Consider the function $f(z)=\frac{z-a}{1-\overline{a}z}$ with $|a|<1$. I need to find $$\int_{|z|=1} \overline{f(z)} f'(z)dz.$$ I tried to do it directly using the parametrization $\gamma(t)=e^{it}$ with $0 \leq t \leq 2 \pi$, but it becomes too tedious and I couldn't get a result. I'd like to know if there's a faster and easier way to do it.
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$f$ is a Möbius transformation which maps the unit disk conformally onto itself. In particular, $f$ maps the unit circle onto itself, so that $$ \overline{f(z)} = \frac{1}{f(z)} $$ for $|z| = 1$. (See for example suppose $|a|<1$, show that $\frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself. .)
This transforms the integral into something that can be computed with the residue theorem (or with the argument principle).
Martin R
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By the argument principle, $\int f'/f = 2 \pi i ( zeros - poles) = 2 \pi i (1-0) = 2 \pi i$. Is that right? – Twnk Sep 06 '20 at 02:10
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@Twink: Yes, exactly. – Martin R Sep 06 '20 at 05:58
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This is just the area theorem: if $f$ is analytic and univalent on the closed disc say and $f(\partial \mathbb D)=J$, the area enclosed by the analytic Jordan curve $J$ is (by Green theorem first and by the fact that $f$ is a parametrization of $J$ hence can be taken as $w$):
$A(interior J)=\frac{1}{2i}\int_J\bar wdw=\frac{1}{2i}\int_{|z|=1}\bar ff'dz$
In this case the $f(\mathbb D)=\mathbb D$ so the area is $\pi$ and the integral is then $2\pi i$
Conrad
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