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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

  • 1
    OP: Whether you are [bdas](http://math.stackexchange.com/q/223713) 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 $U$?2012-12-16

1 Answers 1

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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