# Homework Help: Holomorphic function reduces to a polynomial

1. Sep 25, 2010

### snipez90

1. The problem statement, all variables and given/known data
Let f: C -> C be a holomorphic function such that there is a constant R such that |z| >
R implies |f(z)| > R. Show that f is a polynomial.

2. Relevant equations
Not sure, I pulled this randomly from a complex analysis qualifying exam.

3. The attempt at a solution
So from experience a typical way to show that a holomorphic function is a polynomial is to apply Cauchy estimates (e.g. the immediate estimates from the Cauchy integral formula). However that approach doesn't seem to work here, since we usually have to let the boundary circle in the Cauchy integral formula either get larger and larger or smaller and smaller. To me it's not clear how the given growth condition gives estimates.

I've also thought about the maximum modulus principle, but I don't how to use it well, even if it does apply here. Can someone provide a hint? Thanks in advance.

2. Sep 25, 2010

### Office_Shredder

Staff Emeritus
You might want to look at 1/f(z). It's bounded, except for the possibility of poles inside |z|<R. Try to make a new holomorphic function from this

3. Sep 25, 2010

### snipez90

All right thanks. I did consider 1/f, but erroneously thought of Liouville. I'll try your suggestion.

4. Sep 25, 2010

### Office_Shredder

Staff Emeritus
Well, Liouville will come into play. But first you need to find a slightly different function that's actually bounded