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

I have been trying to solve the following problem:

The transformation $w=e^{i\theta}\frac{z-p}{\bar pz-1}$, where $p$ is constant, maps $|z|<1$ onto

  1. $|w|<1$ if $|p|<1$,
  2. $|w|>1$ if $|p|>1$,
  3. $|w|=1$ if $|p| = 1$,
  4. $|w| = 3$ if $|p| = 0$.

I could not get my calculations right. Please help.

share|cite|improve this question
Please check that my edits have not changed the meaning of your question. Also, you may want to clarify exactly what the question is. Do you mean, which of the four options are true? Is it multiple choice? Or are you trying to show all four? – froggie Nov 27 '12 at 15:57
@froggie It is a multiple choice question.I want to mean that "which of the four options is true?" Thanks for the edits..It has been done perfectly.. – learner Nov 27 '12 at 16:07
@saz I have eliminated the options (3) and (4)..But i could not tell anything about (1) and (2). – learner Nov 27 '12 at 18:57
up vote 1 down vote accepted

We have $$\left|\frac{z-p}{\bar{p} \cdot z-1} \right|^2 = \frac{(z-p) \cdot \overline{(z-p)}}{(\bar{p} \cdot z-1) \cdot \overline{(\bar{p} \cdot z-1)}} \\ = \frac{z \cdot \bar{z}-p \cdot \bar{z} - \bar{p} \cdot z + p \cdot \bar{p}}{z \cdot \bar{z} \cdot \bar{p} \cdot p- \bar{p} \cdot z - p \cdot \bar{z}+1}$$

Hence $$\left|\frac{z-p}{\bar{p} \cdot z-1} \right|^2 < 1 \\ \Leftrightarrow z \cdot \bar{z}-p \cdot \bar{z} - \bar{p} \cdot z + p \cdot \bar{p} < z \cdot \bar{z} \cdot \bar{p} \cdot p- \bar{p} \cdot z - p \cdot \bar{z}+1 \\ \Leftrightarrow |z|^2 \cdot (1-|p|^2) < (1-|p|^2) \\ \Leftrightarrow |z|<1$$

So the first option is true. Similar proof shows that the second one is true (remark: $1-|p|^2 <0$ for $|p|>1$). The third one can't be true since pre-images of closed subsets are closed ($f$ is continuous). The last one is clearly false.

Remark The given mapping is a Möbius transformation. This class of functions has some nice properties...

share|cite|improve this answer
thank you sir. I have got it. – learner Nov 28 '12 at 18:12

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.