While working on complex integration problem I got stuck at the following problem:
$\int \frac{|dz|}{|z-2|^2}$ where $|z| = 1$ is the domain.
The only idea that I am getting is that we can use the fact $z\bar{z} = 1$. Kindly help in solving the above problem. Thanks for giving time.
$\endgroup$ 01 Answer
$\begingroup$Use
$$|z-2|^2 = (z-2) (\bar{z}-2) = |z|^2 - 4 \operatorname{Re}{z} + 4$$
so when $z=e^{i \phi}$, $|z-2|^2 = 5-4 \cos{\phi}$ and $|dz|=d\phi$, or
$$|z-2|^2 = 5-2 z - 2 z^{-1} $$ $$|dz|=-i \frac{dz}{z}$$
The integral is then equal to
$$\int_0^{2 \pi} \frac{d \phi}{5-4 \cos{\phi}} = i \oint_{|z|=1} \frac{dz}{2 z^2-5 z+2}$$
which is easily evaluated by the residue theorem.
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