TITLE: Meromorphic function tending to infinity cannot have poles at all integer points
QUESTION [8 upvotes]: 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.

REPLY [6 votes]: 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 $0$.