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```MATH3403/AS6/sol/ZLC/2021-22
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH3403: Functions of a Complex Variable
Solution to Assignment 6
P (z)
may only have singularities at the zeros of the denominator Q(z) = 1 −
Q(z)
sin(πz/2). Since sin(πz/2) = (exp(πiz/2) − exp(−πiz/2))/2i, we find the singularities
must satisfy exp(πiz/2) = i, so πz/2 = π/2 + 2kπ for some integer k. Therefore, all
possible singularities must have the form z = 1 + 4Z.
1. f (z) =
Moreover, the first order derivative of Q(z) (that is −π/2 cos(πz/2)) is 0 at z = 1 + 4Z,
and the second order derivative (that is (π/2)2 sin(πz/2)) is not 0 at z = 1 + 4Z, so
the denominator Q(z) = 1 − sin(πz/2) has double zeros at z = 1 + 4Z.
For z ̸= 1, −3, P (z) is not 0, while Q(z) has a double zero, so f has double poles at
z = 1 + 4Z for z ̸= 1 or −3.
For z = 1, P (z) has a double zero, the same result holds for Q(z), so z = 1 is an
removable singularity.
For z = −3, the nominator P (z) has a simple zero, while the denominator Q(z) has a
double zero, so z = −3 is a simple pole.
2. Repeat the proof of Riemann Removable Singularity Theorem, Theorem 5.5 in the
lecture note. We only need to modify the estimate on g(z) = zf (z) as below (here we
also take z0 = 0)
|g(z)| ≤ |z| · A|z|−1+ϵ = A|z|ϵ ,
so that limz→0 g(z) = 0.
All other arguments remain unchanged.
3. We first show that the origin is contained in the image. In other words, we need to
prove f has a zero inside the unit disc.
Suppose not, we may consider the analytic function 1/f . We find that |1/f (z)| ≤ 1
whenever |z| = 1 and there exists a point z0 ∈ D(1) such that |1/f (z0 )| > 1. This
So we have proved f has a zero inside the unit disc. Now we want to prove f (z) − w
also has a zero inside the unit disc for every w ∈ D(1). Notice that
|f (z)| ≥ 1 > |w|,
1
z ∈ ∂D(1),
so by Rouche’s theorem, the number of zeros of f (z) − w inside D(1) is the same as
f (z), which is at least one. So we are done.
4. The radius of convergence of f is 1. We consider the point z = reiθ for every θ =
2n
2πp/2k , where p and k are positive integers. Whenever n ≥ k, we have eiθ
= 1. It
follows that
k−1
∞
X
n X
n
iθ
iθ 2
f (re ) =
re
+
r2 ,
n=0
n=k
so that for every integer N > k we have
lim sup |f (reiθ )|
r→1−
≥ lim sup
r→1−
N
X
r
2n
−
k−1
X
reiθ
2n
n=0
n=k
=N − k + 1 −
k−1
X
reiθ
2n
n=0
Since N can be arbitrarily large, we find that |f (reiθ )| → ∞ as r → 1− , for every θ of
the form 2πp/2k .
Finally, if for some η ∈ [0, 1], e2πiη ∈ ∂D(1), then for small ϵ > 0, D(e2πiη , ϵ) ∪ ∂D(1)
k
must contain at least one point of the form e2πip/2 for some positive integers p, k. To
see this, note that the distance between two points on ∂D(1) is always smaller than
the length for the arc joining the two points. Thus it suffices to find positive integers
k
p, k such that the arc length between e2πiη and e2πip/2 is less than ϵ, i.e.
η−
p
≤ ϵ.
2k
p
Notice that 2k + 1 points k divides the interval [0, 1] into 2k small intervals, each of
2
1
length k . So for fixed η, we can always find p such that
2
η−
p
1
≤
.
2k
2k
Since k can be arbitrarily large, we can find desired integers p and k. So we are done.
2
```
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