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Math Help - Complex Analysis, Schwarz lemma

  1. #1
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    Complex Analysis, Schwarz lemma

    Please help me with this one:

     D=\left \{ z:\left | z \right |<1 \right \}

    Let f and  g be an analytical functions \rightarrow D " alt="f, g \rightarrow D " />

    and also f(0)=g(0), f'(0)=g'(0)

    Prove that f(z)=g(z), \forall z\in D.

    Thanks!
    Last edited by sinichko; December 15th 2010 at 12:11 AM.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by sinichko View Post
    Please help me with this one:

     D=\left \{ z:\left | z \right |<1 \right \}

    Let f and  g be an analytical functions \rightarrow D " alt="f, g \rightarrow D " />

    and also f(0)=g(0), f'(0)=g'(0)

    Prove that f(z)=g(z), \forall z\in D.

    Thanks!
    Not true, for example f(z) = z^2,\ g(z) = z^3.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by sinichko View Post
    Please help me with this one:

     D=\left \{ z:\left | z \right |<1 \right \}

    Let f and  g be an analytical functions \rightarrow D " alt="f, g \rightarrow D " />

    and also f(0)=g(0), f'(0)=g'(0)

    Prove that f(z)=g(z), \forall z\in D.

    Thanks!
    That statement is false, choose:

    f(z)=z^2,\;g(z)=z^3

    Fernando Revilla

    Edited: Sorry, I didn't see Opalg's post.
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  4. #4
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    See if you have the statement right.

    Meanwhile, a good strategy for any Schwarz problem is to construct a function satisfying the hypotheses of the lemma. Then everything frequently falls nicely onto your lap.

    You need an analytic function \to D" alt="h \to D" /> sending 0 to 0.

    Try h(z) = \frac{1}{2}\left[f(z) - g(z)\right]..
    (Question: why did I scale by one half?)

    This may not lead where you want (especially since at the moment the conclusion you seek is unclear), but constructing something like this is usually a good first step.

    Verify that h satisfies the hypotheses of the lemma and apply it. See what happens.
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