# Proveing that if a Holomorphic function is one to one on a circle, then it's one to one within the disk created by the circle.

I would like to prove that if a function $f(z)$ is holomorphic on $\overline{D(P,r)}$ and one to one on $\partial D(P,r)$ then $f$ is one to one on $D(P,r)$. I noticed that for $w \in f(D(P,r))$ with $w \notin f(\partial D(P,r))$

$\frac{1}{2\pi i}\int_{D(P,r)}\frac{f'(s)}{f(s) - w}ds = \frac{1}{2\pi i}\int_{\gamma}\frac{1}{s-w}ds$ where $\gamma(\theta) = f(re^{\theta i} + P)$

Then it could be argued that since $f(re^{\theta i} + P)$ is one to one, it can only go around $w$ once.

The problem is actually proveing this(unless there is a better route to proveing it).

Also if we knew beforehand that $f(D(P,r))$ was disjoint from $f(\partial D(P,r))$ we could use a continuity argument where the independent variable was $w$

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Maybe this proof is helpful... – saz Nov 21 '12 at 9:10
... or this one... – saz Nov 22 '12 at 17:01
Herp yes the above solves it, thank you Saz. – Herp Nov 22 '12 at 20:40