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Let $a \in \mathbb{C}$, $|a| < 1$. Also let $f(z) = \dfrac{z - a}{1 - \overline{a}z}$.

I am asked to prove that $|f(z)| < 1$ if $|z| < 1$ and that $|f(z)| = 1$ if $|z| = 1$.

What is a good way to proceed on this problem?

Marcus Emilsson
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  • I tried to post some ideas but, as it is usual in the last few months gods know why, the system like "freezes" and I can't see clearly what I'm typing in MathJax (in the post I'm writing it just appears the usual jibberish of the MathJax symbols) and I just can't go on like that. Sorry. – Timbuc Apr 27 '15 at 08:37
  • $f(z)$ is the limit of a converging geometric series. Maybe that helps. – Sebastian Bechtel Apr 27 '15 at 08:46
  • This has been asked and answered many times. Here are just a few few other relevant ones:first second third fourth. There are many more to be found. – mrf Apr 27 '15 at 08:54

2 Answers2

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We first prove that if $|z|=1$ then $|f(z)|=1$. For we have $$ |f(z)|=|\frac{z-a}{1-\bar az}|=|\frac 1{\bar z}\frac{z-a}{1-\bar az}|=|\frac {z-a}{\bar z-\bar a}|=1 $$ Now apply maximum principle to get $|f(z)|<1$ for $|z|<1$.

k1.M
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$|f(z) |^2 = \dfrac {(z-a)(\bar z -\bar a)}{(1-\bar az)(1-a\bar z) } = \dfrac{ u^2 + v^2 -2 \Re (z \bar a)}{ 1 + u^2 v^2- 2\Re (z\bar a) } $, where $|z| = u , |a|= v$ . Note that $\Re (z \bar a) = \Re (\bar z a) $. So If $|z| =1$, you will get $f(z) =1$. And if $|z|<1$ , $$|f(z)| < 1 \\ \Leftrightarrow u^2 +v^2 < 1 + u^2 v^2\\ \Leftrightarrow (1-u^2)(1-v^2)>0 $$ which is true.

anumosh
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  • Could you please explain why $(1-u^2)(1-v^2) < 0$? If I understand correctly, we are assuming $u < 1$ and $v < 1$, so wouldn't $(1-u^2)(1-v^2) > 0?$ – dp1221 Dec 22 '21 at 19:38
  • yes, you are right. it was a typo. fixed it. – anumosh Jan 20 '22 at 19:59