problems in complex analysis

Sep 2009
173
0
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
Jun 2009
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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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Sep 2009
173
0
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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