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Math Help - complex anaysis help

  1. #1
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    complex anaysis help

    Let w,z \in \mathbb{C}.
    If |w|<1 and |z| \leq 1, show that |\frac{z-w}{1-\bar wz}| \leq 1. When does equality hold?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dori1123 View Post
    Let w,z \in \mathbb{C}.
    If |w|<1 and |z| \leq 1, show that |\frac{z-w}{1-\bar wz}| \leq 1. When does equality hold?
    Rudin, huh? What have you tried?
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    I know |\frac{z-w}{1- \bar wz}|=\frac{|z-w|}{|1-\bar wz|} and |z-w| \leq |z|+|w| \leq 1+1 =2, not sure what to do with |1 - \bar wz| , I'm trying to work this way, not sure if it works... |1 - \bar wz|=|-(\bar wz - 1)|=|(- \bar w)(z - \frac{1}{\bar w})|=|(- \bar w)(z - \frac{w}{w \bar w})|=|\bar w||z - \frac{w}{w \bar w}|=|\bar w||z - \frac{w}{|w|^2}|=...
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    Quote Originally Posted by dori1123 View Post
    Let w,z \in \mathbb{C}.
    If |w|<1 and |z| \leq 1, show that \Bigl|\frac{z-w}{1-\bar wz}\Bigr| \leq 1. When does equality hold?
    You want to show that |z-w|\leqslant|1-\bar wz|. Square both sides and use the fact that |z|^2 = \bar zz. Then what you are trying to prove is equivalent to (z-w)(\bar z -\bar w) \leqslant (1-\bar wz)(1-w\bar z). Multiply out both sides and simplify ... .
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