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The question is from Stein & Shakarchi - Complex Analysis Chapter 2, Exercise 14.

Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ is given by a power series expansion $$f(z)=\sum^\infty_{n=0}a_n z^n$$ in the unit disc $D_1(0)$, then $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$

I solved this problem by using the pole formula $$f(z)=(z-z_0)^{-m}g_0(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_n(z)$$ where $g_0$ and $g_n$ are holomorphic and not zero at $z_0$, and $m$ is a positive integer. These are defined on $D_{1+\epsilon}(0)\subset\Omega$, which contains $z_0$ and $\epsilon$ is sufficiently small. I've got below : $$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}=\lim_{n\to\infty} \dfrac{\frac{f^{(n+1)}(0)}{(n+1)!}}{\frac{f^{(n)}(0)}{n!}}=\lim_{n\to\infty}\dfrac{1}{n+1}\left(\dfrac{-m-n}{0-z_0}+H(0)\right)=\dfrac{1}{z_0}.$$ where $H(z)$ is holomorphic in the disc.

Actually, the pole formula I've used appears in Chapter 3 so I should not use this, but I think it is worth to try. Is there something wrong about this solution?

Very thanks.

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    $\begingroup$ I don't understand the question. If $f$ is given by a power series valid in a nbhd of the closed unit disk then it has no poles on the unit circle. $\endgroup$ Jan 24, 2017 at 20:17
  • $\begingroup$ @juanarroyo OH, I've made a horrible mistake. Thank you for pointing out that. I edited the question. The power series is vaild for $f$ only in the disc. $\endgroup$
    – Jinmu You
    Jan 24, 2017 at 20:29
  • $\begingroup$ Shouldn't that be "$f(z)=(z-z_0)^{-m}g_1(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_{n+1}(z)?$" $\endgroup$
    – zhw.
    Jan 24, 2017 at 20:40
  • $\begingroup$ @zhw. I wrote that two function $g_1$ and $g_2$ for just a notation. Maybe $g_2$ consists of $(z-z_0)$ and $g_1$, but the form will be very complicated. But your notation seems pretty nice. Thanks. $\endgroup$
    – Jinmu You
    Jan 24, 2017 at 20:43

3 Answers 3

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First note that the proof you gave works perfectly well for $f(z)=\frac{1}{z+1} + \frac{1}{z-1}$. Your functions $g_n$ will not be defined in $D_{1+\epsilon}(0)$, but they are defined on an open set containing $0$, which is enough for your proof to go through. This however would yield a contradiction, since then you would have shown that $-1=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=1$.

So what is the flaw in your proof? I see two. First, you assumed that none of the $a_n$ are $0$. Indeed if you look at the Taylor series for the function $f$ I defined, you will see that every even term is $0$. Thus the limit doesn't even exist.

Second, I'm not sure how you got that function $H$. Your proof is missing a lot of details, and when I try to reproduce it instead of $H$ I get

$$\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$$

If you're looking for solutions, this has been asked before.

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  • $\begingroup$ Thanks very much. I know other good sols and I'm trying a different approach. I think $f$ you defined isn't satisfy a condition of the question : it has two poles in $\Omega$ (on the unit circle), so it's not a good counterexample. I didn't consider of $a_n$ being zero, since there was no description about that in the original question. I just have predicted that none of them will be zero since the author didn't mention about that. That maybe a critical point. For $H(z)/(n+1)$, your formula is true and I found there is a logical gap since I assumed $g_n(0)\neq 0$. Thanks. $\endgroup$
    – Jinmu You
    Jan 25, 2017 at 6:47
  • $\begingroup$ "it has two poles in $\omega$": but that's precisely the point. Your proof did not use the fact that $f$ has only one pole, so I can just as well apply it to my $f(z)$ and achieve a contradiction. Part of proving that $lim\frac{a_n}{a_{n+1}} = z_0$ is proving that the limit exists in the first place. $\endgroup$ Jan 25, 2017 at 14:01
  • $\begingroup$ I totally misunderstood that. Thanks for kind explanation! Could I stick to this approach with some modification to prove the question? $\endgroup$
    – Jinmu You
    Jan 25, 2017 at 14:27
  • $\begingroup$ I'm not sure, I thought about it for a bit but I couldn't figure out a way to proceed past $\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$. $\endgroup$ Jan 25, 2017 at 14:39
  • $\begingroup$ I'm stuck too.. OK, Thanks for helping me!! $\endgroup$
    – Jinmu You
    Jan 25, 2017 at 15:00
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I see you are trying a different approach. It's good.
However your argument does not work since you do not use the condition that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle.

Since your argument is a local one, if $g_0(z)$ is defined on $D_1(0)\cup \{|z-z_0|<\delta \}$, holomorphic and not zero at $z_0$, your argument goes and concludes $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\frac{1}{z_0}$.

If your argument is correct, you can prove:

Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for two poles at $z_0$ and $z_1$ on the unit circle. Show that if $f$ is given by a power series expansion $$f(z)=\sum^\infty_{n=0}a_n z^n$$ in the unit disc $D_1(0)$, then $$ \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0\quad \text{and}\quad \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$
Proof. Consider $f(z)=(z-z_0)^{-m}g_0(z)$ on $D_1(0)\cup \{|z-z_0|<\delta \}$. Then $$ \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$ Consider $f(z)=(z-z_1)^{-m}g_1(z)$ on $D_1(0)\cup \{|z-z_1|<\delta \}$. Then $$ \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$

Do you recognize your argument is wrong ?

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  • $\begingroup$ Yes, I got it. Thanks very much! $\endgroup$
    – Jinmu You
    Jan 25, 2017 at 14:29
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Hint: Suppose the pole of order $d\geq 1$. Then $(z-z_0)^d f(z)$ is holomorphic in a disc of radius $1+\epsilon$. Show that: $$ f(z) = \sum_{j=1}^d \frac{p_j}{(z-z_0)^j} + g(z) $$ with $p_d\neq 0$ and $g$ holomorphic in a disc of radius $1+\epsilon$. Now, expand each term in a powerseries at $0$ and show that the series for $p_d(z-z_0)^{-d}$ (a negative binomial expansion) will dominate all other terms and give the provided limit ratio.

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