The line integration of $\int_0^{2\pi} \frac{d \theta}{(z+a)(\bar z+\bar a)}$, where $|z|=1$ and $a \in \mathbb{C}$ with $|a|<1$ [duplicate]

If $a \in \mathrm {C}, \, |a|< 1$ then find the value of $$ I = \frac {1- |a|^2}{π} \int_{\Gamma} \frac{ |dz|} { |z + a| ^2},$$ where $\Gamma$ is the simple closed contour $|z|= 1$ taken with positive orientation.

So, I took $a= 1/2, \, z = e^{i \theta},$ so that $|z|= 1$ and $|dz| = d\theta$

Now,

\begin{align*} I &=\frac {1- |a|^2} {π} \int \limits_{0}^{2π} \frac {d \theta}{ | e^{i\theta} + a|^2} \\ &=\frac {3} {4π} \int \limits_{0}^{2π} \frac {d \theta}{ 1+ 2|a| \cos \theta + |a|^2} \\ &=\frac {3} {4π} \int \limits_{0}^{2π} \frac {d \theta}{5/4+ \cos \theta} \\ &=\frac {3} {4π}\frac {2π}{ \sqrt{ (5/4)^{2} - 1}} \\ &= 2 \end{align*}

Could anyone suggest me more general solution of this problem for any arbitrary $a,\,|a| < 1$?

Solution 1:

Intuitively, the phase of $a$ should not matter because the integral is w.r.t. $|dz|$ i.e. simply rotate the complex plane until the integral starts and ends at the same angle as $a$, and the value will not change. Practically what this is means is that we can take the number to be on the real line with value $|a|$

Once you have the integral on the second line, you can use trig double angle identities:

$$\frac{1-|a|^2}{\pi}\int_\Gamma \frac{|dz|}{|z+a|^2} = \frac{1-|a|^2}{\pi}\int_0^{2\pi}\frac{d\theta}{1+|a|^2+2|a|\cos\theta}$$

$$ = \frac{1-|a|^2}{\pi}\int_0^{2\pi}\frac{d\theta}{(1+|a|^2)\left(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}\right)+2|a|\left(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\right)}$$

$$= \frac{1-|a|^2}{\pi}\int_0^{2\pi}\frac{d\theta}{(1+|a|)^2\cos^2\frac{\theta}{2}+(1-|a|)^2\sin^2\frac{\theta}{2}} = \frac{4}{\pi}\frac{1+|a|}{1-|a|}\int_0^{\pi}\frac{\frac{1}{2}\sec^2\frac{\theta}{2}d\theta}{\left(\frac{1+|a|}{1-|a|}\right)^2+\tan^2\frac{\theta}{2}}$$

$$= \frac{4}{\pi}\tan^{-1}\left(\left(\frac{1-|a|}{1+|a|}\right)\tan\frac{\theta}{2}\right)\Biggr|_0^{\pi^-} = \frac{4}{\pi}\left(\frac{\pi}{2}-0\right)=2$$