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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 contradicts the Maximal Principle. 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