Homework 7
1. Express in the form a + bi.
a. (1 + i)20 .
√ πi
Solution: 1 + i = 2e 4 . Hence (1 + i)20 = 210 e5πi = −210 .
b.
1−2i
.
2+i
Solution:
1−2i
2+i
=
1−2i 2−i
2+i 2−i
=
−5i
5
= −i.
2. Solve z 2 − 4z + (4 + 2i) = 0.
−πi
2
2
Solution:
− 4z + (4 + 2i) = (z − 2)2 + 2i = 0 gives (z√− 2)√
= −2i√= 2e 2 . Hence
√ z −πi
z − 2 = 2e 4 +kπi for k = 0, 1. It follows that z = 2 + 2( 21 2 − 12 i 2) = 3 − i and
√ √
√
z = 2 − 2( 12 2 − 12 i 2) = 1 + i.
3. Describe the sets whose points satisfy the following relations. Which of these sets are
regions (i.e., open and connected sets)?
a. |z + i| ≤ 1.
Solution: |z + i| ≤ 1 ⇐⇒ the distance from z to −i is ≤ 1, i.e., the closed disk
with center −i and radius 1. This set is not open.
z−1 = 1.
b. z+1
z−1 = 1 ⇐⇒ the distance of z to 1 equals the distance to −1,
Solution: z+1
i.e. the perpendicular bisector of the line segment connecting 1 and −1 or the
imaginary axis. This set is not open.
c. |z − 3| > |z − 2|.
Solution: |z − 3| > |z − 2| ⇐⇒ the distance of z to 3 is bigger than the distance
to 2, i.e. all z with Re x < 52 . This set is open and connected.
d.
1
z
= z.
Solution: z1 = z ⇐⇒ |z| = 1. i.e., this set is the unit circle centered at the
origin. This set is not open.
4. Let a ∈ C with |a| < 1, Prove
a−z 1 − az = 1 ⇔ |z| = 1.
a−z = 1 ⇔ |a − z|2 = |1 − az|2 ⇔ (a − z)(a − z) = (1 − az)(1 − az) ⇔
Solution: 1−az
|a|2 −az−az+|z|2 = 1−az−az+|a|2 |z|2 ⇔ (1−|a|2 )|z|2 = 1−|a|2 ⇔ |z|2 = 1 ⇔ |z| = 1.
5. Find all solutions of
a. ez = −i.
π
Solution: ez = −i = e− 2 i if and only if z = − π2 i + 2kπi where k ∈ Z.
b. sin z = 0.
iz
−iz
Solution: sin z = e −e
= 0 if and only if e2iz = 1 if and only if 2iz = 2kπi
2i
with k ∈ Z if and only if z = kπ with k ∈ Z.
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c. Log z = 1 + i.
Solution: Logz = Log|z| + iArgz = 1 + i if and only if Log|z| = 1 and Argz = 1
if and only if z = e(cos 1 + i sin 1).
6.
a. Prove that f (z) = Re z is nowhere complex differentiable.
∂v
= 1 for all z and ∂y
= 0 for all z, so the Cauchy-Riemann equation
Solution: ∂u
∂x
hold nowhere.
b. Prove that f (z) = |z| is nowhere complex differentiable.
Solution: If f is differentiable at z0 , then g(z) = f (z)2 = |z|2 = x2 + y 2 is
differentiable at z0 . One easily sees by the Cauchy-Riemann equations that g is
only differentiable at z = 0. This leaves z = 0 as the only possible point where f
could be differentiable, but one can easily see that the partials of f with respect
to x don’t exist at z = 0 (the limit from the right of the difference quotient is
equal to 1 and from the left equal to −1). Hence f is nowhere differentiable.
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