Originally Posted by
Bruno J. The statement is false as written; what about a constant map? I suppose you are assuming the map to be conformal; if so, then it's true and may be proven like this :
By Riemann's mapping theorem, we have a conformal map $\displaystyle \sigma : G \rightarrow D$ where $\displaystyle D$ is the open unit disc. Consider $\displaystyle F : D\rightarrow D$ given by $\displaystyle F(z) = (\sigma f\sigma^{-1})(z)$. Now $\displaystyle F$ is a conformal map from the unit disc to itself, and therefore has the form $\displaystyle F(z) = \frac{z-a}{1-\overline{a}z}$ for some $\displaystyle |a|<1$. If $\displaystyle |a|>0$, then the fixed points of this transformation are the two solutions of $\displaystyle z^2=\frac{a}{\overline{a}}$, which lie on the boundary of $\displaystyle D$ (and hence they are not fixed points of $\displaystyle F$, whose domain does not include the boundary). So the only possibility is $\displaystyle |a|=0$, i.e. $\displaystyle F(z)=z$ and $\displaystyle f(z)=z$.