Contour integral

Let $a,  b \in \mathbb{C}$ and $|b|<1$. Evaluate the contour integral:

$$\frac{1}{2\pi} \oint \limits_{|z|=1} \left| \frac{z-a}{z-b}\right|^2 \;|{\rm d}z| = \frac{|a-b|^2}{1-|b|^2}+1$$

Solution



$$\begin{aligned}
\frac{1}{2\pi}\oint \limits_{|z|=1}\left | \frac{z-a}{z-b} \right |^2 \left | {\rm d} z\right | &=\frac{1}{2\pi} \oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \bar{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \bar{z}-\bar{b} \right )}\frac{{\rm d}z}{iz}\\
 &= \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \frac{1}{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \frac{1}{z}-\bar{b} \right )}\frac{{\rm d}z}{z}\\
 &= \frac{1}{2\pi i}\oint \limits_{|z|=1}\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\left ( \frac{1}{z-b}- \frac{1}{z} \right )\, {\rm d}z\\
 &=\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\bigg|_{z=b}- \frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )} \bigg|_{z=0} \\
 &= \frac{\left ( a-\bar{b} \right )\left ( \bar{a}-b \right )}{1-b\bar{b}}+1= \frac{\left | a-b \right |^2}{1-|b|^2}+1
\end{aligned}$$