1
2
z=i
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
"
1 2 (x + 1) (x
a)
dx +
ZR
!
Integrate along Z
2,
2
1
(x2
a)
dx !
1 + 1) (x
a)
dx = I:
and let " ! 0+ . This gives
f (z) dz ! 4,
1 + 1) (x Z 1
(x2
a+"
R
Integrate along Z
1 a + i: 2 (a2 + 1) 2 (a2 + 1)
=
i(
0)
a2
1 +1
=
a2 + 1
i:
and let R ! 1. This gives
f (z) dz
(R2
4
1 1) (R
jaj)
R
R2
! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I
a2 + 1
i=2 i
1 a + i ; 2 (a2 + 1) 2 (a2 + 1)
i.e., PV
Z
1 1
(x2
1 + 1) (x
113
a)
dx =
a2
a : +1
VII.6.4 Suppose m 2 and a1 < a2 < < am . By integrating around the boundary of an indented half-disk in the upper half-plane, show that Z 1 1 dx = 0: PV a1 ) (x a2 ) (x am ) 1 (x Solution
VII.6.4 γR
-R
γ1
γ2
γ m - 1 γm
a1
a2
a m - 1 am
R
Set I = PV and integrate f (z) =
(z
Z
1 1
(x
a1 ) (z
a1 ) (x
1 a2 )
(z
1 a2 )
am )
(x
am )
dx
1 k=1 (z
= Qm
ak )
along the semicircular contour indented at z = aj with small semicircles of radii ", see Figure VII.6.4. Residue at a simple pole at zj = aj ; 1 j m, where by Rule 3, 1 k=1 (z
Res Qm
ak )
1 k=1;k6=j (z
; aj = Qm
114
ak )
z=aj
1 k=1;k6=j (aj
= Qm
ak )
:
Integrate along the union of line segments on real axis denoted by let R ! 1 and " ! 0+ . This gives Z =
=
Z
Z
Z
f (z) dz = L
a1 " R
a1 " R
a1 "
1 k=1 (z
Qm
Integrate along
j=1
Integrate along Z
R
1;
j
ak )
dz+
k=1 Z m 1 X ak+1 " ak +"
k=1
f (z) dz+
m,
2; : : :
1 k=1 (z
Qm
R
f (z) dz =
ak )
dz+
ak ) 1 1
Z
R
am +"
dx+
Z
am +"
1 k=1 (x
Qm
i(
0)
j=1
1 Qm k=1;k6=j (aj
ak )
ak )
1 k=1 (x
ak )
Qm
ak ) !
1 k=1 (z
Qm
R
and let " ! 0+ . This gives m X
and
am +"
z = x = dz = dx m X1 Z ak+1 " 1 Qm dx+ ak ) k=1 (x k=1 ak +" Z !
f (z) dz ! R,
Z
ak +"
=
1 Qm k=1 (x
m Z X
f (z) dz+
m X1 Z ak+1 "
L,
dx = I:
:
and let R ! 1, this gives 1 k=1 (z
Qm
f (z) dz R
jak j)
R
Rm
1
! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I
i
m X j=1
1 k=1;k6=j (aj
Qm
ak )
= 0:
Now, note that the the sums of the residues at a simple pole at zj = aj ; 1 j m, are real, and identifying the real part, we have Z 1 1 PV dx = 0: a1 ) (x a2 ) (x am ) 1 (x
115
dz =
dx !
VII.6.5 Show that PV
Z
1 1
xa 1 dx = xb 1
a ; b
cot
b
0 < a < b:
Hint. For b > 1 one can integrate a branch of z a 1 = z b of aperture 2 =b, indented at z = 1 and z = e2 i=b .
1 around a sector
Solution
VII.6.5 (b = 3)
γ5 z2
γ4
γ6 γ8
γ7
γ2 γ1
z1
γ3 R
z3
Set I = PV and integrate
Z
1 1
xa 1 dx xb 1
za 1 jzja 1 ei(a 1) arg z f (z) = b = z 1 jzjb eib arg z 1
along the contour in Figure VII.6.5. We make a branchcut for z a negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1, where by Rule 3, Res
za 1 jzja 1 ei(a 1) arg z ; 1 = zb 1 bz b 1 116
z=1
1 = : b
1
along the
Residue at a simple pole at z2 = e2 Res
za 1 2 ;e zb 1
Integrate along Z
f (z) dz+ 1
Z
=
jzja 1
and
1
Z
i=b
= 3,
e
jzjb eib arg z
Z
1
dz +
1 "
"
2,
Z Integrate along
4,
Z Integrate along
Z
f (z) dz+ 5
jzja 5
5
Z
e bz b
e2
=
1
i=b 2 i(a 1)=b
e
b
z=e2 i=b
:
Z
jzja
1 i(a 1) arg z
e
dz =
z = xe0i dz = dx
= jzjb eib arg z 1 Z R a 1 Z 1 a 1 xa 1 x x dx + dx ! dx = I: b b x 1 1 xb 1 1+" x 0 3
and let " ! 0+ . This gives 2
f (z) dz !
i(
0)
1 b
i : b
=
and let R ! 1. This gives when 0 < a < b f (z) dz 4
and
7,
Ra 1 2 R Rb 1 b
2 bRb
a
! 0:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 7
1 i(a 1) arg z
e
1 i(a 1) arg z
f (z) dz =
1 i(a 1) arg z
Integrate along
, where by Rule 3,
and let R ! 1 and " ! 0+ . This gives
3
=
Z
jzja
i=b
dz +
Z
jzja
1 i(a 1) arg z
e
z = xe2 i=b dz = e2 i=b dx
= b ib arg z jzjb eib arg z 1 1 7 jzj e Z 1+" a 1 2 i(a 1)=b Z " a 1 2 i(a 1)=b x e x e 2 i=b = e dx + e2 i=b dx! b b x 1 x 1 R 1 " Z 0 a 1 2 i(a 1)=b Z 1 a 1 x e x 2 i=b 2 i=b 2 i(a 1)=b ! e dx = e e dx = b x 1 xb 1 1 0 = e2 i=b e2 i(a 1)=b I 117
dz =
Integrate along Z
6
6,
f (z) dz !
Integrate along
8,
Z
and let " ! 0+ . This gives i(
0)
e2
i=b 2 i(a 1)=b
e
=
b
i
i=b 2 i(a 1)=b
e2
e
:
b
and let " ! 0+ . This gives when 0 < a < b f (z) dz 8
2 "a ! 0: b
"a 1 2 " 1 "b b
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that i + 0 e2 b Hence solving for I, I
i 1 + e2 I= b 1 e2
i=b 2 i(a 1)=b
i=b 2 i(a 1)=b
e i=b e2
i(a 1)=b
e
I
i
i 1 + e2 = b 1 e2
e2
e
= 2 i 0:
b
ia=b ia=b
i=b 2 i(a 1)=b
=
cos b sin
a b a b
=
i.e., PV
Z
1 1
xa 1 dx = xb 1
b
cot
118
a ; b
0 < a < b:
b
cot
a ; b
VII.6.6 By integrating a branch of (log z) = z b 1 around an indented sector of aperture 2 =b, show that for b > 1, Z
0
1
2 log x dx = ; xb 1 b2 sin2 ( =b)
PV
Z
1
0
1 xb
1
dx =
b
cot ( =b) :
Solution
VII.6.6 (b = 3)
γ5 z2
γ4
γ6 γ8
γ7
γ2 γ1
z1
γ3 R
z3
Set I=
Z
1
0
and integrate
log x dx xb 1
f (z) =
J = PV
Z
1
0
1 xb
1
dx
log z log jzj + i arg z = b z 1 jzjb eib arg z 1
along the contour in Figure VII.6.5. We make a branchcut for log z along negative imaginary axis, where =2 < arg z < 3 =2. Residue at a simple pole at z1 = 1, where by Rule 3, Res
log z log jzj + i arg z ;1 = b z 1 bz b 1
Residue at a simple pole at z2 = e2
i=b
= 0: z=1
, where by Rule 3,
119
Res
log z 2 ;e zb 1
Integrate along Z
f (z) dz+ 1
=
Z
1
Z
and
3,
=
log jzj + i arg z bz b 1
= z=e2
i=b
2 ie2 b2
i=b
:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
Z
z = xe0i = b ib arg z dz = dx jzjb eib arg z 1 1 3 jzj e Z R Z 1 Z 1 " logx logx logx dx + dx ! dx = I: = b b x 1 1 xb 1 1+" x 0 "
log jzj + i arg z 1
i=b
Integrate along
2,
dz +
log jzj + i arg z
dz =
and let " ! 0+ . This gives Z f (z) dz ! i ( 0) (0) = 0: 2
Integrate along
4,
and let R ! 1. This gives when b > 1 q 2 Z log2 R + 2b 2 R 2 log R f (z) dz ! 0: b R 1 b bRb 1 4
Integrate along
5
Z =
f (z) dz+ 5
Z
Z
and
7,
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 7
log jzj + i arg z
dz +
Z
log jzj + i arg z
dz =
z = xe2 i=b dz = e2 i=b dx
b ib arg z b ib arg z 1 1 5 jzj e 7 jzj e Z 1+" Z " logx + 2 i=b 2 i=b logx + 2 i=b 2 = e dx + e b x 1 xb 1 R 1 " Z 1 Z 0 logx + 2 i=b 2 i=b logx 2 i 2 2 i=b ! e dx = e dx e b b x 1 x 1 b 1 0
=
120
e2
i=b
I
i=b
i=b
dx ! Z 1 0
=
1
xb
2 i 2 e b
1 i=b
dx =
J:
Integrate along Z
6
6,
and let " ! 0+ . This gives
f (z) dz !
i(
i=b
2 ie2 b2
0)
=
2 2 e2 b2
i=b
:
and let " ! 0+ . This gives when b > 1 q 2 Z log2 " + 2b 2 " jlog "j 2 " f (z) dz ! 0: b 1 " b b 8
Integrate along
8,
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that
Multiplying with e
2 i 2 e b i=b , this yields
e2
I +0 2
i=b
I
i=b
J+
2 2 e2 b2
i=b
= 2 i 0:
2 2 2 i J + 2 = 0: b b Separating real and imaginary parts, we get simultaneous equations e
2 i=b
cos
1 I
2
2 b
1 I + 2b2 = 0 sin 2b I 2b J = 0;
i.e.,
I=
Z
0
1
2 log x dx = ; xb 1 b2 sin2 ( =b)
J = PV
Z
0
121
1
1 xb
1
dx =
b
cot ( =b) :
VII.6.7 Suppose that P (z) and Q (z) are polynomials, deg Q (z) deg P (z)+2, and the zeros of Q (z) on the real axis are all simple. Show that
PV
Z
1 1
X X P (z) P (z) P (z) dx = 2 i Res ; zj + i Res ; xk ; Q (z) Q (z) Q (z)
summed over the poles zj of P (z) =Q (z) in the open upper halfplane and the poles xk of P (z) =Q (z) on the real axis. Remark. In other words, the principal value of the integral is 2 i times the sum of the residues in the upper half-plane, where we count the poles on the real axis as being half in and half out of the upper half-plane. Solution
VII.6.7 γR zj - 1
z3
zj
-R
γ1
γ2
a1
a2
z2
z1 γ m - 1 γm a m - 1 am
R
Use a semicircular contour indented at ak = xk with small semicircles of radii ", see Figure VII.6.7. Get Z P (z) P (z) P (z) ! i( 0) Res ; xk = i Res ; xk Q (z) Q (z) Q (z) k The condition deg Q deg P + 2 guarantees that the integral converges at 1: Otherwise we would have to use a residue at 1 (in this case deg Q = deg P + 1). 122
Use the Residue Theorem, take limit, and the result follows Set I = PV and integrate
Z
1 1
f (z) =
P (x) dx Q (x)
P (z) Q (z)
along the contour in Figure VII.6.7. The sum of the residues at the simple poles ak = xk ; 1 axis is m X
m on the real
P (z) ; xk Q (z)
Res
k=1
The sum of the residues at the simple poles zk ; 1 tion contour is j X
k
j inside the integra-
k
P (z) ; zj Q (z)
Res
k=1
Integrate along segments on real axis, and let R ! 1 and " ! 0+ . This gives Z
f (z) dz = =
Z
Z
a1 " R
a1 "
f (z) dz+ R
m X1 Z ak+1 " ak +"
k=1 m X1 Z ak+1 "
P (z) dz+ Q (z) k=1
ak +"
=
a1 " R
P (z) dz+ Q (z)
R
f (z) dz =
am +"
Z
R
am +"
P (z) dz = Q (z)
z = x = dz = dx Z R m X1 Z ak+1 " P (x) P (x) P (x) dx+ dx+ dx ! Q (x) Q (x) am +" Q (x) k=1 ak +" Z 1 P (x) ! dx = I: 1 Q (x) =
Z
f (z) dz+
Z
123
Integrate along This gives
R,
where we have deg Q (z) Z
Integrate along m Z X k=1
1;
k
Rdeg P Rdeg Q
f (z) dz R
2; : : :
m,
f (z) dz !
deg P (z) + 2 and let R ! 1. R
R
! 0:
and let " ! 0+ . This gives i(
0)
m X k=1
Res
P (z) ; xk Q (z)
:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that ! j m X X P (z) P (z) ; xk ; zk I i Res =2 i Res ; Q (z) Q (z) k=1 k=1
i.e.,
PV
Z
1 1
m m X X P (z) P (z) P (x) dx = 2 i Res ; zj + i Res ; xk : Q (x) Q (z) Q (z) j=1 k=1
124
VII.7.1 1 2 3 P
L K LLL
Show that Z
1
0
jsin xj dx = +1: x
Hint: Show that the area under the mth arc of jsin xj =x is Solution
VII.7.1
π
For x in the interval (m
1) +
4
jsin xj
2π
3π
x sin
4π
5π
m
4
, we have
1 =p ; 4 2
and 1 x
1 : m
We estimate the integrand jsin xj x
1 1 p : 2m
We have Z
m
=4
(m 1) + 4
jsin xj dx x
Hence 125
1 1 p 2m
=
C : m
1=m.
Z
0
m
jsin xj dx x
m Z X k=1
k
=4
(k 1) + 4
1 1 jsin xj dx = C 1 + + : : : + x 2 m
! +1
as m ! 1. We have used the fact that the harmonic series 1 + 12 + 13 + 14 + + m1 tends R m+1 dx to in…nity as as m ! 1, by taking the oversum for the integral 1 , x Z m+1 1 dx 1 1 + = log (m + 1) ! +1 1+ + + 2 3 m x 1 as m ! 1.
126
VII.7.2 Show that lim
R!1
Z
R R
x3 sin x dx = : 2e (x2 + 1)2
Solution
VII.7.2 γ2
z1
γ1
-R
R
z2
Set I=
Z
1 1
x3 sin x dx (x2 + 1)2
and integrate f (z) =
z 3 eiz = (z 2 + 1)2 (z
z 3 eiz i)2 (z + i)2
along the contour in Figure VII.7.2. Residue at a double pole at z1 = i, where by Rule 2, Res Integrate along
1
z 3 eiz d z 3 eiz 1 ; i = lim = : 2 2 z!i dz (z + i) 4e (z 2 + 1)
, and let R ! 1. This gives
127
Z
f (z) dz = 1
= !
Z
Z
1
1 1
Z R x3 eix z 3 eiz z = x dz = = 2 dx ! 2 dz = dx (z 2 + 1)2 R (x + 1) Z 1 3 Z 1 3 x3 eix x cos x x sin x dx = dx + i 2 2 2 dx = 2 2 (x2 + 1) 1 (x + 1) 1 (x + 1) Z 1 3 x cos x = 2 dx + iI: 2 1 (x + 1)
Integrate along 2 , and let R ! 1. Use Jordan’s Lemma this gives Z Z R3 R3 iz e jdzj < ! 0: f (z) dz R (R2 1)2 2 (R2 1)2 2 Using the Residue Theorem and letting R ! 1, we obtain that Z 1 3 x cos x 1 dx + iI = 2 i ; 2 2 4e 1 (x + 1) and hence, setting imaginary parts equal, i.e., Z 1 3 x sin x : 2 dx = 2 2e 1 (x + 1)
128
VII.7.3 Evaluate the limits lim
R!1
Z
R R
x sin (ax) dx; x2 + 1
1 < a < 1:
Show that they do not depend continuously on the parameter a. Solution
VII.7.3a z1
γ1
-R
R z2
γ2
Case 1 a < 0: Set I= and integrate f (z) =
Z
1 1
x sin (ax) dx x2 + 1
zeiaz = z2 + 1 (z
zeiaz i) (z + i)
along the contour in Figure VII.7.3a. Residue at a simple pole at z2 = i, where by Rule 3, Res Integrate along
1,
zeiaz zeiaz ; i = z2 + 1 2z
= z= i
and let R ! 1. This gives 129
ea : 2
Z
Z
Z zeiaz z = xe0i f (z) dz = = dz = dz = dx z2 + 1 1 1 Z 1 Z Z 1 x cos (ax) xeiax dx = dx + i ! 2 x2 + 1 1 1 x +1
R R 1
xeiax dx ! x2 + 1
x sin (ax) dx = 2 1 x +1 Z 1 x cos (ax) = dx + iI: x2 + 1 1
Integrate along 2 , and let R ! 1. Use Jordan’s Lemma this gives Z Z R R f (z) dz eiaz jdzj < 2 ! 0: 2 R 1 2 R 1 R 2 Using the Residue Theorem and letting R ! 1, we obtain that Z 1 ea x cos (ax) dx + iI + 0 = 2 i ; x2 + 1 2 1 and hence, setting imaginary parts equal, I=
ea :
Case 2 a = 0: lim
R!1
Case 3 a > 0:
Z
R R
x sin (ax) dx = lim R!1 x2 + 1
130
Z
R R
x2
0 =0 +1
VII.7.3b γ4
z1
γ3
-R
R
z2
Set J= and integrate f (z) =
Z
1 1
x sin (ax) dx x2 + 1
zeiaz = z2 + 1 (z
zeiaz i) (z + i)
along the contour in Figure VII.7.3b Residue at a simple pole at z1 = i, where by Rule 3, Res Integrate along Z
3,
zeiaz zeiaz ; i = z2 + 1 2z
= z=i
Z
4,
2
:
and let R ! 1. This gives
Z zeiaz z = xe0i dz = = f (z) dz = dz = dx z2 + 1 3 3 Z 1 Z 1 Z xeiax x cos (ax) ! dx = dx + i 2 x2 + 1 1 x +1 1
Integrate along
a
e
R R 1
xeiax dx ! x2 + 1
x sin (ax) dx = 2 1 x +1 Z 1 x cos (ax) = dx + iJ: x2 + 1 1
and let R ! 1. Use Jordan’s Lemma this gives 131
Z
f (z) dz 4
R R2
1
Z
4
eiaz jdzj <
R R2
1
R
! 0:
Using the Residue Theorem and letting R ! 1, we obtain that Z 1 x cos (ax) e a dx + iI + 0 = 2 i ; x2 + 1 2 1 and hence, setting imaginary parts equal, I = e a: and we have the solution Z
1 1
8 <
x sin (ax) 0 dx = : x2 + 1
ea ; e a;
Not continuous at a = 0 becauce sin 0 = 0.
132
a < 0; a=0 a > 0:
VII.7.4 By integrating z a 1 eiz around the boundary of a domain in the …rst quadrat bounded by the real and imaginary axes and a quatercircle, show that
lim
R!1
lim
R!1
where
Z
R
xa
1
cos xdx =
(a) cos ( a=2) ;
0 < a < 1;
xa
1
sin xdx =
(a) sin ( a=2) ;
0 < a < 1;
0
Z
R
0
(a) is the gamma function de…ned by Z 1 (a) = ta 1 e t dt: 0
Remark: The formula for the sine integral holds also for this, integrate by parts.
1 < a < 0. To see
Solution
VII.7.4
γ3
γ2 γ1
R
Set I = lim
R!1
Z
R
x
a 1
cos xdx
0
where 0 < a < 1, and integrate 133
J = lim
R!1
Z
0
R
xa
1
sin xdx
f (z) = z a 1 eiz = jzja
1 i(a 1) arg z iz
e
e
along the contour in Figure VII.7.4 Integrate along 1 , and let R ! 1 and " ! 0+ . This gives Z
f (z) dz = 1
Z
z = xe0i jzj e e dz = = dz = dx 1 Z 1 Z 1 Z a 1 ix a 1 ! x e dx = x cos xdx + i a 1 i(a 1) arg z iz
0
0
=
Z
R
0
1
xa
1
xa 1 eix dx ! sin xdx = I + iJ:
0
Integrate along 2 , and let R ! 1. Use Jordan’s Lemma this gives Z Z a 1 f (z) dz R eiz jdzj < Ra 1 ! 0: R1 a 2 2 Integrate along Z
Z
3,
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 3
Z R z = xe i=2 = jzj e e dz = = xa 1 e ia=2 e x dx ! i=2 dz = e dx 0 3 Z 1 Z 1 ! xa 1 e ia=2 e x dx = e ia=2 xa 1 e x dx = e ia=2 (a) = a 1 i(a 1) arg z iz
0
0
=
(a) cos( a=2)
(a) sin ( a=2) :
Using the Residue Theorem and letting R ! 1, we obtain that I + iJ
(a) cos( a=2)
(a) sin ( a=2) = 0:
We equate real and imaginary parts I=
(a) cos( a=2)
J=
(a) sin ( a=2) :
Formulas for the Gamma function, Mathematics Handbook page 278 134
(z + 1) = z (z) Z
R
x
a 1
cos xdx = x
a 1
sin x
0
(a
1)
Z
Z
R 0
R
x
sin xdx = x
a 1
sin x
0
(a
1)
Z
x
sin xdx = R
a 1
R
x
a 2
0
Z
xa
2
sin xdx =
0
lim
R!1
Z
R
xa
0
=
1
1
cos xdx
sin R
R
xa
R
xa
1
cos xdx
xa
1
cos xdx
0
Ra 1 sin R sin xdx = a 1
R
sin xdx
0
Z
0
Z
Z
R 0
R a 2
2
0
R a 2
1) xa
(a
sin R (1 a) R1
1 a
1
Z
0
1 a
a
R
1
(a) cos ( a=2)
(a + 1) cos (a + 1) = a 2 (a) [cos ( a=2) cos ( =2) sin ( a=2) sin ( =2)] = = (a) sin ( a=2) :
sin xdx =
135
VII.7.5 Show that lim
R!1
Z
R
sin x
2
dx = lim
R!1
0
Z
R
cos x
2
0
p dx = p ; 2 2
by integrating eiz around the boundary of the pie-slice domain determined by 0 < arg z < =4 and jzj < R. Remark. These improper integrals are called the Fresnel integrals. The identities can also be deduced from the preceding exercise by changing variable. Solution
VII.7.5
γ3
γ2
γ1
R
Set I = lim
R!1
where
Z
R
sin x
2
dx
J = lim
R!1
0
1 < a < 0, and integrate f (z) = eiz
2
along the contour in Figure VII.7.5 Integrate along 1 , and let R ! 1. This gives
136
Z
0
R
cos x2 dx
Z
Z R 2 z = xe0i eix dx ! = f (z) dz = e dz = dz = dx 0 1 Z 1 Z 1 Z 11 2 sin x2 dx = J + iI: cos x2 dx + i eix dx = ! Z
iz 2
0
0
0
Integrate along 2 as we can parametrize as z = Reit where 0 t and let R ! 1. Use that 4x= sin 2x for 0 x =4, this gives Z
f (z) dz = 2
=
Z
=4
Z
2
eiz dz = 2
eiR
2 e2it
iReit dt
0
=R
Z
z = Reit = dz = iReit dt Z =4 2 R eiR (cos(2t)+i sin(2t)) dt = 0
=4
e
R2 sin 2t
dt
R
Z
=4
e
4R2 t=
ds =
0
0
=
4R
4R 1
Integrate along 3 as we can parametrize as z = xe and let R ! 1. This gives Z
f (z) dz = 3
=4,
Z
iz 2
e dz = 3 Z 1 2 ! e xe 0
z = xe i=4 dz = e i=4 dx i=4
=
h
e
R2
e
i=4
Z
0
4R2 t=
i
=4
= 0
4R
where 0
R
e
x2
e
i=4
! 0: x
R,
dx !
Z 1 + i 1 x2 p e dx = dx = 2 0 Z 1 Z 1 1 1 x2 = p e dx i p e 2 0 2 0
x2
dx:
We will use integral (41) from Mathematics Handbook page 177 with a = 1, r Z 1 1 ax2 e dx = : 2 a 0 and hence
137
Z
f (z) dz = 3
1 p 2
Z
0
1
e
x2
dx
1 ip 2
Z
0
1
e
x2
dx =
p p 2 2
Using the Residue Theorem and letting R ! 1, we obtain that p p J + iI + 0 p i p = 0: 2 2 2 2 We equate real and imaginary parts p p I= p J= p : 2 2 2 2
138
p i p : 2 2
VII.8.1 1 2 3 P L K Evaluate the residue at 1 of the following functions. 3 (c) zz2 +11 (e) q z n e1=z , n = 0; 1; : : : (a) z2z 1 9 1 (f) zz ab (b) (z2 +1) (d) zz6 +11 2
Note. There are two possibilities for (f), one for each branch of the square root. a) z z2
1
=
1 1 1 1 = z 1 z2 z
1+
1 1 + 4 + ::: ; 2 z z
hence Res
z z2
1
;1 =
1:
b) 1 =O (z 2 + 1)2
1 z4
as z ! 1;
hence Res
(z 2
1 ; 1 = 0: + 1)2
c) z3 + 1 z (z 2 1) + z + 1 z 1 = =z+ 2 + 2 = 2 2 z 1 z 1 z 1 z 1 1 1 1 1 1 1 1 =z+ 1 + 2 + 4 + ::: + 2 ; =z+ 1 + 2 z 1 z z z z 1 z 1 z2 hence Res
z3 + 1 ;1 = z2 1 139
1:
d) z9 + 1 z 3 (z 6 1) + z 3 + 1 z3 + 1 3 = = z + ; z6 1 z6 1 z6 1 hence z9 + 1 ; 1 = 0: z6 1
Res e)
z n e1=z = z n 1 +
1 1 1 1 1 + + + + + ::: = 2 3 3 z 2!z 3!z 3!z 4!z 4 1 1 1 = zn + zn 1 + zn 2 + zn 3 + zn 2! 3! 4!
We can see that the coe¢ cient of
1 z
4
+:
is
1 ; (n + 1)! hence Res z n e1=z ; 1 =
1 ; (n + 1)!
n = 0; 1; : : : :
f) r
z z
analytic at w = 0: coe¢ cient of w is g 0 (0) =
a = b 1 2
s
1 aw 1 bw
1 1
a=z = b=z
1=2
b a : 2
140
r
1 1
aw ; bw
a(1 bw)+b(1 aw) (1 bw)2
= w=0
1 p1 2 1
(b
a) =
7.8.2 1 2 3 P L K Show by integrating around the dogbone contour that Z 1 x4 35 p dx = : 128 x (1 x) 0 Solution
VII.8.2
γ2
γ1 γ3
γ4
Set I=
Z
1
0
and integrate z4
f (z) = p
z (1
z)
=p
x4
p
x (1
jzjei arg
along the contour in Figure VII.8.2.
141
x)
z=2
dx
p
z4 j1
zjei arg0 (1
z)=2
VIII.8.2 arg z
arg0 z
y π −π
π
-2
y
π
-1
−π
0
−π
0
0
1
0
2
0
π
x
0
π
-2
π
arg z
π
arg0 (1
π
-1
π
−π
π
-2
-1
−π
0
1
2π
2π
0
2
0
−π
0
0
1
0
2
0
2π
x
0
2π
-2
0
The argument for the function f (z) = p
jzjei arg
z=2
0
pz
-1
2π
2π
0
0
π
1
π
π
2
4
j1 zjei arg0 (1
z)=2
y 3π/2 −π/2
3π/2
-2
−π/2
-1
3π/2
π
−π/2
0
π/2
1
π/2
π/2
2
π/2
x
Residue at a simple pole at z1 = 1 , where by result from Exercise VII.8.13 and Rule 1,
= =
128 (w
Res
1 f w2
1 w
;0 =
9=2
1)
2
Res 4
1 w4
1 q w2 1 1 w
1 w
3
; 05 =
1 1 d4 1 1 p p = lim = w!0 4! dz 4 w5 w 1 w 1 35 35 = = e i=2 = 9=2 i9 arg(w 1)=2 128 128 (jw 1j) e
Res
35
x
y
π
Res [f (z) ; 1] =
2π
z)
y π
0
142
35 i 128
π
x
Integrate along Z =
1,
f (z) dz =
Z
1
1 "
p
"
jzjei arg
Integrate along
Integrate along
=
z=2
2,
Z Z
and let " ! 0+ . This gives
p
z4
dz =
1 "
and let " ! 0+ . This gives
(1 + ")4 p 2 " (1 + ") "
2
x4 p dx = jxjei 0=2 j1 xjei 0=2 x4 p dx = I: x (1 x)
p
zjei arg0 (1 z)=2 " Z 1 Z 1 " x4 p dx ! = x (1 x) 0 "
j1
f (z) dz 3,
Z
2 "9=2 ! 0:
and let " ! 0+ . This gives
f (z) dz =
Z
3
"
1 "
p jzjei arg
Integrate along
z=2
4,
Z
p
z4 j1
zjei arg0 (1 z)=2 Z 1 " x4 p = x (1 "
Z
"
1 p dx = jxjei 0=2 j1 xjei2 =2 1 " Z 1 x4 p dx ! dx = I: x) x (1 x) 0
dz =
p
and let " ! 0+ . This gives "4 p " (1
f (z) dz 4
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0+I +0=2 i
35 i ; 128
i.e., Z
0
1
x4
p
x (1
x)
143
dx =
35 : 128
7.8.3 1 2 3 P L K Fix an integer n, positive or negative. Determine for which complex values of the parameter a the integral Z 1 xn dx a x)1 a 0 x (1 converges, and evaluate it. Solution. ( Obs fel i lösningen argumentet) x 2 R+ 1 x 2 R+ a = Re a + i Im a xa = xRe a+i Im a = xRe a xi Im a The integral is generalized in 0 and 1 " 2 R, x = 0 g (x) = xa1 n f (x) 1 = (1 x) 1 a ! 1, as x ! 0 Rg(x) R" " g (x) dx < 1 , a n < 1 , f (x) dx < 1 , a < n + 1. 0 0 The integral is generalized in 0 and 1 " 2 R, x = 0 1 g (x) = (1 x) 1 a n f (x) = xxa ! 1, as x ! g(x) R1 g (x) dx < 1 , 1 "
1 a<1,
R"
f (x) dx < 1 , a > 0. 0 Thus R1 f (x) dx < 1 , 0 < a < n + 1. 0 Show that xi Im a begränsad, set 2 [arg x] xi Im a = ei Im a(log x+i ) = ei Im a log x = 1 On principal arg 0 Determination for which complex values of the parameter a the integral converges
144
VII.8.3
γ1
γ2
γ3
γ4
Set I=
Z
0
1
xn xa (1
x)1
a dx
and integrate f (z) =
zn z a (1
z)1
a
=
jzja eai arg z j1
along the contour in Figure VII.8.3.
145
zn zj1 a e(1
a)i arg0 (1 z)
VIII.8.3 arg z
arg0 z
y π −π
π
-2
−π
π
-1
y 0
−π
0
0
1
0
2
0
π
x
0
π
-2
π
arg z
π
arg0 (1
π
-1
π
−π
π
-2
−π
-1
0
2π
1
0
2
2π
0
−π
0
0
1
0
2
0
2π
x
0
2π
-2
0
The argument for the function f (z) =
0
-1
2π
2π
0
0
π
1
π
2
π
zn jzja eai arg z j1
1 a (1 a)i arg (1 z) 0 zj e
y 2π − απ −απ
2π − απ
-2
−απ
-1
2π − απ
2π − 2απ
−απ
0
π − απ
1
π − απ
π − απ
2
π − απ
x
Residue at a simple pole at z3 = 1 , where by result from Exercise VII.8.13 and Rule 1,
=
p
1 f Res w2
w2 1 Res ;0 = w(1 + w2 )
Integrate along
1,
x
y
π
Res [f (z) ; 1] =
2π
z)
y π
0
1 w lim
;0 =
w!0
p
jw2
"
1 Res w2 1jei arg(w w2 + 1
2
1 wn 1 wa
1
1)=2
and let R ! 1 and " ! 0+ . This gives 146
=
#
;0 1 1 a w ei
=2
=
= i
π
x
Z =
f (z) dz =
Z
1
jzja eai arg z j1
"
Integrate along Z
2,
=
and let " ! 0+ . This gives (1 + ")4 (1 + ")a (1 (1 + "))1
f (z) dz 4
Integrate along Z
Z 1 " zn zn = dx = zj1 a e(1 a)i arg0 (1 z) jzja eai 0 j1 zj1 a e(1 a)i 0 " Z 1 " zn dx ! = z a (1 z)1 a " Z 1 zn dx = I: ! a z)1 a 0 z (1
1 "
3,
a
2 "
2 "
a
! 0:
and let R ! 1 and " ! 0+ . This gives
f (z) dz =
Z
3
"
1 "
jzja eai arg
Integrate along Z
Z " zn zn = dx = a ai 0 z j1 zj1 a e(1 a)i arg0 (1 z) j1 zj1 a e(1 a)i 2 1 " jzj e Z 1 " zn 2 ia dx ! = e z a (1 z)1 a " Z 1 zn 2 ia ! e dx = e2 ia I: a z)1 a 0 z (1 4,
and let " ! 0+ . This gives
f (z) dz 4
"n "a
(1
1 a
")
2 "
2 "n+1
a
! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I + 0 e2
ia
I +0=2 i
( 1)n 147
(a
1)
(a n!
n)
:
Solve for I, we obtain that ( 1)n 2 i (a I= 1 e2
1) (a n) n! ia
( 1)n 2 i (a 1)n!(a = e ia e ia
n)
= ( 1)n
(a
1)
sin ( a)
i.e., Z
0
1
xn xa (1
1 a dx
x)
= ( 1)n
(a sin ( a)
148
1)
(a n!
n)
;
0 < a < 1; n = 0; 1; 2; : : :
(a n!
n)
;
VII.8.4 Show that Z
1 1
p
1 x2 dx = 1 + x2
p
2
1 :
Solution
VII.8.4
γ2
γ1 γ3
γ4
Set I= and integrate
Z
1 1
p
1 x2 dx 1 + x2
p j1 + zjei arg 1 z2 f (z) = 2 = z +1 along the contour in Figure VII.8.4. p
149
p j1 2 z +1
(1+z)=2
zjei arg0 (1
z)=2
VIII.8.4 arg z
arg0 z
y π −π
π
-2
−π
-1
y
π
0
−π
0
0
1
0
0
2
π
x
0
π
arg (1 + z)
π
-2
π
arg0 (1
-1
π
0
π
2π
−π
π
-2
−π
0
-1
0
1
2π
0
2
2π
x
z)
y π
0
y 0 0
0
1
0
0
2
2π
x
0
0
The argument for the function f (z) = p
jzjei arg
z=2
2π
-2
0
pz
2π
-1
0
2π 0
π
1
π
4
j1 zjei arg0 (1
z)=2
y 3π/2 −π/2
3π/2
-2
−π/2
π
-1
0
π 0
π/2
1
π/2
π/2
2
π/2
x
Residue at a simple pole at z1 = i, where by Rule 3, "p # p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 Res ;i = (z i) (z + i) p p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 = = 2z z=i p p p p j1 + ijei arg(1+i)=2 j1 ijei arg(1 i)=2 j1 + ije(i =4)=2 j1 ije(i7 =4)=2 = 2i 2i pp pp p p i 2 2e 2( 1) 2 i = = = =p 2i 2i 2i 2 150
π
2
π
x
Residue at a simple pole at z2 =
Res
=
"p
p
j1
i, where by Rule 3,
# p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 ; i = z2 + 1 p p j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 = = 2z z= i p p p i arg(1 i)=2 i arg(1+i)=2 ( i =4)=2 ije j1 + ije j1 ije j1 + ije(i =4)=2 = = 2i 2i pp pp p 1 2 2 2 = = = p i: 2i 2i 2
Residue at a simple pole at z3 = 1 , where by result from Exercise VII.8.13 and Rule 1,
Res [f (z) ; 1] = =
w2 1 ;0 = Res w(1 + w2 )
=
1,
lim
;0 =
w!0
p
jw2
Res 4
f (z) dz = Z 1 " p j1 + zjei arg
1 w2
1
3
1 ; 05 = w2 1+ w12
1jei arg(w w2 + 1
(1+z)=2
p
j1 2 z +1
zjei arg0 (1
2
z)=2
dz = !
and let " ! 0+ . This gives q Z 1 (1 ")2 2 " f (z) dz 1 + (1 ")2 2
Integrate along
q
1)=2
=
ei
=2
=
i
and let " ! 0+ . This gives
1
1+"
1 w
1 f w2
p
Integrate along Z
Res
2
Z
Z
1 "
1+"
1 " 1+"
p
p
1 x2 dx = 1 + x2
2,
151
p j1 + xjei 0=2 j1 1 + x2
2 "3=2 ! 0:
I:
xjei2
=2
dx !
Integrate along Z =
3,
and let " ! 0+ . This gives
f (z) dz = Z 1+" p j1 + zjei arg 3
(1+z)=2
p
j1 2 z +1
1 "
zjei arg0 (1
dz =
! and let " ! 0+ . This gives q Z 1 (1 ")2 2 " f (z) dz 1 + (1 ")2 4
Integrate along
Z
z)=2
Z
1 "
p
1 " 1+"
p p j1 + xjei 0=2 j1 x2 + 1
1 x2 dx = 1 + x2
1+"
I:
4,
2 "3=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0
I +0=2 i
1 1 p i+ p i 2 2
i.e., Z
1 1
p
1 x2 dx = 1 + x2
152
p
2
1 :
i
xjei 0=2
dx !
VII.8.5 1 2 3 P L K Show that the sum of the residues of a rational function on the extended complex plane is equal to zero. Solution I
jzj=R
Since
H
=
H
f (z) dz = 2 i I
jzj=R
, we get 2 i
X
Residues in …nite plane;
f (z) dz = 2 i Res [f; 1]
P
Res = 0.
153
VII.8.6 1 2 3 P L K Find the residue of z= (z 2 + 1) at each pole in the extended complex plane, and check that the sum of the residues is zero. Solution Residue at a simple pole at z1 = i, where by Rule 3, Res
z2
z z ;i = +1 2z
z=i
1 = : 2
Residue at a simple pole at z1 = i, where by Rule 3, Res
z2
z z ; i = +1 2z
z= i
1 = : 2
For residue at 1, we have that a
1
z2 1 = lim = 1: 2 z!1 z + 1 z!1 1 + 1=z 2
= lim zf (z) = lim z!1
hence Res
z2
z ;1 = +1
a
1
=
1;
and the sum of the residues is zero in the extended complex plane. (Since
z z 2 +1
=
a
1
z
+O
1 z2
as z ! 1).
154
VII.8.7 1 2 3 P L K Find the sum of the residues of (3z 4 + 2z + 1) = (8z 5 + 5z 2 + 2) at its poles in the (…nite) complex plane. Solution For residue at 1, we have that a
1
3z 5 + 2z 2 + z 3 + 2=z 3 + 1=z 4 3 = lim = ; 5 2 3 4 z!1 8z + 5z + z z!1 8 + 5=z + 1=z 8
= lim zf (z) = lim z!1
hence 3 : 8 We have that the sum of the residues in the complex …nite plane is thus the sum of residues in …nite complex plane is 3=8. Res [f; 1] =
f (z) =
a
1
z
+O
1 z2
as z ! 1, so again
155
Res [f (z) ; 1],
VII.8.8 1 2 3 P L K Fix n 1 and k 0. Find the residue of z k = (z n 1) at 1 by expanding 1= (z n 1) in a Laurent series. Find the residue of z k = (z n 1) at each …nite pole, and verify that the sum of all the residues are zero. Solution. Let f (z) =
f (z) =
zk zn
zk . zn 1
At 1, we have
zk = 1 1
n 1 zn
= zk
n
n
1+z
+z
2n
+ : : : = zk
Then Res [f; 1] = 1 if k mn = 1, thus k = mn m 1, and otherwise Res [f; 1] = 0.
n
+z k
2n
+z k
3n
1 for some integer
Let wj be an nth root of unity, i.e wj = e2 ij=n where 0 j n 1. Then f (z) has simple poles at w0 ; : : : ; wn 1 . Residue at a simple pole at zj = wj , where by Rule 3 and that wjn = 1 Res [f; wj ] =
zk nz n
1
= z=wj
z k+1 nz n
=
wjk+1 wjk+1 = : nwjn n
1
=
z=wj
Let S be the sum of residues, then 1 X k+1 1 S= wj = 1 + w + w2 + n j=0 n n 1
+ wn
1 n
1 wn 1 w
= 0;
unless w = 1. 2 i If w = 1, then wjk+1 = wj = 1 for every j and w = 1 , e n (k+1) = 1 , k+1 = m, thus k = mn 1 for some integer m 1. n Thus we have that the sum of the residues in the complex plane are zero.
156
+: : : :
VII.8.9 1 2 3 P L K Show that f (z) is analytic at 1, then Res [f (z) ; 1] =
lim z (f (z)
z!1
f (1)) :
Solution Since f (z) is analaytic at 1, we have a 1 +O z
1 z2
:
a 1 +O z
1 z2
a0
=a
f (1)) =
a
f (z) = a0 + Then
z (f (z)
f (1)) = z a0 +
1
+O
as z ! 1. From this follows that lim z (f (z)
z!1
157
1
= Res [f (z) ; 1] :
1 z
!a
1
VII.8.10 1 2 3 P L K Let D be an exterior domain. Suppose that f (z) is analytic on D [ @D and at 1. Show that Z f( ) 1 d = f ( ) f (1) ; z 2 D: 2 i @D z Solution. Cauchy integral formula for exterior domain. If D is an exterior domaina, f (z) is analytic in D and att 1, then Z f( ) f( ) 1 d = f (z) + Res ; 1 = f (z) f (1) 2 i @D z z IIII
Res
f( ) ;1 = z
lim
!1
f( ) = z
f (1) :
Or f ( ) = a0 + f( ) = z 1 2 i
Z
@D
f( ) 1 d = f (z) + z 2 i
a
1
a0 z
+
a
+O
2 2
+ ::: 1 2
Z
f( ) d = z jzj=R Z 1 a0 = f (z) d = f (z) 2 i jzj=R z
158
a0 :
VII.8.11 1 2 3 P L K If f (z) is not integrable at 1, we de…ne the principal value Z 1 Z R PV f (x) dx = lim PV f (x) dx: R!1
1
R
Show that
PV
Z
1 1
8 Im a > 0; < i ; 1 Im a = 0; dx = 0; : x a i ; Im a < 0:
Solution Set a = a1 + a2 i, where a1 and a2 is …xed. We compute the principal value for the following 3 cases. Case 1, Im a < 0 , a2 < 0. Integrate and let R ! 1. This gives Z
R R
dx x
= log
=
Z
R
dx = [log (x a1 a2 i
a R x R a1 a2 i + i arg (R R a1 a2 i
a1
a1 a2 i)
R a1 a2 i R a1 a2 i
a2 i)]RR = log i arg ( R !0+i 0
=
a2 i) !
a1
i =
i :
Case 2, Im a = 0 , a2 = 0. Integrate and let R ! 1 and " ! 0+ . This gives Z
R R
dx x
a
=
Z
= log "
R R
Z R dx dx + = [log jx a1 j]a1R " +[log jx x a x a 1 1 R a1 +" R a1 log jR + a1 j + log jR a1 j log " = log ! 0: R + a1
dx = x a1
Z
a1 "
Case 3, Im a > 0 , a2 > 0. Integrate and let R ! 1. This gives 159
a1 j]R a1 +" =
Z
R R
dx x
= log
=
Z
R
dx = [log (x a1 a2 i
a R x R a1 a2 i + i arg (R R a1 a2 i
a1
a1 a2 i)
a2 i)]RR = log i arg ( R !0+i 2
We have that PV
Z
1 1
8 Im a > 0; < i ; 1 Im a = 0; dx = 0; : x a i ; Im a < 0:
160
a1
R a1 a2 i R a1 a2 i a2 i) ! i =i :
=
VII.8.12 1 2 3 P L K Suppose that P (z) and Q (z) are polynomials such that the degree of Q (z) is strictly greater than the degree of P (z). Suppose that the zeros x1 ; : : : ; xm of Q (z) on the real axis are all simple, and set x0 = 1. Show that PV
Z
1 1
X X P (x) P (x) P (x) dx = 2 i Res ; zj + i Res ; zk ; Q (x) Q (x) Q (x)
summed over the poles zj of P (z) =Q (z) in the open upper halfplane and summed over the xk : s including 1. Hint. Use the preceding exercise. See also Exercise 6.7 Solution. By VII.8.11, formula holds for
1 z a
By Exercise b.7 (p. 215.) it holds if h i P (x) deg Q deg P + 2;in initial case Res Q(x) ; 1 = 0. By limiticity, it hods form sums, hence whenever deg Q deg P + 1.
161
VII.8.13 1 2 3 P L K Show that the analytic di¤erential f (z) dz transforms the change of variable w = 1=z to f (1=w) dw=w2 . Show that the residue of f (z) at z = 1 coincides with that of f (1=w) =w2 at w = 0. Solution P We have that Res [f (z) ; 1] = a1 if f (z) = 11 aj z j , jzj > R. We transform f (z) dz by the change of variable z = w1 , then dz = and f (z) dz = f w1 w12 dw. Now we take the residue for f (1=w) dw=w2 at w = 0, we have that
f
1 w
1 = w2
1 1 X 1 X aj ( aj ) w = w2 1 wj 1
a 4 w2
=
a 3w
j 2
a
= a 1w
2
1
a0 w
thus Res
f
1 w
1 ;0 = w2
a
1
We have that Res [f (z) ; 1] =
a1 = Res
162
f
1 dw, w2
1 w
1 ;0 : w2
2
;
VII.9.1 Show using residue theory that Z
1
0
p 3
1 x2
x3
dx =
2
p 3
3
:
Solution Set I=
Z
1
p 3
0
and integrate f (z) =
Z
0
1
p 3
1 z2
z3
1 x2
dx = q 3
x3
jzj2 e2i arg
along the contour in Figure VII.9.1.
163
dx
1 p z=3 3 j1
zjei arg0 (1
z)=3
VIII.9.1 arg z
arg0 z
y π −π
π
-2
−π
π
-1
y 0
−π
0
0
1
0
2
0
π
x
0
π
-2
π
arg z
π
arg0 (1
π
-1
0
π
2π
−π
π
-2
−π
-1
2π
0
2
0
−π
0
0
1
0
2
0
2π
x
0
2π
-2
0
The argument for the function f (z) = p 3
jzj2 e2i arg
0
p1
z=3 3
-1
2π
2π
0
0
j1 zjei arg0 (1
π
1
π
π
2
z)=3
y 4π/3 −2π/3
4π/3
-2
−2π/3
-1
4π/3
2π/3
−2π/3
0
π/3
1
π/3
π/3
2
π/3
x
Residue at a simple pole at z1 = 1 , where by result from Exercise VII.8.13 and Rule 1,
= Integrate along
Res 1,
x
y
π
Res [f (z) ; 1] =
2π
z)
y π
0
1
Res
1 f w2
1 w
;0 =
1 1 p ;0 = 3 w w 1
lim p 3
w!0
2
Res 4 jw
and let " ! 0+ . This gives 164
1 q w2 3
1 1 w2
1
1
1jei arg(w 1)=3
1 w
=
3
; 05 = e
i=3
π
x
Z
f (z) dz = 1
=
Z
1 "
q
=
Z
1 "
1 p jxj2 e2i 0=3 3 j1
q 3
"
Integrate along
2,
Z Integrate along Z
jzj2 e2i arg
3
"
xjei 2 =3
dz =
zjei arg0 (1 Z 2 i=3 dx = e
z)=3 1 "
p 3
"
!e
2 i=3
Z
0
1
p 3
1 x2
1 x2
x3
x3
dx = e
dx ! 2 i=3
I:
and let " ! 0+ . This gives f (z) dz
4
3,
1 p z=3 3 j1
"4 p " (1
")
2 "
2 "9=2 ! 0:
and let " ! 0+ . This gives
f (z) dz = 3
=
Z
"
1 "
=
Z
"
1 "
q 3
jzj2 e2i arg
1 q p 3 jxj2 e2i 0=3 3 j1
Integrate along
4,
Z
dz = zjei arg0 (1 z)=3 Z 1 " 1 p dx = dx ! 3 2 3 x x " i 0=3 xje Z 1 1 p ! dx = 3 x2 x3 0
I:
and let " ! 0+ . This gives f (z) dz
4
1 p z=3 3 j1
"4 p " (1
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 165
e
2 i=3
I +0
I +0=2 i
e
i=3
:
Solve for I, we obtain that 2 ie I= 1 e
i=3
2 i = i=3 e e
2 i=3
i=3
=
sin ( =3)
i.e., Z
0
1
p 3
1 x2
x3
dx =
166
2
p 3
3
:
=
2
p 3
3
;
VII.9.2 Show using residue theory that Z 1 1 p x x2 0
1
dx =
2
:
Solution
VII.9.2 γ2
γ4
γ3
-R
γ5
γ1 γ8
γ7
R
γ6
Set I=
Z
1
0
and integrate 1 f (z) = p z z2
1 x x2 p
1
dx
1 p = p i arg (z+1)=2 1 z jz + 1je 0 jz
along the contour in Figure VII.9.2.
167
1jei arg
(z 1)=2
VIII.9.2 arg0 z
arg z
y π π
π
-2
π
π
-1
y 0
π
2π
0
1
2π
0
2
π
x
2π
−π
arg0 (z + 1)
π
-2
arg0 (z
−π
π
-1
−π
π
π
-2
π
0
-1
0
0
2
0
2π
0
1
2π
0
2
π
x
2π
−π
The argument for the function f (z) = p z
π
-2
−π
π
-1
π
−π
0
1
−π
1p jz+1jei arg0 (z+1)=2 jz 1jei arg
(z
0
2
0
1)=2
y π 0
π
-2
0
π/2
-1
π/2
π/2 π/2
0
1
π
0
2
π
x
Residue at a simple pole at z1 = i, where by Rule 3, "
1 p Res p z jz + 1jei arg(z+1)=2 jz
Integrate along
1jei arg(z
1 p p jz + 1jei arg(z+1)=2 jz
1,
0
x
y 0
2π
0
1
1)
y π
0
1)=2
#
;0 =
1jei arg(z
=
1)=2 z=0
and let R ! 1 and " ! 0+ . This gives
168
1 e
i=2
=
i
0
x
Z
f (z) dz = 1
=
Z
R
1+"
z
=
p Z
jz +
1 p
1jei arg0 (z+1)=2
R
1+"
x
p
1
jx + Z R =
1jei 0=2
1+"
1 x x2 p
1jei arg
jz
p
1jei 0=2
jx 1
dx ! !
Integrate along
2,
Z Integrate along Z
(z 1)=2
Z
dz =
dx =
1
1
1 x x2 p
1
dx = I:
and let " ! 0+ . This gives f (z) dz 2
1 R R2 p
1
R
R
! 0:
and let R ! 1 and " ! 0+ . This gives
3,
f (z) dz = 3
=
Z
1 "
1 p p dz = i arg (z+1)=2 z jz + 1je 0 jz 1jei arg (z 1)=2 R Z 1 " 1 p p dx = = i =2 x jx + 1je jx 1jei =2 R Z R Z R 1 1 x= t p p dx = = dt = I: 2 2 dx = dt 1 1 1 " x x 1 t t
Integrate along
4,
Z Integrate along
and let " ! 0+ . This gives f (z) dz
4
5,
"4 p " (1
")
2 "
2 "9=2 ! 0:
and let R ! 1 and " ! 0+ . This gives 169
Z
f (z) dz = 5
=
Z
R 1 "
= Z Integrate along
Z
Z
Z
R
7,
jz +
1 " R
1 "
6,
Integrate along
z
p
1 p jz
1jei arg0 (z+1)=2
p x jx + 1jei
1 p x x2
1
=2
dx =
1 p
jx
1jei arg 1jei (
x= dx =
t dt
(z 1)=2
)=2
=
Z
dz =
dx =
1
R
1 p 2 t t
1
dt = I:
and let " ! 0+ . This gives f (z) dz 6
1 R R2 p
1
R
R
! 0:
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 7
=
Z
1+"
R
Integrate along
1 p dz = z jz + 1jei arg0 (z+1)=2 jz 1jei arg (z 1)=2 Z 1+" 1 p p = dx = x jx + 1jei 2 =2 jx 1jei 0=2 R Z R 1 p = dx ! 2 1 1+" x x Z 1 1 p ! dx = I: x x2 1 1 p
8,
Z
and let " ! 0+ . This gives f (z) dz
4
"4 p " (1
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that 170
I + 0 + I + 0 + I + 0 + I + 0 = 2 i ( i) ; i.e., Z
0
1
1 x x2 p
171
1
dx =
2
:
VII.9.3 Show using residue theory that Z 1 1 p 2 1 1 (x + 1)
x2
dx = p
2
Solution
VII.9.3
γ8
γ7 γ5
-R
γ4
γ1
γ2
γ3
R
γ6
Set I=
Z
1 1
1 p (x2 + 1) 1
x2
dx
and integrate
f (z) =
1 p (z 2 + 1) 1
x2
=
(z
1 p p i) (z + i) j1 + zjei arg (1+z)=2 j1
along the contour in Figure VII.9.3.
172
zjei arg0 (1
z)=2
VIII.9.3 arg z
arg0 z
y π −π
π
-2
−π
-1
y
π
0
−π
0
0
1
0
0
2
π
x
0
π
-2
π
arg (1 + z)
arg0 (1
π
-1
π
0
π
2π
−π
π
-2
−π
-1
1
2π
0
2
x
2π
z)
y π
0
y
0
0
0
0
0
1
0
0
2
2π
x
0
2π
-2
0
The argument for the function f (z) =
(z i)(z+i)
p
0
2π
2π
0
0
-1
1 j1+zjei arg
(1+z)=2
p
π
1
π
π
2
j1 zjei arg0 (1
x
π
z)=2
y 3π/2 −π/2
3π/2
-2
−π/2
π
-1
π
0
0
π/2
1
π/2
π/2
2
π/2
x
Residue at a simple pole at z1 = i, where by Rule 3,
Res
"
(z
i) (z + i)
p
1 1
p
j1
zjei arg(1
z)=2
;i =
p = j1 + zjei arg(1+z)=2 j1 zjei arg(1 z)=2 z=i 1 1 p p p = p i arg(1+i)=2 i arg(1 i)=2 (i =4)=2 (i + i) j1 + zje j1 zje 2i j1 + zje j1 zje(i7 1 1 1 1 p = p i = pp pp = p = 2i 2( 1) 2 2i 2 2 2i 2 2e i (z + i)
p
j1 + zjei arg(1+z)=2
#
173
=4)=2
Residue at a simple pole at z2 =
Res
"
(z
i) =
1
p
j1 + zjei arg(1+z)=2
(z
i)
p
j1 +
p i) j1 + zjei arg(1
Integrate along Z
=
zjei arg(1
j1
i)=2
=
zjei arg(1+z)=2 p j1
p
j1
z)=2
#
; i =
zjei arg(1
zjei arg(1+i)=2
=
1 pp pp = 2i 2 2e0i
=
z)=2
p
z= i
1
p 2i j1 + j1 1 1 1 p = p = p i 2i 2 2 2i 2 2 zje( i =4)=2
zje(i
and let " ! 0+ . This gives
f (z) dz = 1
= Z
1,
p
1
1
( i
i, where by Rule 3,
1 " 1+"
Z
1 "
1
p
p dz = j1 + zjei arg (1+z)=2 j1 zjei arg0 (1 z)=2 Z 1 " 1 1 p p p dx = 2 1 (z 2 + 1) j1 + xjei0=2 j1 xjei2 =2 1+" (x + 1) Z 1 1 p dx = 2 1 x2 1 (x + 1) 1+"
(z 2 + 1)
Integrate along
2,
Z Integrate along
and let " ! 0+ . This gives f (z) dz
4
3,
"4 p " (1
")
2 "
and let " ! 0+ . This gives
174
2 "9=2 ! 0:
x2 I:
dx !
=4)=2
=
Z
f (z) dz = 1
= =
Z
1+"
(z 2
1 "
Z
+ 1)
1+"
(z 2
1 "
+ 1)
Integrate along
p
4,
1
p
j1 +
zjei arg (1+z)=2
j1 + 1
xjei 0=2
p
j1
p
xjei 0=2 !
Z
j1
dx =
zjei arg0 (1 Z 1 " 1+"
1 1
(x2
z)=2
(x2
1 p + 1) 1
dz =
1 p + 1) 1 x2
dx =
x2
dx !
I:
and let " ! 0+ . This gives Z f (z) dz ! 0: 4
Integrate along Z
=
and let R ! 1 and " ! 0+ . This gives
f (z) dz = 5
= Z
5,
R 1 "
Z
R 1 "
1 p p (z 2 + 1) j1 + zjei arg(1+z)=2 j1
(z 2 + 1)
Integrate along Z Integrate along
1
p
j1 + xjei(
6,
and let R ! 1. This gives
f (z) dz 6
7,
dz = zjei arg(1 z)=2 Z R 1 p p dx ! dx = 2 )=2 1 x2 ( i) j1 xjei0=2 1 " (x + 1) Z 1 1 p ! dx = J: 2 (x + 1) 1 x2 ( i) 1
(R2
1 p 1) R2
1
2 R
2 ! 0: R3
and let R ! 1 and " ! 0+ . This gives
175
Z
f (z) dz = 7
= =
Z
1 " R
Z
1 "
(z 2
R
(z 2
+ 1)
Integrate along
j1 +
1
p
j1 +
xjei =2
zjei arg(1+z)=2
p
j1 !
Z
p j1
xjei2 =2
dx =
1
(x2
1
zjei arg(1 Z R
1 "
1 p + 1) 1
z)=2
(x2
dz = 1 p + 1) 1
x2 ( i)
dx =
x2 ( i) J:
and let " ! 0+ . This gives
8,
Z
+ 1)
1
p
"4 p " (1
f (z) dz 4
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that I +0
I
J +0+J +0=2 i
1 1 p i+ p i 2 2 2 2
i.e., Z
1 1
(x2
1 p + 1) 1
176
x2
dx = p : 2
dx !
VII.9.4 Show using residue theory that Z 1 1 2 q dx = p : 3 1 3 (1 x) (1 + x)2 Solution Set I=
Z
and integrate
f (z) = q 3
1 (1
z) (1 + z)2
1 1
q 3 (1
= p 3
j1
1
dx 2
x) (1 + x)
zjei arg0 (1
along the contour in Figure VII.9.4.
177
1 q z)=3 j1 + zj2 e2i arg
(1+z)=3
VIII.9.4 arg0 z
arg z
y
y π π
π
-2
arg0 (1
π
π
-1
0
π
2π
0
1
2π
π
0
2
x
2π
π
-2
−π
z)
−π
π
-1
−π
0
2π
-2
0
0
0
1
0
0
2
0
x
arg (1 + z)
y 2π
0
-1
y
2π
2π
0
0
π
1
π
π
2
π
x
π
−π
The argument for the function f (z) = p 3
π
-2
j1 zjei arg0 (1
-1
−π
z)=3
1 p
0
0
0
0
j1+zj2 e2i arg
0
1
0
0
2
(1+z)=3
y 4π/3 −2π/3
4π/3
-2
−2π/3
2π/3
-1
0
2π/3 0
π/3
1
π/3
π/3
2
π/3
x
Residue at a simple pole at z1 = 1 , where by result from Exercise VII.8.13 and Rule 1,
178
0
x
Res [f (z) ; 1] =
1 1 1 ; 0 = Res 4 2 q f 2 w w w 3 1 3 2 1 1 ; 05 = Res 4 q w 3 2 (w 1) (w + 1)
Res
= 2
lim 4 p
w!0
Integrate along Z
=
3
jw
1jei arg(w
1)=3
1 q
jw + 1j2 e2i arg(w+1)=3
and let " ! 0+ . This gives
1 1 w
3
5=
1+
1 2 w
e
i=3
3
; 05 =
f (z) dz = 1
= Z
1,
2
Z
1 " 1+"
1 " 1+"
p 3 j1
xjei2
Integrate along
2,
Z Integrate along
p 3
dz = j1 + zj2 e2i arg (1+z)=3 Z 1 " 1 1 2 i=3 q q dx = e dx ! 1+" 3 (1 =3 j1 + xj2 e2i 0=3 x) (1 + x)2 Z 1 1 2 i=3 q !e dx = e 2 i=3 I: 2 3 1 (1 x) (1 + x)
j1
3,
zjei arg0 (1
z)=3
and let " ! 0+ . This gives f (z) dz
4
1 q
"4 p " (1
")
2 "
and let " ! 0+ . This gives
179
2 "9=2 ! 0:
Z
f (z) dz = 3
=
1+"
1 "
1+"
1 q dz = 1 " j1 zjei arg0 (1 z)=3 j1 + zj2 e2i arg (1+z)=3 Z 1 " 1 1 q q dx ! dx = p 2 2i 0=3 2 3 3 1+" i 0=3 j1 xje j1 + xj e (1 x) (1 + x) Z 1 1 q ! dx = I: 2 1 3 (1 x) (1 + x)
= Z
Z
p 3
Integrate along
4,
Z
and let " ! 0+ . This gives "4 p " (1
f (z) dz 4
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that e
2 i=3
I +0
I +0=2 i
e
i=3
:
Solve for I, we obtain that I=
i=3
2 ie e
2 i=3
1
=
e
i=3
2 i e
i=3
=
2 =p ; sin ( =3) 3
i.e., Z
1 1
1
q 3
(1
2 dx = p : 3 x) (1 + x)2
180
VII.9.5 Show using residue theory that Z 1 1 q 0 (x + 1) 4 x (1
dx = p : 4 2 x) 3
Solution Set I=
Z
0
and integrate
f (z) =
1 q (z + 1) 4 z (1
1
1 q (x + 1) 4 x (1
= z)3
dx x)
p (z + 1) 4 jzjei arg
along the contour in Figure VII.9.5.
181
3
z=4
1 q 4
j1
zj3 e3i arg0 (1
z)=4
VIII.9.5 arg z
arg0 z
y π −π
π
-2
−π
π
-1
y 0
−π
0
0
1
0
0
2
π
x
0
π
arg z
π
-2
π
-1
π
arg0 (1
π
−π
π
-2
−π
-1
2π
0
1
2π
0
2
2π
x
z)
y π
0
y
π
0
−π
0
0
1
0
0
2
2π
x
0
The argument for the function f (z) =
0
p
2π
-2
(z+1) 4 jzjei arg
-1
0
1 p
z=4 4
−π/4
7π/4
-2
Residue at a simple pole at z1 =
−π/4
-1
7π/4
3π/2
−π/4
3π/4
3π/4
1
3π/4
3π/4
2
1, where by Rule 3,
182
3π/4
2π
0
0
π
1
j1 zj3 e3i arg0 (1
y 7π/4
2π
x
π
z)=4
π
2
π
x
2
Res 4
(z + 1)
1 q 4 i arg(z)=4 jzje j1
p 4
zj3 e3i arg(1 z)=4
1 q jzjei arg(z)=4 4 j1
p 4
3
; 15 = =
zj3 e3i arg(1
z)=4 z= 1
1
q = p 4 4 i arg(z)=4 jzje j1
= zj3 e3i arg(1 z)=4
1 1 1 1 1 1 p + ip = p = p +i p 4 4 i=4 8e 8 2 2 2 2 242 Residue at a simple pole at z2 = 1 , where by result from Exercise VII.8.13 and Rule 1, = p 4
Res [f (z) ; 1] =
Res
1 f w2
1 w
2
Res 4
;0 =
=
2
lim 4
w!0
1 w2
(1 + w)
1 w
+1
1 q 4
(w
1 q
1 w
4
3
1)
Residue at a simple pole at z2 = i, where by Rule 3, Integrate along 1 , and let R ! 1 and " ! 0+ . This gives Z
; 05 =
; 05 = 0
f (z) dz = 1
=
Z
"
=
1 3
1 3 w
3
Z
"
1 "
1 "
(z + 1)
p 4
jzjei arg
1 q p (x + 1) 4 jxjei 0=4 4 j1
1 q z=4 4 j1
dz = zj3 e3i arg0 (1 z)=4
dx = i
xj3 e3i 2
183
"
=4
!i
Z
Z
0
1
1 "
1 q (x + 1) 4 x (1
1 q (x + 1) 4 x (1
x)3
dx = iI:
3
x)
dx !
Integrate along
2,
Z Integrate along Z
3
=
4
3,
f (z) dz = Z " =
(z + 1) 1
p 4
jzjei arg z=4
q p 4 4 i 0=4 (x + 1) jxje j1
Integrate along
4,
Z
2 "
")
2 "9=2 ! 0:
and let R ! 1 and " ! 0+ . This gives
"
1 "
"4 p " (1
f (z) dz
1 "
Z
and let " ! 0+ . This gives
1 q 4
dz = zj3 e3i arg0 (1 z)=4
j1
dx =
xj3 e3i 0=4
"
Z
!
Z
1
0
1 "
1 q (x + 1) 4 x (1
1 q (x + 1) 4 x (1
dx =
3
x)
dx !
I:
3
x)
and let " ! 0+ . This gives f (z) dz
4
"4 p " (1
")
2 "
2 "9=2 ! 0:
Using the Residue Theorem and letting R ! 1 and " ! 0+ , we obtain that iI + 0
I +0=2 i
1 1 p +i p +0 ; 4 2 2 242
i.e., Z
0
1
1 q (x + 1) 4 x (1
184
dx = p : 4 2 x) 3
VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8
1
VIII.1.1 1 2 3 P
L K LLL Show that z 4 + 2z 2 z + 1 has exactly one root in each quadrant.
Solution
VIII.1.1
Set p (z) = z 4 +2z 2 z +1, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.1. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, 4 4it 2 2it it and our function p (z) becomes p (t) = R e + 2R e Re + 1. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it . We …nd that the real part have a zero at t = 1, and the imaginary part have a zero for t = 0. We make the following table and do the sketches,
2
t 1 1 0
Re z + 0 1
Im z
arg z 0
VIII.1.1 (R = 4)
VIII.1.1 (R = 4) 4 3
2
0
2
0
1 -2 -1 -1
1
2
-2
Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) have a root in the fourth quadrant too. 2 Becauce that p (x) = (x2 + 1) x > 0 it is clear that p (z) have no real roots and p (z) must have on has one zero in each quadrant.
3
3
4
VIII.1.2 1 2 3 P L K Find the number of zeros of the polynomial p (z) = z 4 +z 3 +4z 2 +3z+2 in each quadrant.
Solution
VIII.1.2
Set p (z) = z 4 + z 3 + 4z 2 + 3z + 2, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.2. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, and our function p (z) becomes p (t) = t4 +t3 +4t2 +3t+2 > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and 4 4it 3 3it 2 2it it our function p (z) becomes p (t) = R e + R e + 4R e + 3Re + 2. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 + 3it + 2 = t4 4t2 + 2 + i ( t3 + 3t) . We …nd that the real part have a zero
4
p p p p at t = 2 2 and t = 2 + 2, and the imaginary part have a zero for p t = 0 and t = 3. We make the following table and do the sketches, t 1 p
p 2 + 2 p p3 p 2 2 0
Re z + 0 0 +
Im z
arg z 0
VIII.1.2 (R = 10)
VIII.1.2 (R = 10) 4
2
0 + 0
2 3 2
-4
2
-2
2 -2 -4
Thus when we move from iR to 0, the argument for p (z) on this line segment on the imaginary axis changes so that, 4 arg p (z) 2 . Now we have that the total change of argument for p (z) is 0, so we have no zero in the 1st quadrant. Since p (z) has real coe¢ cients, the roots come in conjugate pairs (since there are nor real roots), therefore there are no roots in the 4th quadras as well. By symmetry, there are 2 roots each in the 2nd and 3rd quadrants.
5
4
VIII.1.3 1 2 3 P L K Find the number of zeros of the polynomial p (z) = z 6 + 4z 4 + z 3 + 2z 2 + z + 5 in the …rst quadrant fRe z > 0; Im z > 0g. Solution
VIII.1.3
Set p (z) = z 6 + 4z 4 + z 3 + 2z 2 + z + 5, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.3. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, and our function p (z) becomes p (t) = t6 + 4t4 + t3 + 2t2 + t + 5 > 0. Since p (t) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and our function p (z) becomes p (t) = R6 e6it +4R4 e4it +3R3 e3it +2R2 e2it +Reit +5. Since the variation in argument is determined by the dominating term R6 e6it for R large the change of argument on this arc is 6 times the variation in argument for the arc, and thus 4 arg p (z) 6 2 = 3 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)6 + 4 (it)4 + (it)3 + 2 (it)2 + it + 5 = t6 + 4t4 2t2 + 5 + i ( t3 + t) . We …nd that the real part have a zero at t 1; 95, and the imaginary part have a zero for t = 0 and t = 1. We make the following table and do the sketches, 6
t 1 1; 95 1 0
Re z
Im z
arg z
VIII.1.3 (R = 3)
VIII.1.3 (R = 3)
1000
0 + +
2
0 0
-2
500
0 0
2
4
-2 -4
-1000 -500 -500
500 1000 -6
Thus when we move from iR to 0, the argument for p (z) on this line segment on the imaginary axis changes so that, 4 arg p (z) . Now we have that the total change of argument for p (z) is 4 , so we have that p (z) has exactly two zeros in the …rst quadrat. (None on real or imaginary axis)
7
6
8
10
VIII.1.4 1 2 3 P
L K LLL Find the number of zeros of the polynomial p (z) = z 9 +2z 5 2z 4 +z+3 in the right half-plane. Solution
VIII.1.4
Set p (z) = z 9 + 2z 5 2z 4 + z + 3, and compute 4 arg p (z) along the three segments in the sector path in …gure VIII.1.4. First of all p (z) have no real zeros. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, and our function p (z) becomes p (t) = t9 + 2t5 2t4 + t + 3. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and 9 9it 5 5it 4 4it it our function p (z) becomes p (t) = R e + 2R e 2R e + Re + 3. Since the variation in argument is determined by the dominating term R9 e9it for R large the change of argument on this arc is 9 times the variation in argument for the arc, and thus 4 arg p (z) 9 2 = 92 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)9 + 2 (it)5 2 (it)4 + 2 it + q 3 = 2t4 + 3 + iy (y 4 + 1) . We …nd that the real part have a zero at t = 4 32 , and the imaginary part have only a zero for t = 0. We make the following table and do the sketches, 8
t 1 q 4
0
Re z 3 2
0 +
Im z + + 0
arg z
VIII.1.4 (R = 2)
VIII.1.4 (R = 2)
600 2
30
400
2
200
20
0 -600-400-200 -200
200 400 600
10
-400 -600
-20
-10
Thus when we move from iR to 0, the argument is changed from =2 to 0, so the charge of argument on this on the imaginary axis is , thus 4 arg p (z) = =2. Now we have that the total change of argument for p (z) is 4 , so we have exactly two zeros in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has four zeros in the right half-plane.
9
10
20
VIII.1.5 1 2 3 P L K For a …xed real number , …nd the number of zeros z 4 +z 3 +4z 2 + z+3 satisfying Re z < 0. (Your answer depends on .) Solution
VIII.1.5
Set p (z) = z 4 + z 3 + 4z 2 + z + 3, and compute 4 arg p (z) along the three segments in the contuour in …gure VIII.1.5. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + t3 + 4t2 + at + 3 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and 4 4it 3 3it 2 2it it our function p (z) becomes p (t) = R e + R e + 4R e + aRe + 3. Since the variation in argument is determined by the dominating term R4 e4it for R large (independent of the value of a) the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 + ait + 3 = (t2 1) (t2 p3) + i ( t3 + at). We …nd that the real part have a zero at t = 1 and t = p3 and the imaginary part have a zero for t = 0 and and a second root t = a if a 0. We make the following table and do the sketches, 10
a<0 t 1 p 3 1 0
Re z + 0 0 +
Im z
0
0
a=0 t 1 p 3 1 0
Re z + 0 0 +
Im z
arg z 0
0
VIII.1.5 (a = -4, R = 10)
VIII.1.5 (a = -4, R = 10)
VIII.1.5 (a = 0, R = 10)
VIII.1.5 (a = 0, R = 10)
VIII.1.5 (a = 1/2, R = 10)
VIII.1.5 (a = 1/2, R = 10)
arg z 0 2 2
2 2
0
0
Im z
arg z 0 2 2
0 0
0 0
11
a=1 t 1 p 3 1 0
Re z + 0 0 +
1
a=3 t 1 p 3 1 0
Re z + 0 0 +
Im z
VIII.1.5 (a = 1, R = 10)
VIII.1.5 (a = 1, R = 10)
VIII.1.5 (a = 2, R = 10)
VIII.1.5 (a = 2, R = 10)
VIII.1.5 (a = 3, R = 10)
VIII.1.5 (a = 3, R = 10)
arg z 0 2
0 0
Im z
0
arg z 0 2
0 + 0
Im z 0 + 0
3 2
2
arg z 0 2
0
12
a>3 t 1 p pa 3 1 0
VIII.1.5 (a = 9, R = 10) Re z + + 0 0 +
Im z 0 + + 0
VIII.1.5 (a = 9, R = 10)
arg z 0 0 2 2
0
Thus when we move from iR to 0, the argument for p (z) only change if 1 < a < 3 and we have that 4 arg p (z) 2 , for all other value on a we have that 4 arg p (z) 0. Remark that we can se that for a = 1 and a = 3 we have zeros on the imaginary axis. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. If 1 3, the total change of argument for p (z) is 0 and no zeros on the imaginary axis, thus we must have four zeros in the open left half-plane. If < 1 and > 3, the total charge of argument for p (z) is 2 and no zeros of the imaginary axis, thus we must have two zeros in the open left half-plane. If = 1 and = 3, the total charge of argument for p (z) is 2 and we have two zeros on the imaginary axis, thus we must have remaining two zeros in the open left half-plane.
13
VIII.1.6 1 2 3 P L K For a …xed real number , …nd the number of solutions of z 5 +2z 3 z 2 +z = satisfying Re z > 0. Solution
VIII.1.6
Set p (z) = z 5 + 2z 3 z 2 + z , and compute 4 arg p (z) along the two segments in the contuour in …gure VIII.1.6. The arc 1 from iR to iR we parametizize as 2 : z = Reit , =2 t =2, and our function p (z) becomes p (t) = R5 e5it +2R3 e3it R2 e2it +Reit . Since the variation in argument is determined by the dominating term R5 e5it for R large (independent of the value of ) the change of argument on this arc is 5 times the variation in argument for the arc, and thus 4 arg p (z) 5 2 = 52 . The positive imaginary axis 2 from iR to iR we parametizize as 2 : z = it; R t R, and our function p (z) becomes f (t) = (it)5 + 2 (it)3 2 (it)2 +pit = t2 p + it (t2 1) . We …nd that the real part have a zero at and t = and the imaginary part have a zero for t = 1, t = 0 t= and t = 1. We make the following table and do the sketch,
14
VIII.1.6 (a = 4, R = 4)
t 1
Re z
Im z
arg z
VIII.1.5 (a = 2, R = 10) t 1 p 3 p a 1 0
Re Im + 0 0 0 0 + 3 0
Re z
Im z
VIII.1.5 (a = 2, R = 10)
arg 0 2 3 2
2
VIII.1.5 (a = 0, R = 10)
t 1
VIII.1.6 (a = 4, R = 4)
arg z
15
VIII.1.5 (a = 0, R = 10)
Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) = 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant.
16
VIII.1.7 1 2 3 P L K For a …xed complex number , show that if m and n are large integers, then the equation ez = z + has exactly m + n solutions in the horizontal strip f 2 im < Im z < 2 ing. Solution
VIII.1.7
Set p (z) = z 4 +2z 2 z +1, and compute 4 arg p (z) along the three segments in the contuour in …gure VIII.1.7. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and our function p (z) becomes p (t) = R4 e4it + 2R2 e2it Reit + 1. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it . We …nd that the real part have a zero at t = 1, and the imaginary part have a zero for t = 0 on the positive imaginary axis. We make the following table and do the sketch,
17
VIII.1.7 t 1 1 0
Re Im + 0 + 0
arg 0
20 10
2
0
-20
-10 -10
10
20
-20
Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) = 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant.
18
VIII.1.8 1 2 3 P
L K LLL Show that if Re > 1, then the equation ez = z + solution in the left half-plane.
has exactly one
Solution
VIII.1.8
Set p (z) = z 4 +2z 2 z +1, and compute 4 arg p (z) along the three segments in the contuour in …gure VIII.1.8. The positive real axis 1 from 0 to R we parametizize as x : z = t; 0 t R, 2 and our function p (z) becomes p (t) = t4 + 2t2 t + 1 = (t2 + 1) t > 0. Since p (x) > 0 for t 0, their is no change of argument on this line segment on the real axis, thus 4 arg p (z) = 0. The arc 2 from R to iR we parametizize as 2 : z = Reit , 0 t =2, and our function p (z) becomes p (t) = R4 e4it + 2R2 e2it Reit + 1. Since the variation in argument is determined by the dominating term R4 e4it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p (z) 4 2 = 2 . The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z = it; 0 t R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 it + 1 2 = (t2 1) it . We …nd that the real part have a zero at t = 1, and the imaginary part have a zero for t = 0 on the positive imaginary axis. We make the following table and do the sketch,
19
VIII.1.8 t 1 1 0
Re Im + 0 + 0
arg 0
20 10
2
0
-20
-10 -10
10
20
-20
Thus when we move from iR to 0, the argument for p (z) remain in the 4-th quadrant except for touching the imaginary axis at i and terminating at 1, and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p (z) = 0. Now we have that the total change of argument for p (z) is 2 , so we have exactly one zero in the …rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p (z) has one zero in each quadrant. Set f (z) = ez z , Re > 1, and get f (iy) = cos y Re + i (sin y y) i Im . Thus when we move from f ( iR) to f (iR), f (iy) remains in the left half-plane and the change in arg f is . When we move from f (iR) to f ( iR) along f Rei , 2 < < 32 , we cross the real axis from the 4-th to the 1 st quadrant, thus the change in arg f is again . (May also use Rouche’s Theorem.)
t 1 q 4
0
Re Im + 3 2
0 +
+ +
arg
VIII.1.4 (R = 2)
2
100
2
50
VIII.1.4 (R = 2) 30 20
0 -100
-50
50
100
10
-50 -100
20
-20
-10
10
20
t 1 q 4
0
Re Im + 3 2
0 +
+ +
arg
VIII.1.4 (R = 2)
2
400
2
200
VIII.1.4 (R = 2) 30 20
0 -400 -200
200
400
10
-200 -20
-400
(it)4
(it)3 + 13 (it)2
it +36 = t4 + it3
21
13t2
it + 36
-10
10
20
VIII.1.9 1 2 3 P L K Show that if f (z) is analytic in a domain D, and if is a closed curve in D such that the values of f (z) on lie in the slit plane Cn ( 1; 0] then the increase in the argument of f (z) around is zero. Solution
22
VIII.2.1 1 2 3 P
L K LLL Show that 2z 5 + 6z 1 has one root in the interval 0 < x < 1 and four roots in the annulus f1 < jzj < 2g. Solution Set …rst p (z) = f1 (z) + h1 (z), where f1 (z) = 2z 5 and h1 (z) = 6z 1. On the circle jzj = 2 we have, jf1 (z)j = j2z 5 j = 2 25 = 64 and jh1 (z)j = j6z 1j 13, then jf1 (z)j > jh1 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has all 5 roots in the disk jzj = 2. Set now p (z) = f2 (z) + h2 (z), where f2 (z) = 6z and h2 (z) = 2z 5 1. On the circle jzj = 1 we have, jf2 (z)j = j6zj = 6 and jh2 (z)j = j2z 5 1j 3, then jf2 (z)j > jh2 (z)j on jzj = 1 so by Rouche’s Theorem p (z) has 1 roots in the disk jzj = 1. It follows that p (z) = 2z 5 + 6z 1 has 5 1 = 4 roots in the annulus 1 < jzj < 2. As any complex roots come in conjugate pairs that have the same magnitude, the root in the disk jzj = 1 must be real, we have that p (0) = 1 and p (1) = 7, thus this root must be in the interval 0 < x < 1.
23
VIII.2.2 1 2 3 P
L K LLL How many roots does z 9 + z 5 fjzj = 1g and fjzj = 2g?
8z 3 + 2z + 1 have between the circles
Solution Set …rst p (z) = f1 (z)+h1 (z), where f1 (z) = z 9 and h1 (z) = z 5 8z 3 +2z+1. On the circle jzj = 2 we have, jf1 (z)j = jz 9 j = jzj9 = 29 = 512 and jh1 (z)j = jz 5 8z 3 + 2z + 1j 101, then jf1 (z)j > jh1 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has 9 roots in the disk jzj = 2. Set now p (z) = f2 (z) + h2 (z), where f2 (z) = z 9 + z 5 + 2z + 1 and h2 (z) = 8z 3 . On the circle jzj = 1 we have, jf2 (z)j = j 8z 3 j = 8 and jh2 (z)j = jz 9 + z 5 + 2z + 1j 5, then jf2 (z)j > jh2 (z)j on jzj = 2 so by Rouche’s Theorem p (z) has 3 roots in the disk jzj = 1. Thus it follows that p (z) = z 9 + z 5 8z 3 + 2z + 1 has 9 3 = 7 roots in the annulus 1 < jzj < 2.
24
VIII.2.3 1 2 3 P
L K LLL Show that if m and n are positive integers, the the polynomial z2 + 2! has exactly n zeros in the unit disk. p (z) = 1 + z +
+
zm + 3z n m!
Solution m 2 Set p (z) = f (z) + h (z), where f (z) = 3z n and h (z) = 1 + z + z2! + + zm! . On the circle jzj = 1 we have, 2 m jf (z)j = j3z n j = 3 jzjn = 3 1n = 3 and jh (z)j = 1 + z + z2! + + zm! P1 1 k=0 k! = e, then jf (z)j > jh (z)j on jzj = 1 so by Rouche’s Theorem 2 m p (z) = 1 + z + z2! + + zm! + 3z n has exactly n roots in the unit disk jzj = 1.
25
VIII.2.4 1 2 3 P
L K LLL Fix a complex number such that j j < 1. For n 1, show that n z (z 1) e has n zeros satisfying jz 1j < 1 and no other zeros in the right half-plane. Determine the multiplicity of the zeros. Solution Set p (z) = f (z) + h (z), where f (z) = (z 1)n ez and h (z) = . On the i circle jz 1j = 1 which we parametrize by z = 1 + e , 0 2 we have, n 1+i n z i 1+cos 1 jf (z)j = j(z 1) e j = 1 + e e =e 1, then 0 2 and jh (z)j = j j = j j < 1 from the text, then jf (z)j > jh (z)j on jz 1j = 1 so by Rouche’s Theorem p (z) = (z 1)n ez has n roots in the disk jz 1j = 1. We have that f 0 (z) = n (z 1)n 1 ez + (z 1)n ez = (n + z 1) (z 1)n 1 ez , thus the zeros of of f 0 (z) are at z = 1 and at z = 1 n, where the only zeros in the right half-plane are at z = 1. Since f (1) = , the zeros of f (z) must be simple unless = 0, in which case we ge a zero of order n at z = 1. *********** Här är lite konstigt den derivarande satsen, hur funkar den med nollställen ***********
26
VIII.2.5 1 2 3 P L K For a …xed satisfying j j < 1, show that (z 1)n ez + (z + 1)n has n zeros in the right half-plane, which are all simple if 6= 0. Solution
27
VIII.2.6 1 2 3 P L K Let p (z) = z 6 + 9z 4 + z 3 + 2z + 4 be the polynomial treated in the example in this section. (a) Determine which quadrants contain the four zeros of p (z) that lie inside the unit circle. (b) Determine which quadrants contain the two zeros of p (z) that lie outside the unit circle. (c) Show that the two zeros of p (z) that lie outside the unit circle satisfy fjz 3ij < 1=10g. Solution
28
VIII.2.7 1 2 3 P
L K LLL Let f (z) and g (z) be analytic functions on the bounded domain D that extend continuously to @D and satisfy jf (z) + g (z)j < jf (z)j + jg (z)j on @D. Show that f (z) and g (z) have the same number of zeros in D, counting multiplicity. Remark. This is a variant of Rouche’s theorem, in which the hypotheses are symmetric in f (z) and g (z). Rouche’s theorem is obtained by setting h (z) = f (z) g (z). For the solution of the exercise, see Exercise 9 in the preceding section. Solution
29
VIII.2.8 1 2 3 P L K Let D be a bounded domain, and let f (z) and h (z) be meromorphic functions on D that extend to be analytic on @D. Suppose that jh (z)j < jf (z)j on @D. Show by example that f (z) and f (z) + h (z) can have di¤erent numbers of zeros on D. What can be said about f (z) and f (z) + g (z)? Prove your assertion. Solution
30
VIII.2.9 1 2 3 P L K Let f (z) be a continuously di¤erentiable function on a domain D. Suppose that for all complex constants a and b, the increase in the argument of f (z)+ az +b around any small circle in D on which f (z)+az +b 6= 0 is nonnegative. Show that f (z) is analytic. Solution
31
VIII.3.1 1 2 3 P
L K LLL Let ffk (z)g be a sequence of analytic functions on D that converges normally to f (z), and suppose that f (z) has a zero of order N at z0 2 D. Use Rouche’s theorem to show that there exists > 0 such that for k large, fk (z) has exactly N zeros counting multiplicity on the disk fjz z0 j < g. Solution
32
VIII.3.2 1 2 3 P
L K LLL Let S be the family of univalent functions f (z) de…ned on the open unit disk fjzj < 1g that satisfy f (0) = 0 and f 0 (0) = 1. Show that S is closed under normal convergence, that is, if a sequence in S converges normally to f (z), then f 2 S. Remark. It is also true, but more di¢ cult to prove, that S is a compact family of analytic functions, that is, every sequence in S has a normally convergent subsequence. Solution
33
VIII.4.1 1 2 3 P
L K LLL Suppose D is a bounded domain with piecewise smooth boundary. Let f (z) be meromorphic and g (z) analytic on D. Suppose that both f (z) and g (z) extend analytically across the boundary of D, and that f (z) 6= 0 on @D. Show that 1 2 i
I
@D
X f 0 (z) dz = mj g (zj ) ; f (z) j=1 n
g (z)
where z1 ; : : : ; zn are the zeros and poles of f (z), and mj is the order of f (z) at zj . Solution
34
VIII.4.2 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane that is doubly periodic. Suppose that the zeros and the poles of f (z) are at the points z1 ; : : : ; zn and at their translates by periods of f (z), and suppose no zj is a translate of f (z) at zj . Show P by a period of another zk . Let mj be the order that mj zj is a period of f (z). Hint. Integrate zf 0 (z) =f (z) around the boundary of the fundamental parallelogram P constructed in Section VI.5. Solution
35
VIII.4.3 1 2 3 P
L K LLL Let ffk (z)g be a sequence of analytic functions on a domain D that converges normally to f (z). Suppose that fk (z) attains each value w at most m times (counting multiplicity) in D. Show that either f (z) is constant, or f (z) attains each value w at most m times in D. Solution
36
VIII.4.4 1 2 3 P L K Let f (z) be an analytic function on the open unit disk D = fjzj < 1g. Suppose there is an annulus U = fr < jzj < 1g such that the restriction of f (z) to U is one-to-one. Show that f (z) is one-to-one on D. Solution
37
VIII.4.5 1 2 3 P L K Let f (z) = p (z) =q (z) be a rational function, where p (z) and q (z) are polynomials that are relatively prime (no common zeros). We de…ne the degree of f (z) to be the larger of the degrees of p (z) and q (z). Denote the degree of f (z) by d. (a) Show that each value w 2 C, w 6= f (1), is assumed d times by f (z) on C. (b) Show that f (z) attains each value w 2 C d times on C (as always, counting multiplicity). Solution
38
VIII.4.6 1 2 3 P L K Let f (z) be a meromorphic function on the complex plane, and suppose there is an integer m such that f 1 (w) has at most m points for all w 2 C. Show that f (z) is a rational function.
39
VIII.4.7 1 2 3 P L K Let F (z; w) be a continuous function of z and w that depends analytically on z for each …xed w, and let F1 (z; w) denote the derivative of F (z; w) with respect to z. Suppose F (z0 ; w0 ) = 0, and F1 (z0 ; w0 ) 6= 0. Choose such that F (x; w0 ) 6= 0 for 0 < jz z0 j . (a) Show that there exists > 0 such that if jw w0 j < , there is a unique z = g (w) satisfying jz z0 j < and F (z; w) = 0. (b) Show that Z F1 ( ; w) 1 d ; jw w0 j < : g (w) = 2 i j z0 j= F ( ; w) (c) Suppose further that F (z; w) is analytic in w for each …xed z, and let F2 (z; w) denote the derivative of F (z; w) with respect to w. Show that g (w) is analytic, and g 0 (w) =
F2 (g (w) ; w) =F1 (g (w) ; w) :
(d) Derive the inverse function theorem given in this section, together with the formula for the derivative of the inverse function, as a corollary of (a), (b), and (c). Remark. this is the implicit function theorem for analytic functions. Note that a speci…c formula is given for the function g (w) de…ned implicitely by F (g (w) ; w) = 0.
40
VIII.4.8 1 2 3 P L K Let D be a bounded domain, and let f (z) be a continuous function on D[@D that is analytic on D. Show that @ (f (D)) f (@D), that is, the boundary of the open set f (D) is contained in the image under f (z) of the boundary of D. Solution
41
VIII.5.1 1 2 3 P L K Find the critical points and critical values of f (z) = z + 1=z. Sketch the curves where f (z) is real. Sketch the regions where Im f (z) > 0 and where Im f (z) < 0. Solution.
42
VIII.5.2 1 2 3 P L K Suppose g (z) is analytic at z = 0, with power series g (z) = 2 + iz 4 + O (z 5 ). Sketch and label the curves passing through z = 0 where Re g (z) = 2 and Im g (z) = 0. Solution
43
VIII.5.3 1 2 3 P L K Find the critical points and critical values of f (z) = z 2 + 1. Sketch the set of points z such that jf (z)j 1, and locate the critical points of f (z) on the sketch. Solution
44
VIII.5.4 1 2 3 P L K Suppose that f (z) is analytic at z0 . Show that if the set of z such that Re f (z) = Re f (z0 ) consists of just one curve passing through z0 , then f 0 (z0 ) 6= 0. Show also that if the set of z such that jf (z)j = jf (z0 )j consists of just one curve passing through z0 , then f 0 (z0 ) 6= 0. Solution
45
VIII.5.5 1 2 3 P L K How many critical points, counting multiplicity, does a polynomial of degree m have in the complex plane? Justify your answer. Solution
46
VIII.5.6 1 2 3 P L K Find and plot the critical points and critical values of f (z) = z 2 + 1 and of 2 its iterates f (f (z)) = (z 2 + 1) + 1 and f (f (f (z))). Suggestion. Use the chain rule. Solution
47
VIII.5.7 1 2 3 P L K Let f (z) be a polynomial of degree m. How many (…nite) critical points does the N fold iterate f f (N times) have? Describe them in terms of the critical points of f (z). Solution
48
VIII.5.8 1 2 3 P L K We de…ne a pole of f (z) to be a critical point of f (z) of order k if z0 is a critical point of 1=f (z) of order k. We de…ne z = 1 to be a critical point of f (z) of order k if w = 0 is a critical point of g (w) = f (1=w) of order k. Show that with this de…nition, a point z0 2 C is a critical point of order k for a meromorphic function f (z) if and only if there are open sets U containing z0 and V containing w0 = f (z0 ) such that each w 2 v, w 6= w0 , has exactly k + 1 preimages in U . Remark. We say that f (z) is a (k + 1) sheeted covering of f 1 (V n fw0 g) \ U over V n fw0 g. Solution
49
VIII.5.9 1 2 3 P L K Show that a polynomial of degree m, regarded as a meromorphic function on C , has a critical point of order m 1 at z0 = 1. Solution
50
VIII.5.10 1 2 3 P L K Locate the critical points ana critical values in the extended complex plane of the polynomial f (z) = z 4 2z 2 . Determine the order of each critical point. Sketch the set of points z such that Im z 0. Solution
51
VIII.5.11 1 2 3 P L K Show that if f is a rational function, and if g is a fractional linear transformation, then f and g f have the some critical points in the extended complex plane C . What can be said about the critical values of g f ? What can be said about the critical points and critical values f g? Solution
52
VIII.5.12 1 2 3 P L K Let f (z) = p (z) =q (z) be a rational function of degree d, so that p (z) and q (z) are reactively prime, and d is the larger of the degrees of p (z) and q (z). (See Exercise 4.5.) Show that f (z) has 2d 2 critical points, counting multiplicity, in the extended complex plane C . Hint if deg p 6= deg q, then the number of critical points of f (z) in the …nite plane C is deg (qp0 q 0 p) = deg p + deg q 1, while the order of the critical points at 1 is jdeg p deg qj 1. Solution
53
VIII.5.13 1 2 3 P L K Show that the set of solution points (w; z) of the equation z 2 2 (cos w)+1 = 0 consists of the graphs of two entire functions z1 (w) and z2 (w) of w. Specify the entire functions, and determine where their graphs meet. Remark. The solutions set forms a reducible one-dimensional analytic variety in C2 . Solution
54
VIII.5.14 1 2 3 P L K Let a0 (w) ; : : : ; am 1 (w) be analytic in a neighborhood of w = 0 and vanish at w = 0. Consider the monic polynomial in z whose coe¢ cients are analytic functions in w, P (z; w) = z m + am
1
(w) z m
1
+
+ a0 (w) ;
jwj < :
Suppose that for each …xed w, 0 < jwj < , there are m distinct solutions of P (z; w) = 0. (a) Show that the m roots of the equation P (z; w) = 0 determine analytic functions z1 (w) ; : : : ; zm (w) in the slit disk fjwj < n ( ; 0]g. Hint. Use the implicit function theorem (Exercise 4.7). (b) Glue together branch cuts to form an m sheeted (possibly disconnected) surface over the punctured disk f0 < jwj < g on which the branches zj (w) determine a continuous function. (c) Suppose that the indices are arranged so that for some …xed k, 1 k n, the continuation of zj (w) once around w = 0 is zj+1 (w) for 1 j k 1, while the continuation of zk (w) once around w = 0 is z1 (w). Show that Q (z; w) = (z z1 (w)) (z zk (w)) determines a polynomial in z whose coe¢ cients are analytic functions of w for jwj < . Show further that the polynomial Q (z; w) is an irreducible factor of P (z; w), and that all irreducible factors of P (z; w) arise from subsets of the zj (w)’s in this way. (d) Show that if f (z) is an analytic function that has a zero of order m at z = 0, and z1 (w) ; : : : ; zm (w) are solutions of the equation w = f (z), then the polynomial P (z; w) = (z z1 (w)) (z zm (w)) is irreducible. Solution
55
VIII.5.15 1 2 3 P L K Consider monic polynomial in z of the form P (z; w) = z m + am
1
(w) +
+ a0 (w) ;
where the functions a0 (w) ; : : : ; am 1 (w) are de…ned and meromorphic in some disk centered at w = 0. Let P0 (z; w) and P1 (z; w) be two such polynomials, and consider the following algorithm. Using the division algorithm, …nd polynomials A2 (z; w) and P2 (z; w) such that P0 (z; w) = A2 (z; w) P1 (z; w)+ P2 (z; w) and the degree of P2 (z; w) is less then the degree of P1 (z; w). Continue in this fashion, …nding polynomials Aj+1 (z; w) and Pj+1 (z; w) such that Pj
1
(z; w) = Aj+1 (z; w) Pj (z; w) + Pj+1 (z; w)
and deg Pj+1 (z; w) < deg Pj (z; w), until eventually we reach Pl (z; w) = 0;
Pl
1
(z; w) = Al (z; w) Pl (z; w) :
Let D (z; w) be the monic polynomials in z obtained by dividing Pl (z; w) by the coe¢ cient of the highest power of z. (a) Show that D (z; w) is the greatest common divisor of P0 (z; w) and P1 (z; w), in the sense that D (z; w) divides both P0 (z; w) and P1 (z; w), and each polynomial that divides both P0 (z; w) and P1 (z; w) also divides D (z; w). (b) Show that there are polynomials A (z; w) and B (z; w) such that D = AP0 +BP1 . (c) Show, that if P0 (z; w) and P1 (z; w) are relatively prime (that is D (z; w) = 1), then there is " > 0 such that for each …xed w, 0 < jwj < ", the polynomials P0 (z; w) and P1 (z; w) have no common zeros. (d) Show that any polynomial P (z; w) as above can be factored as a product of irreducible polynomials, and the factorization is unique up to the order of the factors and multiplication of a factor by a meromorphic function in w. (e)
56
Show that if the coe¢ cients of P (z; w) are analytic at w = 0, then the irreducible factors of P (z; w) can be chosen so that their coe¢ cients are analytic at w = 0. (f) Show that if P (z; w) is irreducible, then there is " > 0 such that for each …xed w, 0 < jwj < ", the roots of P (z; w) are distinct. (g) Show that the results of Exercise 14(a)-(c) hold without the supposition that the solutions of P (z; w) = 0 are distinct. Solution
57
VIII.6.1 1 2 3 P L K Sketch the closed path (t) = eit sin (2t), 0 t 2t, and determine the winding number W ( ; ) for each point not on the path. Solution
58
VIII.6.2 1 2 3 P L K Sketch the closed path (t) = e 2it cos t, 0 t 2 , and determine the winding number W ( ; ) for each point not on the path Solution
59
VIII.6.3 1 2 3 P L K Let f (z) be analytic on an open set containing a closed path , and suppose f (z) 6= 0 on . Show that the increase in arg f (z) around is 2 W (f ; 0). Solution
60
VIII.6.4 1 2 3 P L K Let D be a domain, and suppose z0 and z1 lie in the same connected component of CnD. (a) Show that the increase in the argument of f (z) = (z z0 ) (z z1 ) around any closed curve in D is an even multiple of 2 . (b) Show that (z z0 ) (z z1 ) has an analytic square in D. (c) Show by example that (z z0 ) (z z1 ) does not unnecessarily have an analytic cube in D. Solution
61
VIII.6.5 1 2 3 P L K Show that if is a piecewise smooth closed curve in the complex plane, with trace , and if z0 62 , then Z 1 dz = 0; n 2: (z z0 )n Solution
62
VIII.6.6 1 2 3 P L K Let be a closed path in a domain D such that W ( ; ) = 0 for all 62 D. Suppose that f (z) is analytic on D except possibly at a …nite number of isolated singularities z1 ; : : : ; zm 2 Dn . Show that Z X f (z) dz = 2 i W ( ; zk ) Res [f; zk ] : Hint. Consider the Laurent decomposition at each zk , and use Exercise 5.
Solution
63
VIII.6.7 1 2 3 P L K Evaluate 1 2 i
Z
dz z (z 2
1)
;
where is the closed path indicated in the …gure. Hint. Either use Exercise 6, or proceed directly with partial fractions.
64
VIII.6.8 1 2 3 P L K Let (t) and (t), a t b, be closed paths. (a) Show that if 2 C does not lie on the straight line segment between (t) and (t), for a t b, then W ( ; ) = W ( ; ). (b) Show that if j (t) (t)j < j (t)j for a a t b, then W ( ; ) = W ( ; ). Solution
65
VIII.6.9 1 2 3 P L K Let f (z) be a continuous complex-valued function on the complex plane such that f (z) is analytic for jzj < 1, f (z) 6= 0 for jzj 1, and f (z) ! 1 as z ! 1. Show that f (z) 6= 0 for jzj < 1. Solution
66
VIII.6.10 1 2 3 P L K Let K be a nonempty closed bounded subset of the complex plane, and let f (z) be a continuous complex-valued function on the complex plane that is analytic on CnK and at 1. Show that every value attained by f (z) on C = C [ f1g is attained by f (z) somewhere on K, that is f (C ) = f (K). Solution
67
VIII.7.1 1 2 3 P L K Let f (z) be an entire function, and suppose g ( ) is analytic for in the open upper hand lower half-planes and across the interval ( 1; 1) on the real line. Suppose that Z 1 f (x) dx = g ( ) 1 z for in the upper half-plane. What is the value of the integral when the lower half-plane? Justify your answer carefully. Solution
68
is in
VIII.7.2 1 2 3 P L K Show that Z
1 1
dx x
= Log
1 +1
;
2 Cn [ 1; +1] :
(Note that we use the principal branch of the logarithm here.) Reconcile this result with your solution to Exercise 1. Solution
69
VIII.7.3 1 2 3 P L K Find the Cauchy integrals of the following functions around the unit circle = fjzj =g, positively oriented. (c) x = Re z (d) y = Im (z) (a) z (b) z1 Solution
70
VIII.7.4 1 2 3 P L K Suppose f (z) is analytic on an annulus f < jzj < g, and let f (z) = f0 (z)+ f1 (z) be the Laurent decomposition of f (z). (See Section VI.1.) Fix r between and , and let F ( ) be the Cauchy integral of f (z) around the circle jzj = r. Show that f0 ( ) = F ( ) for j j < r, and f1 ( ) = F ( ) for j j > r. Show further that f0 ( ) = F ( ) and f1 ( ) = F+ ( ). Remark The formula f (z) = f0 (z) + f1 (z) re‡ects the jump theorem for the Cauchy integral of f (z) around circles jzj = r. Solution
71
VIII.7.5 1 2 3 P L K Let be a piecewise smooth curve, and let F ( ) be the Cauchy integral (7.1) of a continuous function f (z) on . Show that if g (z) is a smooth function on the complex plane that is zero o¤ some bounded set, then Z ZZ @g F (z) dx dy: g (z) f (z) dz = 2i C @z Hint. Recall Pompeiu’s formula (Section IV.8).
72
VIII.7.6 1 2 3 P L K Determine whether the point z lie inside or outside. Explain. Solution
73
VIII.7.7 1 2 3 P L K A simple arc in C is the image of a continuous one-to-one function (t) from a closed interval [a; b] to the complex plane. Show that a simple arc in C has a connected complement, that is Cn is connected. You may use the Tietze extension theorem, that a continuous real-valued function on a closed subset of the complex plane can be extended to a continuous realvalued function on the entire complex plane. Hint. Suppose z0 belongs to a bounded complement of Cn . Find at continuous determination h (z) of log (z z0 ) on , extend h (z) to a continuous function on C , and de…ne f (z) = z z0 on the component Cn containing z0 , and f (z) = eh(z) on the remainder of C n . Consider the increase in the argument of f (z) around circles centered at z0 . Prove the Jordan curve theorem for a simple closed curve by …lling in the following proof outline. (a) Show that each component Cn has boundary . Hint. For z0 = (t0 ) 2 , apply the preceding exercise to the simple arc n (I), where I is a small open parameter interval containing t0 . (b) Prove the Jordan curve theorem in the case where contains a straight line segment. (c) Show that for any z0 = (t0 ) 2 , any small disk D0 containing z0 , and component U of Cn , there are points z1 = (t1 ) and z2 = (t2 ) such that the image of the parameter segment between t1 and t2 is contained in D0 and such that z1 and z2 can be joined by a broken line segment in U \ D0 . (d) With notation as in (b), let be the simple curve obtained by replacing the segment of in D0 between z0 and z1 by the broken line segment in U \ D0 between them, and let be the simple closed curve in D0 obtained by the following the segment of in D0 from z0 to z1 and returning to z0 along the broken line segment. Show that W ( ; ) = 0 and W ( ; ) = W ( ; ) for 2 Cn , 6= D0 . (e)
74
Using (b) and (d), show that Cn has at least two components and that W ( ; ) = 1 for in each bounded component of Cn . (f) By taking U in (c) to be a bounded component of Cn , show that W ( ; ) = 0 for in any other component of Cn . Solution
75
VIII.8.1 1 2 3 P L K Which of the following domains in C are simply connected? Justify your answers. (a) D = fIm z > 0g n [0; i], the upper half-plane with a vertical slit from 0 to i. (b) D = fIm z > 0g n [i; 2i], the upper half-plane with a vertical slit from i to 2i. (c) D = Cn [0; +1], the complex plane slit along the positive real axis. (d) D = Cn [ 1; 1], the complex plane with an interval deleted. Solution
76
VIII.8.2 1 2 3 P L K Show that a domain D in the extended complex plane C = C [ f1g is simply connected if and only if its complement C nD is connected. Hint. If D 6= C , move a point in the complement of D to 1. If D = C , …rst deform a given closed path to one that does not cover the sphere, then deform it to a point by pulling along arcs of great circles. Solution
77
VIII.8.3 1 2 3 P L K Which of the following domains in C are simply connected? Justify your answers. (a) D = C n [ 1; 1], the extended plane with an interval deleted, (b) D = C n f 1; 0; 1g, the thrice-punctured sphere. Solution
78
VIII.8.4 1 2 3 P L K Show that a domain D in the complex plane is simply connected if and only if any analytic function f (z) on D does not vanish at any point of D has an analytic logarithm on D. Hint if f (z) 6= 0 on D, consider the function Z z 0 f (z) G (z) = dw: z0 f (w) Solution
79
VIII.8.5 1 2 3 P L K Show that a domain D is simply connected if and only if any analytic function f (z) on D that does not vanish at any point of D has an analytic square root on D. Show that this occurs if and only if for any point z0 62 D the function z z0 has an analytic square root on D. Solution
80
VIII.8.6 1 2 3 P L K Show that a domain D is simply connected if and only if each continuous function f (z) on D that does not vanish at any point D has continuous logarithm on D. Solution
81
VIII.8.7 1 2 3 P L K Let E be a closed connected subset of the extended complex plane C . Show that each connected component C nE is simply connected. Solution
82
VIII.8.8 1 2 3 P L K Show that simple connectivity is a "topological property" that is, if U and V are domains, and ' is a continuous map of U onto V such that ' 1 is also continuous, then U is simply connected if and only if V is simply connected. Solution
83
VIII.8.9 1 2 3 P L K Suppose that f (z) is analytic on a domain D, and f 0 (z) has no zeros on D. Suppose also that f (D) is simply connected, and that there is a branch g (w) of f 1 that is analytic on w0 = f (z0 ) and that can be continued analytically along any path in f (D) starting at w0 . Show that f (z) is one-to-one on D. Solution
84
VIII.8.10 1 2 3 P L K We P de…ne an integral 1 cycle in D to be an expression of the form = kj j , where 1 ; : : : ; m are closed paths in D and k1 ; : : : ; km are integers. P We de…ne the winding number of about to be W ( ; ) = kj W j ; , 2 CnD. Show that if h ( ) is a continuous integer-valued function on C nD such that h (1) = 0, then there is an integral 1-cycle on D such that W ( ; ) = h ( ) for all 2 CnD. Solution
85
VIII.8.11 1 2 3 P L K An integral 1-cycle is homologous to zero i D if W ( ; ) = 0 whenever 62 D. Let U be a bounded domain whose boundary consists of a …nite number of piecewise smooth closed curves 1 ; : : : ; m , oriented positively with respect to U , such that P U together with its boundary is contained in D. Show that the 1-cycle @U = j is homologous to zero in D. Solution
86
VIII.8.12 1 2 3 P L K Let D be a domain in C such that C nD consists of m + 1 disjoint closed connected sets. Show that there are m piecewise smooth closed curves 1 ; : : : ; m such thatPevery integral 1-cycle can be expressed uniquely in the form = 0 + kj j , where the kj ’s are integers and 0 is homologous to zero in D. Remark. The j ’s form a homology basis for D. Solution
87
