MSO202A COMPLEX ANALYSIS
Assignment 2
Exercise Problems:
1. Let z = x + iy and f (z) = x2 − y 2 − 2y + i(2x − 2xy). Write f (z) as a function of z
and z.
z+z z−z
Proof: Using x = 2
,y = 2i
we get f (z) = z + 2iz.
2. Verify Cauchy-Riemann equation for z 2 , z 3 .
Proof: For z 2 , u = x2 − y 2 , v = 2xy ⇒ ux = 2x, uy = −2y, vx = 2y, vy = 2x.
Similarly for z 3 .
z+z z−z ∂ 1 ∂
3. Using the relations x = ,y = and the chain rule show that = ( −
2 2i ∂z 2 ∂x
∂ ∂ 1 ∂ ∂
i ); = ( + i ).
∂y ∂z 2 ∂x ∂y
Proof: Straight forward.
|w−z|
4. Let z, w ∈ C, |z|, |w| < 1 and zw 6= 1. Prove that |1−wz|
< 1. Further, show that the
equality holds if either |z| = 1 or |w| = 1.
Proof: Suffices to show that |w − z|2 < |1 − wz|2 , i.e. ww + zz − wz − zw <
1 − wz − wz + wwzz. Since (1 − zz)(1 − ww) > 0, the above is true.
In case of equality, we see that either (1 − zz) or (1 − ww) is zero. Hence, in this
case either |z| = 1 or |w| = 1.
5. Determine
P zn all z ∈P
C for which each
P of the following
P power series is convergent.
zn zn 1 1
a) n2 b) n! c) 2n d) 2n zn .
Proof:
(a) Here an+1
an
→ 1 ⇒ R = 1. The series converges for |z| < 1 and diverges for
|z| > 1. For |z| = 1, by Comparison test it follows that the series converges
n
since |z|
n2
= n12 .
an+1
(b) As an
→ 0 ⇒ R = ∞ and so the series converges for all z.
an+1
(c) As an
→ 12 ⇒ R = 2. The series converges for |z| < 2 and diverges for |z| > 2.
Also it diverges for |z| = 2 as the n-th term sequence does not converge to zero.
(d) Let w = z1 , where z 6= 0 and apply previous solution to conclude that the series
converges for |z| > 1/2, and diverges for all other values.
1
6. Find all z ∈ C such that |ez | ≤ 1.
Proof: For z = x + iy, |ez | = ex ≤ 1 ⇔ x ≤ 0.
7. Show that the CR-equations in polar form are given by: ur = 1r vθ and uθ = −rvr .
Proof: Expressing x, y in polar co-ordinates we have
x = r cos θ, y = r sin θ.
So,
∂ ∂ ∂
u = ux x + uy y = ux cos θ + uy sin θ;
∂r ∂r ∂r
∂ ∂ ∂
v = vx x + vy y = vx cos θ + vy sin θ;
∂r ∂r ∂r
∂ ∂ ∂
u = ux x + uy y = r(−ux sin θ + uy cos θ),
∂θ ∂θ ∂θ
∂ ∂ ∂
v = vx x + vy y = r(−vx sin θ + vy cos θ).
∂θ ∂θ ∂θ
Now it is easy to see that the CR-equations hold if and only if ur = 1r vθ and uθ = −rvr .
Problem for Tutorial:
1. Let D = {z ∈ C : |z| ≤ 1}. For a fixed w in D, with |w| < 1, define the mapping
w−z
F : z 7→ 1−wz . Show that
(a) F is a map from D to itself;
(b) F (0) = w and F (w) = 0;
(c) |F (z)| = 1 if |z| = 1;
(d) F : D → D is bijective.
Proof:
(a) Since |w| < 1, |w−1 | > 1 while |z| ≤ 1 for all z ∈ D, so z w̄ 6= 1∀z ∈ D. Thus (a)
follows by applying Ex. 4 above.
(b) direct verification.
(c) Since |w| < 1 it follows once again from Ex. 4 that |F (z)| = 1 only of |z| = 1.
(d) Check that F ◦ F (z) = z.
2
n
P
2. Let R be the radiusPof convergence of n an z . For a fixed k ∈ N, find the radius of
k n kn
P
convergence of (a) an z , (b) an z .
k
1√ 1√ 1
= Rk (b) an (z k )kn is convergent (resp.
P
Proof: (a) n k
= n
lim sup |an | lim sup |an |
divergent) for |z| < R (resp. |z| > R); take w = z 1/k then an wkn converges (resp.
P
1 1
diverges) whenever |w| < R k (resp. |w| > R k )∗ .
∂
3. (a) Show that f satisfies the CR-equations if and only if ∂z f = 0. (Recall from Ex. 3
∂
above that ∂z = 21 ( ∂x
∂ ∂
+ i ∂y ).) Moreover, if f is analytic then f 0 (z) = ∂z
∂
f.
Proof: (a) Let f = u + iv. We have ∂∂z̄ f = 21 [(ux + ivx ) + i(uy + ivy )]. Thus, CR-
∂
equations hold iff ∂z f = 0. Also, f 0 (z) = ux +ivx while ∂z
∂
f = 12 [(ux +ivx )−i(uy +ivy )].
Applying CR-equations we get f 0 (z) = ∂z ∂
f.
4. Consider the following functions
(a)
xy(x + iy)
if x + iy 6= 0
f (x + iy) = x2 + y 2
0 if x + iy = 0
p
(b) f (x + iy) = |xy|
Show that f satisfies the CR-equations but it is not differentiable at the origin.
Proof:
x2 y xy 2
(a) u(x, y) = x2 +y 2
and v(x, y) = x2 +y 2
. So
u(x, 0) − u(0, 0) u(0, y) − u(0, 0)
ux (0, 0) = lim = 0, uy (0, 0) = lim = 0;
x→0 x y→0 y
v(x, 0) − v(0, 0) v(0, y) − v(0, 0)
vx (0, 0) = lim = 0, vy (0, 0) = lim = 0.
x→0 x y→0 y
Thus the CR-equations are satisfied. However, along the x-axis, f takes the
value 0. So, limh→0 f (h+i0)−f
h
(0)
is 0, while
f (h + hi) − f (0) (h3 + ih3 ) 1
lim = lim 2 2
= .
h(1+i)→0 h + hi h(1+i)→0 (h + h )(h + hi) 2
(b) ux (0, 0) = 0 = uy (0, 0); vx (0, 0) = vy (0, 0), hence CR equations are satisfied.
limh+i.0→0 f (h)−f
h
(0)
is 0, while limh(1+i)→0 f (h+hi)−f
h+hi
(0) 1
= 1+i , hence f is not dif-
ferentiable.
∗
Note that |x|k ≤ |y|k ⇐⇒ |x| ≤ |y|
