# 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 $\displaystyle \sigma : G \rightarrow D$ 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 $\displaystyle f(z)=z$.

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 $\displaystyle \sigma : G \rightarrow D$ 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 $\displaystyle f(z)=z$.
Thank u