Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 is open unit disk. – abdakchi Dec 16 '12 at 13:17
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.