complex integration $\frac{1-|a|^{2}}{\pi}\int_{L}\frac{|dz|}{{|z+a|}^{2}}$ How to solve complex integration $\frac{1-|a|^{2}}{\pi}\int_{L}\frac{|dz|}{{|z+a|}^{2}}$, where $L$ is the simple closed curve $|z|=1$ and $a\in\mathbb{C}$ with $|a|<1.$ $z=e^{it}$ is the parametrization of the circle $|z|=1.$If we put values it becomes as $\frac{1-|a|^{2}}{\pi}\int_{0}^{\pi}\frac{|dt|}{{|e^{it}+a|}^{2}}$. Now i don't know to proceed further. Please help. Thanks a lot.
 A: It's most convenient to use
$$\lvert dz\rvert = \frac{dz}{iz}$$
on the unit circle. With that, the integral becomes
$$\frac{1 - \lvert a\rvert^2}{\pi} \int_L \frac{\lvert dz\rvert}{\lvert z+a\rvert^2} = \frac{1 - \lvert a\rvert^2}{\pi i} \int_{\lvert z\rvert = 1} \frac{dz}{(z + a)(\overline{z} + \overline{a})z}  = \frac{1 - \lvert a\rvert^2}{\pi i} \int_{\lvert z\rvert = 1} \frac{dz}{(z + a)(1 + \overline{a}z)},$$
and applying Cauchy's integral formula to $f(z) = \frac{1}{1 + \overline{a}z}$ finishes it.
A: 
Your direct approach in the OP is quite tractable.  We need only to complete the integration. 

Proceeding, we can write the integral of interest as
$$\begin{align}\oint_{|z|=1}\frac{1}{|z+a|^2}|dz|&=\int_0^{2\pi}\frac{1}{|e^{it}+a|^2}\,dt\\\\
&=\int_0^{2\pi}\frac{1}{1+|a|^2+2|a|\cos(t+\arg(a))}\,dt \tag1\\\\
&=2\int_0^{\pi}\frac{1}{1+|a|^2+2|a|\cos(t)}\,dt \tag2\\\\
\end{align}$$
where we exploited the $2\pi$-periodicity and evenness of the integrand in going from $(1)$ to $(2)$.
We now evaluate the integral on the right-hand side of $(1)$ using the classical Weierstrass Substitution $x=\tan(t/2)$ in $(2)$.  This substitution leads to
$$\begin{align}
2\int_0^{\pi}\frac{1}{1+|a|^2+2|a|\cos(t)}\,dt&=4\int_0^{\infty}\frac{1}{(1+|a|)^2+(1-|a|)^2x^2}\,dx \\\\
&=\frac{2\pi}{1-|a|^2} \tag 3
\end{align}$$
Finally, using $(3)$ in $(1)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\frac{1-|a|^2}{\pi}\oint_{|z|=1}\frac{1}{|z+a|^2}|dz|=2}$$
And we're done!
