Holomorphic function reduces to a polynomial

[snipez90]

Homework Statement

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.

Homework Equations

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

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.

 

[Office_Shredder]

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

 

[snipez90]

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

 

[Office_Shredder]

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