Physics 451
Fall 2004 Homework Assignment #9 — Solutions
Textbook problems: Ch. 6: 6.1.3, 6.1.7, 6.2.5, 6.2.6, 6.3.3, 6.4.3, 6.4.4 Chapter 6 6.1.3 Prove algebraically that |z1 | − |z2 | ≤ |z1 + z2 | ≤ |z1 | + |z2 | Interpret this result in terms of vectors. Prove that p |z − 1| < | z 2 − 1| < |z + 1|, for
<(z) > 0
We start by evaluating |z1 + z2 |2 |z1 + z2 |2 = (z1 + z2 )(z1∗ + z2∗ ) = |z1 |2 + |z2 |2 + z1 z2∗ + z1∗ z2 = |z1 |2 + |z2 |2 + (z1 z2∗ ) + (z1 z2∗ )∗ = |z1 |2 + |z2 |2 + 2<(z1 z2∗ )
(1)
We now put a bound on the real part of z1 z2∗ . First note that, for any complex quantity ζ, we have |ζ|2 = (<ζ)2 + (=ζ)2 ≥ (<ζ)2 . Taking a square root gives |ζ| ≥ |<ζ| or −|ζ| ≤ <ζ ≤ |ζ|. For the present case (where ζ = z1 z2∗ ) this gives −|z1 ||z2 | ≤ <(z1 z2∗ ) ≤ |z1 ||z2 |. Using this inequality in (1), we obtain |z1 |2 + |z2 |2 − 2|z1 ||z2 | ≤ |z1 + z2 |2 ≤ |z1 |2 + |z2 |2 + 2|z1 ||z2 | or (|z1 | − |z2 |)2 ≤ |z1 + z2 |2 ≤ (|z1 | + |z2 |)2 Taking the square root then proves the triangle inequality. The reason this is called the triangle inequality is that, in terms of vectors, we can think of z1 , z2 and z1 + z2 as the three sides of a triangle
z2 z 1+
z2
z1
Then the third side (|z1 + z2 |) of a triangle can be no longer than the sum of the lengths of the other two sides (|z1 | + |z2 |) nor shorter than the difference of lengths (|z1 | − |z2 |).
Finally, for the second inequality, we start by proving that |z + 1|2 = |z|2 + 1 + 2 |z − 1|2 for 0. This implies that |z + 1| > |z − 1| for 0. The picture here is that if z is on the right half of the complex plane then it is closer to the point 1 than the point −1. z
z−
z +1
1
z
Given this result, it is simple to see that |z − 1|2 < |z − 1||z + 1| < |z + 1|2 or, by taking a square root p |z − 1| < | (z − 1)(z + 1)| < |z + 1| which is what we set out to prove. 6.1.7 Prove that N −1 X x sin(N x/2) cos(N − 1) a) cos nx = sin x/2 2 n=0 b)
N −1 X
sin nx =
n=0
x sin(N x/2) sin(N − 1) sin x/2 2
We may solve parts a) and b) simultaneously by taking the complex combination
S=
N −1 X
cos nx + i
n=0
N −1 X n=0
sin nx =
N −1 X
(cos nx + i sin nx) =
n=0
N −1 X
einx
n=0
The real part of S gives part a) and the imaginary part of S gives part b). When written in this fashion, we see that S is a terminating geometric series with ratio r = eix . Thus N −1 X
1
1
1
1 − rN 1 − eN ix e 2 N ix (e 2 N ix − e− 2 N ix ) S= r = = = 1 1 1 1−r 1 − eix e 2 ix (e 2 ix − e− 2 ix ) n=0 n
We performed the last step in order to ‘balance’ positive and negative exponentials inside the parentheses. This is so that we may relate both the numerator and denominator to sin α = (eiα − e−iα )/2i. The result is 1
S = e 2 (N −1)ix
sin(N x/2) sin(N x/2) = cos 12 (N − 1)x + i sin 12 (N − 1)x sin x/2 sin x/2
It should now be apparent that the real and imaginary parts are indeed the solutions to parts a) and b). 6.2.5 Find the analytic function w(z) = u(x, y) + iv(x, y) a) if u(x, y) = x3 − 3xy 2 We use the Cauchy-Riemann relations ∂u ∂v =− = 6xy ∂x ∂y ∂v ∂u = = 3x2 − 3y 2 ∂y ∂x
⇒
v = 3x2 y + C(y)
⇒
v = 3x2 y − y 3 + D(x)
In order for these two expressions to agree, the functions C(y) and D(x) must have the form C(y) = −y 3 + c and D(x) = c where c is an arbitrary constant. As a result, we find that v(x, y) = 3x2 y − y 3 + c, or w(z) = (x3 − 3xy 2 ) + i(3x2 y − y 3 ) + ic = z 3 + ic The constant c is unimportant. b) v(x, y) = e−y sin x As above, we have ∂u ∂v = = −e−y sin x ∂x ∂y ∂u ∂v =− = −e−y cos x ∂y ∂x
⇒
u = e−y cos x + C(y)
⇒
u = e−y cos x + D(x)
Thus we must have C(y) = D(x) = c with c a constant. The complex function w(z) is w(z) = c + e−y cos x + ie−y sin x = c + e−y (cos x + i sin x) = c + eix−y = c + eiz
6.2.6 If there is some common region in which w1 = u(x, y) + iv(x, y) and w2 = w1∗ = u(x, y) − iv(x, y) are both analytic, prove that u(x, y) and v(x, y) are constants. If u + iv and u − iv are both analytic, then they must both satisfy the CauchyRiemann equations. This corresponds to ∂v ∂u = , ∂x ∂y
∂u ∂v =− ∂y ∂x
(from u + iv) and ∂v ∂u =− , ∂x ∂y
∂u ∂v = ∂y ∂x
(from u − iv). Clearly this indicates that ∂u ∂u = = 0, ∂x ∂y
∂v ∂v = =0 ∂x ∂y
Since all partial derivatives vanish, u and v can only be constants. 6.3.3 Verify that Z
1+i
z ∗ dz
0
depends on the path by evaluating the integral for the two paths shown in Fig. 6.10. y 2 2
1+ i
z
1 x
1
We perform this integral as a two-dimensional line integral Z
Z
∗
z dz =
(x − iy)(dx + idy)
For path 1, we first integrate along the x-axis (y = 0; dy = 0) and then along the y-axis (x = 1; dx = 0) Z 0
1+i
Z 1 (x − iy) dx + (x − iy) idy z dz = x=1 y=0 0 0 Z 1 Z 1 1 1 = xdx + (i + y)dy = 21 x2 + (iy + 21 y 2 ) = 1 + i ∗
Z
0
1
0
0
0
Similarly, for path 2, we find Z 1 Z 1+i ∗ (x − iy) z dz = 0
0
Z
1
Z idy + x=0
Z ydy +
= 0
0
1
0
1
(x − iy)
dx y=1
1 1 (x − i)dx = 21 y 2 + ( 21 x2 − ix) = 1 − i 0
0
So we see explicitly that the integral depends on the path taken (1 + i 6= 1 − i). H 6.4.3 Solve Exercise 6.3.4 [ C dz/(z 2 +z) where C is a circle defined by |z| > 1] by separating the integrand into partial fractions and then applying Cauchy’s integral theorem for multiply connected regions. Note that, by applying Cauchy’s integral formula to a constant function f (z) = 1, we may derive the useful expression I dz = 2πi (2) C z − z0 provided point z0 is contained inside the contour C (it is zero otherwise). Then, using partial fractions, we see that I I I I I dz 1 1 dz dz dz = = − dz = − 2 z z+1 z+1 C z(z + 1) C C z C z +z Since C is a circle of radius greater than one, it encompasses both points z0 = 0 and z0 = −1. Thus, using (2), we find I dz = 2πi − 2πi = 0 2 C z +z Note that, if the radius of C is less than one, we would have encircled only the pole at z0 = 0. The result would then have been 2πi instead of zero. 6.4.4 Evaluate
I C
dz −1
z2
where C is the circle |z| = 2. Again, we use partial fractions and (2) I I I dz dz 1/2 1/2 = = − dz 2 z−1 z+1 C z −1 C (z + 1)(z − 1) C I I dz 1 dz 1 = − = πi − πi = 0 2 C z−1 2 C z+1 Here it is important that the contour of radius 2 encircles both points z0 = −1 and z0 = 1.