Prove that if the only singularities of a function are poles then the function must be rational.
So originally this was an iff statement, but I have solved the other direction of the proof. I'm just not quite sure how to prove it going this direction.
I'm thinking to assume the only singularities of a function are poles, but I'm not sure how to work this down to being only for rational functions.
I toyed with the idea of contradiction, but still wasn't quite sure how I should go about proving this.
I'm going to assume that your function is $f:\bar{\mathbb C}\to\bar{\mathbb C}$, otherwise the statement isn't true (singularities at $\infty$ need to be considered, otherwise $\exp(z)$ would be a counterexample). I'm also assuming that the singularities are isolated. Then there are only finitely many of them, since $\bar{\mathbb C}$ is compact and an infinite set in a compact space always has accumulation points.
First, we consider all the poles $z_1,\dots,z_n$ outside of $\infty$. Let $h_1,\dots h_1$ be the parts of the Laurent decomposition with negative exponents, extended to all of $\bar{\mathbb C}$. These are known to be rational because only finitely many terms in the fractional part of the Laurent series are non-zero, so they are a finite sum of rational functions, and thus rational. Now consider
$$g(z)=f(z)-\sum_{k=1}^n h_k(z).$$
This function has no poles except possibly at $\infty$. All other singularities are removable. So if we restrict this function to $\mathbb C$, it's entire. So if it has no pole at $\infty$, then it is constant and thus rational due to Liouville's theorem. If it has a pole at $\infty$, then it is a polynomial and thus rational. So in any case, $g$ is rational.
Now since
$$f=g+\sum_{k=1}^n h_k$$
is a finite sum of rational functions it is rational itself.