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This is my first question in mathSE, hope that it is suitable here!

I'm currently self-studying complex analysis using the book by Stein & Shakarchi, and this is one of the exercises (p.67, Q14) that I have no idea where to start.

Suppose $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ has the power series expansion $\sum_{n=0}^\infty a_n z^n$ in the open unit disc, then

$\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0$.

If the limit is taking on $|\frac{a_n}{a_{n+1}}|$ and assume the limit exists, by the radius of convergence we know that the answer is $1$. But what can we say about the limit of the coefficient ratio, which is a pure complex number? I've tried to expand the limit directly by definition, with no luck. And I couldn't see how we can apply any of the standard theorems in complex analysis.

I hope to get some initial directions about how we can start thinking on the problem, rather than a full answer. Thank you for the help!

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Hsien-Chih Chang 張顯之
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2 Answers2

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Hint: Assume that you have a simple pole at $z = z_0$, where $|z_0| = 1$ and try to prove it. In particular, take $f(z) = \frac{g(z)}{z-z_0}$ where $g(z)$ is holomorphic on $\Omega$. Prove the result for this case. (Expand $\frac{1}{z-z_0}$ about $z=0$ and do some manipulations). Now the same idea can be extended for higher order poles.

EDIT: For a simple pole, $f(z) = \frac{g(z)}{z-z_0} = \displaystyle \sum_{n=0}^{\infty} a_n z^n$. Since $g(z)$ is holomorphic, $g(z) = \displaystyle \sum_{n=0}^{\infty} b_n z^n$. So $\displaystyle \sum_{n=0}^{\infty} b_n z^n = (z-z_0) \displaystyle \sum_{n=0}^{\infty} a_n z^n \Rightarrow b_{n+1} = a_n - z_0 a_{n+1}$.

Now what can we say about $\displaystyle \lim_{n \rightarrow \infty} b_n$ and $\displaystyle \lim_{n \rightarrow \infty} a_n$?

(Note: $g(z)$ holomorphic on $\Omega$ whereas $f(z)$ is holomorphic except at $z_0$, a point on the unit disc).

This same idea will work for higher order poles as well.

  • @Sivaram: I must have gone astray somewhere. I tried to bound the limit $\lim_{n\to\infty} b_n/a_n$ and represent $a_n$ in terms of $b_n$. But my question is, while I tempt to prove that $a_n$ is bounded below, $a_n$ can converge to zero if $f(z) = \sum_{n=0}^{\infty} z^n/n$, right? Some of my reasoning must be wrong, but I do not know which one... – Hsien-Chih Chang 張顯之 Jan 19 '11 at 02:42
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    @Hsien-Chih Chang: I was thinking on the following lines. Since $g(z)$ is holomorphic, $b_n \rightarrow 0$ and since $f(z)$ is not holomorphic at $z_0$, $a_n$ doesn't tend to zero. –  Jan 19 '11 at 02:45
  • @Sivaram: Sorry if this is obvious, and I must misunderstand something really fundamental. So for $f(z) = \sum_{n=0}^{\infty} z^n/n$ with $a_n = 1/n$ and $z_0 = 1$, because $a_n$ goes to zero, $f$ must be violating one the conditions of my question, that is $f$ is holomorphic on $\Omega$ which contains the closed unit disc except $z_0$ with power series expansion on the open unit disc. But $f(z)$ is in fact $\log(z-1)$, which seems to satisfy the conditions. – Hsien-Chih Chang 張顯之 Jan 19 '11 at 03:27
  • @Hsien-Chih Chang: My argument is not complete yet. I need to give it further thought though. I think since the closed unit disk is in an open set $\Omega$ and $z_0$ is the only singularity, $a_n$ cannot tend to zero. As for your example, the series is $-\log(1-z)$ only for $|z| < 1$ and is not $-\log(1-z)$ on the entire $\Omega$. So the example you have given doesn't satisfy the premise of the problem. I think you could argue using the fact that the series is valid on $\Omega \backslash \bar{D}$. I will get back to you sometime tomorrow in case you have not figured it out by then. –  Jan 19 '11 at 03:50
  • @Sivaram: Thank you very much for your explanation, I really appreciate your help! I'll give it a harder try on the problem, and give comments here whether it succeed or not. – Hsien-Chih Chang 張顯之 Jan 19 '11 at 03:57
  • @Sivaram: Finally got it! $a_n$ will not converge to zero indeed. Nice hint(s)! – Hsien-Chih Chang 張顯之 Jan 19 '11 at 16:13
  • @Hsien-Chih Chang: So how did you argue that $a_n$ doesn't converge to zero? –  Jan 20 '11 at 03:45
  • @Sivaram: It follows the lines your provided. For a $(k+1)$-pole, by calculating the convolution of the power series of $g(z)$ and $(z-z_0)^{k+1}$ and comparing the coefficients to $a_n$, and after some manipulations we have $a_n = -z_0^{-n-1} g^{(k)}(z_0)$. Using the fact that $g^{(k)}(z_0) \neq 0$ and we're done. – Hsien-Chih Chang 張顯之 Jan 20 '11 at 06:53
  • @Hsien-Chih Chang: Oh Ok. Nice one. I was thinking if we could directly argue that $\lim a_n$ has to be non-zero given that $f$ is analytic in a small neighborhood around $z_0$ but not analytic at $z_0$. But anyways good one. –  Jan 20 '11 at 06:57
  • @Sivaram: I post another question which is related to your approach. – Hsien-Chih Chang 張顯之 Jan 20 '11 at 08:23
  • @Hsien-ChihChang張顯之 Hello! Sorry that I didn't get it, why $\lim_{n \to \infty} a_n$ is not zero. Could you explain more? And in fact we need $\liminf_{n \to \infty} a_n \neq 0$ here. How to get it? Thank you! – Edward Wang Dec 30 '18 at 08:30
  • I honestly fail to see how the same idea will work for higher poles. Can you elaborate? I managed to prove the fact for simple poles, but can't seem to find a generalisation here. – Sellerie May 29 '21 at 09:32
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Let's try another way to solve this problem.

Construct a contour, which consists of two parts. The first part is a circle and is a little bit larger than the unit circle, except near the point $z_0$. We call it $C_1$, and make it have absolute value strictly larger than $1+\delta$ for some $delta$. The second part is a small circle with radius $\epsilon$.

Suppose $z_0$ as pole have degree k, then $f(\zeta)=\frac{g(z)}{(z-z_0)^k}$, where $g(z)$ is holomorhpic.

For $C_1$, we have that $|\int_{C_1}\frac{g(\zeta)}{\zeta^n}\frac{1}{(\zeta-z_0)^k}d\zeta|\leq \frac{1}{\epsilon^k}\int_{C_1}\frac{M}{(1+\delta)^n}d\zeta \to 0$

For $C_\epsilon$, we have $$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+1}}e^{-i\theta k} d\theta$$

In the same way,

$$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} d\theta$$

By multiplying the second one with $z_0$, and computing the difference, we get,

$$ \Delta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} \epsilon e^{i\theta}d\theta \to 0$$, as $\epsilon \to 0$

which means

$$ \frac{\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta}{z_0\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta} \to 1 $$ as $\epsilon \to 0$

Combining all of these and Cauchy's integral formuals that $a_0=f(0)=\frac{1}{2\pi i}\int_C\frac{f(\zeta)}{\zeta}d\zeta$ and $n!a_n=f^{(n)}(0)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{\zeta^{n+1}}d\zeta$, we split $C$ as $C_1$ and $C_\epsilon$, we onle need to choose carefully the $\epsilon$'s and $\delta$'s to complete our proof.

Xiaochuan
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