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

(a) $|f(e^{i\theta})|=1$ for $0≤\theta≤ 2\pi $
(b) $f$ maps $\{z\in\mathbb{C}\mid|z|\leq1\}$ onto itself
(c) $f$ maps $\{z\in\mathbb{C}\mid|z|\leq 1\}$ into itself
(d) $f$ is one-one

How should I able to solve this problem. Can anyone help me please

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What is $D$? The unit circle? –  Nameless Dec 16 '12 at 12:57
    
yes it is....... –  abdakchi Dec 16 '12 at 13:05
    
The unit circle or the unit disc? –  Nameless Dec 16 '12 at 13:08
    
extremely sorry.it is open unit disk. –  abdakchi Dec 16 '12 at 13:17
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OP: Whether you are bdas or not, your definition of $f$ is plagued by the same problem than theirs: what is the image of $1/\bar a$ when $1/\bar a$ is in $U$? –  Did Dec 16 '12 at 13:36
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1 Answer

For the 1st $$\left|f(e^{i\theta})\right|=\frac{\left|e^{i\omega}(e^{i\theta}-a)\right|}{\left|1-\bar{a}e^{i\theta}\right|}=\frac{\left|e^{i\theta}-a\right|}{\left|1-\bar{a}e^{i\theta}\right|}=\frac{\left|e^{i\theta}-a\right|}{\left|1-ae^{-i\theta}\right|}=\frac{\left|e^{i\theta}-a\right|}{\left|e^{i\theta}-a\right|}=1$$ For the 4th, $$f(z_1)=f(z_2)\Rightarrow \frac{z_1-a}{1-\bar{a}z_1}=\frac{z_2-a}{1-\bar{a}z_2}\Rightarrow ...\Rightarrow (z_1-z_2)(1-\left|a\right|^2)=0$$ Since $\left|a\right|<1$, $z_1=z_2$ and so $f$ is 1-1. I am not sure what you are asking in the 3rd.

EDIT: As did pointed out in the comment section, $f$ is not well defined

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