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Math Help - problems in complex analysis

  1. #1
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    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
    Last edited by mr fantastic; May 28th 2010 at 04:46 AM. Reason: Deleted duplicate question.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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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 \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.
    Last edited by mr fantastic; May 28th 2010 at 04:46 AM.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    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
    Last edited by mr fantastic; May 28th 2010 at 04:46 AM.
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