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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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Chih: You do not even know that the limit of the absolute value is 1, since you do not know the limit exists (so you do not know the ration test applies). –  Andres Caicedo Jan 18 '11 at 1:57
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@Andres: Ah, you are right, the ratio test is weaker than the root test... I've modified the statement accordingly. Thanks! –  Hsien-Chih Chang 張顯之 Jan 18 '11 at 2:02
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2 Answers

up vote 6 down vote accepted

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.

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@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 2:42
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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. –  user17762 Jan 19 '11 at 2:45
    
@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 3:27
    
@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. –  user17762 Jan 19 '11 at 3:50
    
@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 3:57
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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.

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