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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.

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    @saz I have eliminated the options (3) and (4)..But i could not tell anything about (1) and (2).2012-11-27

1 Answers 1

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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...

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    thank you sir. I have got it.2012-11-28