complex integration $\frac{1-|a|^{2}}{\pi}\int_{L}\frac{|dz|}{{|z+a|}^{2}}$

Solution 1:

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.

Solution 2:

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!