If $a \in \mathbb{C}$ with $|a| < 1$, then the values of $\displaystyle \frac{(1-|a|^2)}{\pi} \int_{\gamma} \frac{|dz|}{|z+a|^2}$, where $\gamma$ is the simple closed curve $|z| = 1$ taken with the positive orientation. What will be the integration?
I know this theorem: $$\int_{\gamma} f |dz| = \int_{a}^{b} f(\gamma(t))d|\gamma|(t)$$ where $f$ is continuous on ${\gamma}$.
How do I use this theorem?
On the positive oriented contour over the unit circle, we have $\displaystyle\;|dz| = \frac{dz}{iz}$ and $$\frac{1}{|z+a|^2} = \frac{1}{(z+a)(\bar{z}+\bar{a})} = \frac{1}{(z+a)(\frac{1}{z} + \bar{a})} = \frac{z}{\bar{a}}\frac{1}{(z+a)(z+\frac{1}{\bar{a}})} $$ This means $$\mathcal{I} \stackrel{def}{=} \frac{1-|a|^2}{\pi}\oint_{|z|=1} \frac{|dz|}{|z+a|^2} = \frac{1-|a|^2}{\pi\bar{a}i}\oint_{|z|=1} \frac{dz}{(z+a)(z+\frac{1}{\bar{a}})} $$ The last integral is an ordinary contour integral. Since the integrand has only one root $-a$ inside the unit circle, we can evaluate it by taking the residue at $-a$. The end result is $$\mathcal{I} = \frac{1-|a|^2}{\pi\bar{a}i}\times \frac{2\pi i}{-a + \frac{1}{\bar{a}}} = 2$$