TITLE: Prove that a given meromorphic function is rational
QUESTION [6 upvotes]: I'm doing some exercises in complex analysis, and I've reached one I simply can't figure out on my own, which is why I'm hoping for some help.
The exercise:
We assume that $h:\Bbb C\to \Bbb C \cup \{\infty\}$ is meromorphic with finitely many poles $z_1,...z_n$ and assume that there exist $k\gt0$, $N\in\Bbb N$ and $R\gt0$ such that $|h(z)|\le k|z|^N$ for $|z|\gt R$. Prove that h is a rational function.
What I've been thinking so far:
The definition of a rational function is, that you have to be able to write it on the form $f(z)=p(z)/q(z)$, where $p,q\in\Bbb C[z], q\neq 0$. So I guess I have to show that my function $h$ can be written this way too? I have, however not yet been able to figure out a way to do this which makes sense.
Earlier we've done an exercise, in which we've proven that if we let f be an entire function and we assume that $|f(z)|\le A+B|z|^n$ for $z\in\Bbb C$, where $A,B\ge0$ and $n\in\Bbb N$, then f is a polynomial of degree $\le n$. I've been thinking this might be useful, although a meromorphic function isn't entire.
Thank you for your time.

REPLY [4 votes]: Since $h$ has a finite number of poles, $h(z)\cdot (z-z_1)\cdots (z-z_n)$ is an entire function. Let $w_1,\ldots,w_n$ (this could be an empty list) be the zeros of $h(z)\cdot (z-z_1)\cdots (z-z_n)$. There must be a non-zero entire function $f$ such that $$ h(z)\cdot (z-z_1)\cdots (z-z_n)=(w-w_1)\cdots (w-w_n)\cdot f(z) .$$
We show that $f$ must be a constant and the problem will be solved. 
By assumption $$|(w-w_1)\cdots (w-w_n)\cdot f(z)|\le k|z|^N\cdot |(z-z_1)\cdots (z-z_n)|\le j|z|^M$$ for some $j$ and $M$. 
Thus $(w-w_1)\cdots (w-w_n)\cdot f(z)$ must be a polynomial and consequently $f$ is a non-zero polynomial (i.e. a constant).