# Analytic functions with nonessential singularity at infinity must be a polynomial

This is an exercise from Alhfors Complex Analysis book- to show that an analytic function with a nonessential singularity at infinity must be a polynomial. It seems like it should probably be pretty straight forward, but I must be missing something. If it has a removable singularity at infinity than it extends to an analytic function on the Riemann sphere, and so must be constant by Liouville's theorem. What if there is a pole at infinity though? This was homework some time ago, and I never finished it :/ but have been thinking about it again recently. Thanks :)

## 2 Answers

Another hint: look at the function $f(\frac{1}{z})$ at z = 0, it has a nonessential singularity at 0...

• I think I have it now. Funny thing is I tried looking at f(1/z) before and for some reason missed it. I think I have it now. Thank you :) – MJoszef Mar 29 '11 at 1:16
• How do I write the proof formally? suppose f is not a polynomial. So positive power of $z$ is non-termnating. consider $f(1/z)$ The negative power is non-terminating. So has an essential singularity at $z=0$. So, $f$ has essential singularity at $f$. – Unknown x May 22 at 2:17

Hint: consider the Laurent series in the annulus $0 < |z| < \infty$.

• Ahlfors book hasn't introduced Laurent series yet, so I was looking for a way along another path- hopefully. Thank you though. – MJoszef Mar 29 '11 at 1:18