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I was trying to solve problems from Gamelin's complex analysis book, and I came across the following question:

Suppose $f(z) = \sum_{k} a_k z^k$ is analytic for $|z| < R$, and suppose that $f$ extends to be meromorphic for $|z| < R+\epsilon,$ with only one pole $z_0$ on the circle $|z|=R$. Prove that $\frac {a_k}{a_{k+1}}\rightarrow z_0$ as $k\rightarrow \infty$.

Now it's clear that $\lim_{k\rightarrow \infty}|a_k/a_{k+1}| = |z_0| = R,$ as the limit is the radius of convergence for the power series of $f$, and $f$ extends to be analytic only upto $|z|< R$. But I can't figure out how to prove the actual statement. I was trying to work with the Laurent series of $f$ near $z_0$, but I got stuck. Any help would be appreciated.

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Hajime_Saito
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  • Subtract the principal part $h$ of $f$ in $z_0$. Then look at the Taylor expansions of $f - h$ and $h$ with centre $0$. – Daniel Fischer Aug 05 '15 at 14:09
  • @DanielFischer:I still can't get it. Can you kindly explain more, maybe in the form of an answer? I think I have some doubts which will be cleared by understanding the solution. – Hajime_Saito Aug 05 '15 at 14:25

1 Answers1

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Let $h$ be the principal part of $f$ in $z_0$. Then $f-h$ is holomorphic on the disk $\lvert z\rvert < R+\epsilon$ with Taylor series

$$(f-h)(z) = \sum_{n = 0}^\infty c_n z^n.$$

For the principal part we have

$$h(z) = \sum_{k = 1}^m \frac{b_k}{(z - z_0)^k},$$

where $m$ is the order of the pole. Now

\begin{align} \frac{1}{(z-z_0)^k} &= \frac{(-1)^{k-1}}{(k-1)!}\biggl(\frac{d}{dz}\biggr)^{k-1}\frac{1}{z-z_0}\\ &= \frac{(-1)^{k-1}}{(k-1)!}\biggl(\frac{d}{dz}\biggr)^{k-1} \frac{-1}{z_0}\sum_{n = 0}^\infty \frac{z^n}{z_0^n}\\ &= \frac{(-1)^k}{(k-1)!} \sum_{n = 0}^\infty \frac{1}{z_0^{n+1}} \biggl(\frac{d}{dz}\biggr)^{k-1}z^n\\ &= \frac{(-1)^k}{(k-1)!} \sum_{n = k-1}^\infty \frac{1}{z_0^{n+1}} \frac{n!}{(n+1-k)!}z^{n+1-k}\\ &= (-1)^k \sum_{r = 0}^\infty \binom{r+k-1}{k-1}\frac{z^r}{z_0^{k+r}}, \end{align}

so

$$h(z) = \sum_{r = 0}^\infty \underbrace{\Biggl(\sum_{k = 1}^m (-1)^kb_k\binom{r+k-1}{k-1}\frac{1}{z_0^{k}}\Biggr)}_{d_r}\biggl(\frac{z}{z_0}\biggr)^r.$$

Then we have

$$a_n = c_n + \frac{d_n}{z_0^n}$$

and

$$\frac{a_n}{a_{n+1}} = \frac{c_nz_0^{n} + d_n}{c_{n+1}z_0^{n+1}+ d_{n+1}}\cdot z_0.$$

Now we know that $c_n z_0^n \to 0$ since $f-h$ converges in $z_0$, so the assertion follows if we can show that $\liminf\limits_{n\to\infty} \lvert d_n\rvert \geqslant \delta > 0$ and $\dfrac{d_n}{d_{n+1}} \to 1$. If $m = 1$, then we have

$$d_n = -\frac{b_1}{z_0}$$

independent of $n$, so the conditions hold. Let's now suppose $m > 1$.

For fixed $k$ we have

$$\binom{n+k}{k} \sim \frac{n^k}{k!},$$

so

$$d_n = (-1)^m b_m \binom{n+m-1}{m-1}z_0^{-m} + O(n^{m-2}) \sim (-1)^m \frac{b_m}{z_0^m (m-1)!} n^{m-1},$$

which shows a) $\lvert d_n\rvert \to +\infty$ if $m > 1$ and

$$\frac{d_n}{d_{n+1}} \sim \biggl(\frac{n}{n+1}\biggr)^{m-1} \to 1,$$

so the conditions also hold for $m > 1$.

Daniel Fischer
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