# Thread: problems in complex analysis

1. ## problems in complex analysis

Hi All,
I need the solution of this question:

Let G ≠C be simply connected domainin C. Let f:G→G be holomorphic and fixes two Points. Show that f(z) = z for all z in G.(Where c is the set of complex numbers)

Thanks all

2. 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 $\sigma : G \rightarrow D$ where $D$ is the open unit disc. Consider $F : D\rightarrow D$ given by $F(z) = (\sigma f\sigma^{-1})(z)$. Now $F$ is a conformal map from the unit disc to itself, and therefore has the form $F(z) = \frac{z-a}{1-\overline{a}z}$ for some $|a|<1$. If $|a|>0$, then the fixed points of this transformation are the two solutions of $z^2=\frac{a}{\overline{a}}$, which lie on the boundary of $D$ (and hence they are not fixed points of $F$, whose domain does not include the boundary). So the only possibility is $|a|=0$, i.e. $F(z)=z$ and $f(z)=z$.

3. 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 $\sigma : G \rightarrow D$ where $D$ is the open unit disc. Consider $F : D\rightarrow D$ given by $F(z) = (\sigma f\sigma^{-1})(z)$. Now $F$ is a conformal map from the unit disc to itself, and therefore has the form $F(z) = \frac{z-a}{1-\overline{a}z}$ for some $|a|<1$. If $|a|>0$, then the fixed points of this transformation are the two solutions of $z^2=\frac{a}{\overline{a}}$, which lie on the boundary of $D$ (and hence they are not fixed points of $F$, whose domain does not include the boundary). So the only possibility is $|a|=0$, i.e. $F(z)=z$ and $f(z)=z$.
Thank u