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Let $f$ be a meromorphic function on $\mathbb{C}$ for which $|f(z)|\to\infty$ as $|z|\to\infty$. Show that $f$ cannot have poles at all integer points.

I know that we can construct a homeomorphism from the extended complex plane $\mathbb{C} \cup \infty$ to the sphere $S^2$ by stereographic projection, and further that the sphere is compact. So, if a meromorphic function has infinitely many disjoint poles (i.e at the integers), perhaps we can construct a cover without a finite subcover. I am however getting confused as to the details, in particular where $\infty$ comes in. Many thanks for you help.

1 Answers1

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Let $R>0$ be such that $|f(z)|\ge1$ if $|z|>R$. Then $g(z)=1/f(z)$ is meromorphic and bounded on the punctured disk $\{0<|z|<1/R\}$. It can be extended to the disk as an analytic function. Since $f(z)\to\infty$, we must have $g(0)=0$. If all integers were poles of $f$, then $g(1/n)=0$ for all $n\in\mathbf{Z}$, $n\ne0$. This implies that $g$ is identically equal to $0$.

Julián Aguirre
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  • Thank you, this makes sense, and is an easier way than the whole "Riemann sphere is compact" thing. –  Apr 11 '13 at 15:42
  • I have a question to your solution i.e. shouldn't it be $ g(z)=1/f(1/z)$? If yes , then how do you know that $g(z)$ is meromorphic? – luke12 May 06 '13 at 15:58
  • @luke12: Yes, it should be. It's meromorphic for $z \neq 0$ because $f(z)$ is meromorphic for $z \neq \infty$, and since $1/f(1/z) \to 0$ as $|z| \to 0$ we can extend it to an analytic function at $z=0$. – Antonio Vargas May 06 '13 at 16:18
  • @luke12: In the future it would be best no not ask questions as "Answers". If you need to you can ask a new "Question" and link to the Answer you have a question about. I'll be flagging this Answer for deletion. – Antonio Vargas May 06 '13 at 16:20
  • @AntonioVargas: Ok, sure. Next time I'll do that. – luke12 May 06 '13 at 16:31