# A transformation problem related to complex analysis

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 $|p| = 0$.

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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

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.