# problems in complex analysis

#### raed

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

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#### Bruno J.

MHF Hall of Honor
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 $$\displaystyle f(z)=z$$.

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#### raed

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

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