# No essential singularity at infinity implies polynomial [duplicate]

This question already has an answer here:

Suppose $f(z)$ is holomorphic in the whole plane, and that $f(z)$ does not have an essential singularity at $\infty$. Prove that $f(z)$ is a polynomial.

Since $f(z)$ does not have an essential singularity at $\infty$, it has either a removable singularity or a pole. If it has a removable singularity, we can define $f(\infty)$ in such a way that $f$ is holomorphic in the extended plane. Having a pole at point $a$ usually means that we can write $f(z)=\dfrac{g(z)}{(z-a)^k}$ where $g(z)$ is holomorphic. I'm not entirely sure what a pole at infinity means in this case.

## marked as duplicate by Potato, Dominic Michaelis, Cameron Buie, Davide Giraudo, Lord_FarinOct 18 '13 at 20:28

• Hint: $f(1/z)$ has a non-essential singularity at zero. – Prahlad Vaidyanathan Oct 18 '13 at 18:55
• A pole at infinity means we can write $f(z) = z^k\cdot g(z)$ where $g$ is holomorphic in a neighbourhood of $\infty$. – Daniel Fischer Oct 18 '13 at 18:58