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Let $U$ be an open subset of $C$ containing $D=\{ z\in \Bbb C :|z|\leqslant 1\}$ and let $f\colon U\rightarrow C$ defined by $f(z)=e^{i\theta}\frac {z-a}{1-\overline az}$ for $a\in D$ and $0\leqslant \theta \leqslant 2\pi$. Which of the following statements are true?

  1. $|f(e^{i\gamma})|=1$ for $0\leqslant \gamma \leqslant 2\pi$.
  2. $f$ maps $\{z\in C :|z|\leq 1\}$ onto itself.
  3. $f$ maps $\{z\in C :|z|\leq1\}$ into itself.
  4. $f$ is one to one.

Obviously $f$ is not analytic in $U$. Please help how to solve.

share|improve this question
    
Why don't you think $f$ is analytic in $U$? –  Robert Israel Oct 29 '12 at 18:49
    
A more descriptive title might attract more people to your problem. –  Pragabhava Oct 29 '12 at 19:09
    
@bdas:See my answer and do not worry about the down vote. I answered the question based on the assumptions you have been given in the question. –  Mhenni Benghorbal Nov 2 '12 at 16:44
1  
@Mhenni Benghorbal : thank you –  bdas Nov 4 '12 at 12:24
    
@bdas: You are welcome. –  Mhenni Benghorbal Nov 4 '12 at 20:27

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