James Stankowicz - 132A - Due: Jan. 16, 2013
Homework #1 Solutions Problem 1 x + iy .
Write each of the following complex numbers in the form
Solution 1 Part a.
√ 3 1 + 2i = √ 2 √ 3 √ = 1 + 3 2i + 3 2i + 2i =
3
1
= 1 + 3 · 2 2 i − 6 − i2 2 = 1
= −5 + 2 2 (3i − i2) = √ = −5 + i 2 . Note: the rst line used the binomial expansion:
n
(a + b) =
n X n
k
k=0 with
n = 3.
xn−k y k ,
√ 3 1 + 2i by hand: √ √ 2 √ 3 1 + 2i = 1 + 2i 1 + 2i + · · · .
It's also possible (although longer) to expand
Part b.
2
(1 − i) = (1 + i) 3
(1 − i) 1 3 = (1 − i) = (1 + i) (1 − i) 2 1 2 3 = 1 + 3 (−i) + 3 (−i) + (−i) = 2 1 = 1 − 3i − 3 − i3 = 2 1 = (−2 − 2i) = −1 − i . 2 =
Part c.
√ 3 √ 1 + 2i = −5 + i 2. Then: √ √ √ 3 √ 2−i 2−i 1 + 2i = −5 + i 2 2 2 = (1 + i) (1 + i) √ √ √ √ −5 + i 2 2−i −5 2 + 5i + i2 + 2 = = = 2i 1 + 2i + i2 1 √ −4 2 + 7i = = 2i √ = 72 + i2 2 .
First, recognize the result from part a where
Problem 2 [BC 5#11] Solve the equation
z2 + z + 1 = 0
for
z = (x, y)
by writing
(x, y) (x, y) + (x, y) + (1, 0) = (0, 0) and then solving a pair of simultaneous equations in the given equation to show that
x
and
y 6= 0. 1
y.
Suggestion: Use the fact that no real number
x
satis es
James Stankowicz
132
Due: Jan. 16, 2014
Solution 2 By expanding.
(0, 0) = (x, y) (x, y) + (x, y) + (1, 0) = = x2 − y 2 , 2xy + (x + 1, y) = = x2 − y 2 + x + 1, 2xy + y , arrive at the two equations
0 = 2xy + y, 0 = x2 − y 2 + x + 1. Solve the rst for
x: 2xy = −y ⇒ x = − 21 .
The second is then straightforward using the solution for
−
0=
1 2
2
x:
1 − y2 + − +1= 2
1 1 − y2 − + 1 ⇒ 4 2 √ 3 y 2 = ⇒ y = ± 23 . 4 =
Compiling the answer:
z = (x, y) =
Verify
√
− 12 , ±
√
3 2
=
1 2
−1 +
√ 3i .
Problem 3 [BC 13#4] 2z ≥ |Re {z}| + |Im {z}|.
Suggestion: Reduce this inequality to
2
(|x| − |y|) ≥ 0.
Solution The
3
2 |z| ≥ |Re {z}| + |Im {z}|. Let q z = x + iy, for x, y ∈ R. This means |Re {z}| = |x| and p 2 2 2 2 modulus of z is |z| = x + y = |x| + |y| . There is nothing profound about the last equality;
The goal is to verify
|Im {z}| = |y|.
√
it only says that some number squared is the same as squaring the absolute value of the number. Make sure this makes sense to you!The goal is now to show
√
2 |z| ≥ |Re {z}| + |Im {z}| ⇒ √ q 2 2 2 |x| + |y| ≥ |x| + |y| .
Clean this up by squaring both sides, then see if the inequality can be manipulated into an obviously true inequality:
2 2 2 2 |x| + |y| ≥ (|x| + |y|) . Rewrite the right hand side:
2
2
2
(|x| + |y|) = |x| + 2 |x| |y| + |y| . Then the inequality is
2
2
2
2
2 |x| + 2 |y| ≥ |x| + 2 |x| |y| + |y| . Move everything to the left side of the inequality:
2
2
|x| − 2 |x| |y| + |y| ≥ 0 ⇒ 2
(|x| − |y|) ≥ 0. This statement is true because the quantity in parentheses is always some real number, so that its square is always non-negative. As a rigorous mathematical treatment, this is somewhat backwards. A formal proof would start from the statement
2
(|x| − |y|) ≥ 0, then perform all the operations in reverse to arrive at the desired result
√
2 |z| ≥ |Re {z}| + |Im {z}| . 2
James Stankowicz
132
Due: Jan. 16, 2014
Doing so is a bit tricky, particularly when trying to go from
2
2
|x| − 2 |x| |y| + |y| ≥ 0 to
2
2
2
2
2 |x| + 2 |y| ≥ |x| + 2 |x| |y| + |y| , which is why the reverse engineering in this solution is easier.
Problem 4 [BC 13#5] In each case, sketch the set of points determined by the given condition.
Solution 4 (a)
|z − 1 + i| = 1
|z + i| ≤ 3
(b)
y
(c)
|z − 4 i| ≥ 4
y
y
x
r=1
x (1, − i)
(0, 4 i) (0, − i)
r=4
r=3
x
Problem 5 Write each of the following complex numbers in the form
reiθ .
Solution 5 Part a. Again utilize the result of problem 1a:
√
The number the negative
√ −5 + i 2 x-axis;
√
√ 3 2i =
r 2
= −5 + i 2 = =
1+
(−5) + √
27ei(π−arctan(
is in quadrant II (negative
x,
√ 2 √ 2 ei arctan(− 2/5) =
√ √ = 3 3ei(π−arctan( 2/5)) .
2/5))
positive
y ),
this gives a corresponding angle of Arg (z)
It is also possible to try writing
1+
√
2i =
√
and forms an angle of
= π − arctan
3ei arctan(
√
2)
√ 2 5
arctan
√ 2 5
with respect to
.
;
this result, however, is somewhat muddled when trying to simplify, since it is not clear how to treat the
3 arctan
√ 2
that appears as the argument of the exponent. 3
φ =
James Stankowicz
132
Due: Jan. 16, 2014
Part b.
−i i =− 2 + 2i 2
1 1+i
1−i 1−i
=−
i
2
(1 − i) 1 = − (1 + i) = 2 4
3π 1 1 √ 3 3π (−1 − i) = · 2e−i 4 = 2− 2 e−i 4 . 4 4 π The number −1 − i is in quadrant III (negative x, negative y ), and forms an angle of 4 with respect to the negative 3π x-axis; this gives a corresponding angle of Arg (z) = − 4 . Another way to approach this problem is to convert both i and −2 − 2i to polar form, then manipulate the numbers that way (as in problem 6 of this assignment).
=
Part c.
1 7
(1 − i) = (1 − i)
−7
=
√
2e−iπ/4
= −7
= 2− 2 ei7π/4 = 7
= 2− 2 e−iπ/4 . 7
The number
1−i
is in quadrant IV (positive
exponentiating, the
7π/4
x,
negative
y ),
and forms an angle of
that comes out is equivalent, modulo
2π ,
to
−iπ/4.
π 4 with respect to the
x-axis.
After
Problem 6 [BC 24#5b,5d] By writing the individual factors on the left in exponential form, performing the needed operations, and nally changing back to rectangular coordinates, show that: b: d:
5i/ (2 + i) = 1 + 2i √ √ −10 = 2−11 −1 + 3i . 1 + 3i
Solution 6 Part b.
5i = (2 + i) =√ =
√
5eiπ/2 5ei arctan(1/2)
5ei arctan(2) =
=
√
5ei( 2 −arctan( 2 )) = π
1
q 2 2 (1) + (2) ei arctan(2/1) = = 1 + 2i.
π (2) may be understood geometrically. The angle π/2 is straight up along the 2 − arctan 2 to arctan y -axis. Geometrically, π2 − arctan 12 then says to go straight along the y -axis, and then subtract from that angle 1 the angle of arctan 2 (that's the triangle with opposite length 1 and adjacent length 2). The picture: Converting
1
y Want this angle!
θ=
π 2
1
2 arctan 1/2 4
x
James Stankowicz
132
Part d.
Due: Jan. 16, 2014
√ −10 10π 4π π −10 1+i 3 = 2−10 e−i 3 = 2−10 e−i 3 = = 2ei 3 √ 2π = 2−11 · 2ei 3 = 2−11 −1 + i 3 4π 2π − 10π 3 = − 3 = 3 . These angles are equivalent mod (2π) for those who speak math. Of these 2π mod (2π) angles, 3 is the only one that meets the criteria to be in Arg (z). Further, the angle 2π 3 π with angle 3 with respect to the negative x-axis. Drawing the corresponding right triangle gives
Note that for angles three equivalent is in quadrant II, the
x
and
y
components.
Problem 7 [BC 24#9] Establish the identity
1 + z + z2 + · · · + zn =
1 − z n+1 , (z 6= 1) , 1−z
and then use it to derive Lagrange's trigonometric identity:
1 + cos (θ) + cos (2θ) + · · · + cos (nθ) = . Suggestion: As for the rst identity, write the second identity, write
z=
1 sin ((2n + 1) θ/2) + , (0 < θ < 2π) . 2 2 sin (θ/2)
S = 1 + z + z2 + · · · + zn,
eiθ in the rst one.
and consider the dierence
S − zS .
To derive
Solution 7 Starting as hinted:
S − zS = (1 + z + · · · + z n ) − z + z 2 + · · · + z n + z n+1 = = 1 − z n+1 , so that
By plugging in
z = eiθ ,
(1 − z) S = 1 − z n+1 ⇒ 1 − z n+1 . S= 1−z the left hand side is
S = 1 + eiθ + ei(2θ) + · · · + ei(nθ) = = 1 + (cos (θ) + i sin (θ)) + · · · + (cos (nθ) + i sin (nθ)) = which has real part Re {S}
= 1 + cos (θ) + · · · + cos (nθ) .
This means Re {S} will be equal to the real part of the other side of the expression:
Re
1 − z n+1 1−z
= Re
1 − ei(n+1)θ 1 − eiθ
To brute force the issue, attempt to write the denominator in the form through in both numerator and denominator by
e−iθ/2 :
sin
. θ 2
=
1 2i
−iθ/2 1 − ei(n+1)θ e Re = 1 − eiθ e−iθ/2 −iθ/2 e − ei(n+1/2)θ = Re = e−iθ/2 − eiθ/2 −iθ/2 e − ei(n+1/2)θ = Re = −2i sin (θ/2) n o 1 −iθ/2 = Re i e − ei(n+1/2)θ = 2 sin (θ/2) 1 θ θ = Re i cos − i sin − 2 sin (θ/2) 2 2 1 1 − cos n+ θ + i sin n+ θ = 2 2 1 1 = sin (θ/2) + sin n+ θ = 2 sin (θ/2) 2
5
eiθ/2 − e−iθ/2
by multiplying
James Stankowicz
132
1 2
=
+
Due: Jan. 16, 2014
sin((2n+1)θ/2) 2 sin(θ/2)
.
Problem 8 Find all values of the following roots and write them in the form Part a: Part b: Part c:
x + iy :
1 5
(32i) 1 83 1 (3 + 4i) 2
Solution 8 Part a. 1
(32i) 5 = 15 = 2 ei(π/2+2nπ) = = 2ei( 10 + 5 πn) = 2ei(π(1+4n)/10) π
Consider dierent possible
n
2
values.
n=0
:
2eiπ/10 = 2 cos
n=1
:
2eiπ/2 = 0 + 2i ,
n=2
:
2ei9π/10 = 2 cos
n=3
:
2ei13π/10 = 2ei−7π/10 = = −2 cos 7π 10 + i sin
π 10
9π 10
+ i sin
π 10
9π 10
+ i sin
7π 10
n=4
:
2eiπ17/10 = 2e−iπ3/10 = 7π , = 2 cos 3π 10 − i sin 10
n=5
:
2eiπ21/10 = 2eiπ/10 = π = 2 cos 10 + i sin
π 10
,
,
,
,
at which point the solutions repeat cyclically.
Part b.
8 3 = 8ei2πn 1
Consider dierent possible values of
31
= 2ei2πn/3
n:
n=0 n=1
n=2
2ei2π0/3 = 2 , 2π 2π : 2ei2π/3 = 2 cos + i sin = 3 3 √ = −1 + i 3 , √ : 2ei4π/3 = 2e−i2π/3 = −1 − i 3 ,
:
at which point the solutions repeat cyclically.
Part c. 1
(3 + 4i) 2 = 21 = 5ei(arctan(4/3)+2πn) = =
√
5ei( 2 arctan(4/3)+nπ) 1
6
James Stankowicz
132
Consider dierent possible values for
n=0
n=1
Due: Jan. 16, 2014
n:
√
5ei( 2 arctan ( 3 )) = √ 1 4 1 4 = 5 cos arctan + i sin arctan , 2 3 2 3 √ √ i 1 arctan 4 +π ( 3 ) ) = 5eiπ e 2i arctan ( 34 ) = 5e ( 2 : √ 1 4 1 4 = 5 − sin arctan + i cos arctan , 2 3 2 3
:
1
4
at which point the solutions repeat cyclically.
Problem 9 [BC 31#8a,8b] Part a: Prove that the usual formula solves the quadratic equation
az 2 + bz + c = 0, (a 6= 0) when the coe cients
a, b,
and
c
are complex numbers. Speci cally, by completing the square on the left-hand
side, derive the quadratic formula
where both square roots are to be
1 −b + b2 − 4ac 2 z= , 2a 2 considered when b − 4ac 6= 0.
Part b: Use the result in part (a) to nd the roots of the equation
z 2 + 2z + (1 − i) = 0.
Solution 9 Part a.
The rst step in completing the square is factoring out the overall factor on the quadratic term:
c b c b ⇒ 0 = z2 + z + , 0 = a z2 + z + a a a a since
a 6= 0.
Then add and subtract a term so that a perfect square appears:
0=
2 ! 2 1 b 1 b c b − + = z + z+ a 4 a 4 a a 2 2 b 1 b c = z+ − + , 2a 4 a a 2
where the parentheses in the rst line contain the same terms as the parentheses in the second line. Then solve for
2 2 b 1 b c z+ = − ⇒ 2a 4 a a s r 2 b 1 b c b2 − 4ac 1p 2 z+ =± − =± =± b − 4ac ⇒ 2 2a 4 a a 4a 2a √ −b ± b2 − 4ac z= . 2a Part b.
Comparing to
z 2 + 2z + (1 − i) = 0,
obtain:
a = 1, b = 2, c = (1 − i) . Then
p 1 2 ± 4 − 4 (1 − i) = 2 p 1 =− 2 ± 2 1 − (1 − i) = 2 √ =− 1± i .
z=−
This can be cleaned up:
21 z = − 1 ± eiπ/2 . 7
z:
James Stankowicz
132
Due: Jan. 16, 2014
The last term is a bit hairy to deal with:
eiπ/2
21
π 12 = ei( 2 +2πn) =
= eiπ(n+ 4 ) , 1
where
n = 0, ±1, ±2, . . ..
For dierent
n's:
π 1 : ei 4 = √ (1 + i) , 2 1 i 54 π n=1 : e = − √ (1 + i) = −eiπ/4 , 2 result in the expression for z : 1 z = − 1 ± √ (1 + i) . 2
n=0
with repeats after this. Using this
Problem 10 [BC p.35 Problem 5] Let
S
be the open set consisting of all points
z
such that
|z| < 1
or
|z =2| < 1.
State why
S
is not connected.
Solution 10 The region S is not connected because it is impossible to connect the two points z± = 1 ± ∈ S , for ∈ R and 0 < 1 with any polygonal line that remains entirely in S . In English, the point z− is just to the left of z = 1 on the x-axis, and z+ is just to the right of z0 = 1. The only point where the two circles de ning S are almost connected is 0 at z = 1, however neither of the two circles cover the point z = 1. The region S = S ∪ {z = 1} would be (one possible) connected region.
8
