# Prove that a given meromorphic function is rational

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.

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Hint: You're right, but you can find a polynomial to multiply by that will "cancel" all the poles and leave you with an entire function. –  Ted Shifrin May 24 '13 at 20:04

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).

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This is "morally" correct, but not correct as it stands. The first sentence needs to be repaired, with corresponding changes throughout. –  Ted Shifrin May 25 '13 at 12:50
Not sure what you mean. Do mean I am using "pole" too slangishly? There are functions with an infinite number of poles. Also, there are functions with uncountable singularities. –  Bobby Ocean May 26 '13 at 5:05
Multiplying by linear terms only kills single poles. If you have $1/z^2$, for example, you need to multiply by $z^2$, not just $z$. If you multiply by $\prod(z-z_j)^{n_j}$ instead of $\prod(z-z_j)$, where $n_j$ is the order of the pole, you do get an analytic function, and you can proceed as you indicated. –  Potato May 26 '13 at 5:22
Yes of course; Sorry I assumed the list $z_1,\ldots,z_n$ included repeating the poles, which is not a standard assumption. I should have said, let $z_1,\ldots,z_n$ be the poles including multiplicity. –  Bobby Ocean May 27 '13 at 17:30