Evaluate $\int_{|z|=1}\frac{|dz|}{|z-a|^2}$ using Cauchy's Theorem.

Question

Question: Evaluate $\int_{|z|=1}\frac{|dz|}{|z-a|^2}$ using Cauchy's Theorem.

My attempt: So, Cauchy's theorem for derivatives tells us that if $f$ is holomorphic in an open set $\Omega$, and $D$ is a disc with boundary $C$ such that $D$ and $C$ are contained in $\Omega$, we have, for any point $z \in D$,

$$f^{(n)}(z) = \frac{n!}{2\pi i}\int_C\frac{\zeta}{(\zeta - z)^{n+1}}\ d\zeta.$$

Let $f(z) = |z|$, then $f'(z) = z/|z|$, thus we have: $$f'(a) = \frac{a}{|a|}=\frac{1}{2\pi i}\int_{|z|= 1}\frac{|z|}{|z-a|^2}\ dz.$$

this is really close to the integral I want to evaluate, but it obviously isn't quite right. I'm also getting confused because of the absolute value of $|dz|$ in the problem statement - how can I handle that?

Thanks.

Answer 1

Note that for $|z|^2=z\bar z=1$, $\bar z=z^{-1}$ and $|dz|=\frac{1}{iz}dz$. Then, we have

$$\oint_{|z|=1}\frac{1}{|z-a|^2}\,|dz|=\oint_{|z|=1}\frac{1}{i(z-a)(1-az)}\,dz$$

Can you proceed from here?