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Let $U$ be an open set of $\mathbb{C}$ containing $D=\{z\in \mathbb{C}: |z|<1\}$ and let $f:U\to \mathbb{C}$ be map defined by $f(z)= e^{iψ} \frac{z-a}{1-\overline{a}z}$ for $a\in D$ and $ψ\in[0,2\pi]$. Which of the following statements are true?

  1. $|f(e^{iθ})| =1$ for $0<θ<2\pi$ .
  2. $f$ maps $\{z\in\mathbb{C}: |z|<1\}$ onto itself.
  3. $f$ maps $\{z\in \mathbb{C}: |z|\le 1\}$ into itself.
  4. $f$ is one-one.

I am stuck on this problem . Can anyone help me please.................

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i don't know where to begin.......... –  pankaj Dec 18 '12 at 13:54
    
please guide me.......... –  pankaj Dec 18 '12 at 14:27
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1 Answer 1

If you consider $|a|<1$, the functions of the form $f$ restricted to $D$ describe all of the automorphisms of the unit disk.

If you think about what this means, the boundary of the unit disk (the unit circle) is exactly where $|z|=1$. This should answer problem 1 when you think about the norm of $e^{i \theta}$. In other words, if $|z|=1$, then $|f|=1$.

For 2 and 3, recall that since $f$ restricted to $D$ is an automorphism, it takes domain onto the domain. just think what's in $D$.

For the last part, remember that $f$ takes $D$ onto $D$ and has an inverse since it is a conformal mapping. You can even explicitly state inverse and its domain.

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still not getting sir........... –  pankaj Dec 18 '12 at 13:47
    
Sir,give me some more hints for this question.. –  pankaj Dec 18 '12 at 14:17
    
automorphisms of the unit disk??????means............ –  pankaj Dec 19 '12 at 13:14
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@pankaj: It looks like jshin is answering using language and terminology you are unfamiliar with. Please take a look at our FAQ on homework questions to see how you can improve your questions in a way that makes it easier for others to help you. –  Willie Wong Dec 19 '12 at 13:21
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