If $a\in\Bbb C $ with $\vert a \vert\lt 1 $ then,
what is the value of
\begin{equation}
\frac {(1-\vert a \vert ^2)}{\pi}\int_C \frac {\vert dz \vert}{\vert z+a \vert^2}
\end{equation}
where C is simple closed curve $\vert z \vert =1 $ taken with positive orientation.
I tried this problem but I am not getting integral of mod function and how to solve this problem. Please guide me.
$$ \begin{align} \frac{1-|a|^2}\pi\int_C\frac{|\mathrm{d}z|}{|z+a|^2} &=\frac{1-|a|^2}\pi\int_0^{2\pi}\frac{\mathrm{d}\theta}{(|a|+\cos(\theta))^2+\sin^2(\theta)}\tag{1}\\ &=\frac{1-|a|^2}\pi\int_0^{2\pi}\frac{\mathrm{d}\theta}{|a|^2+1+2|a|\cos(\theta)}\tag{2}\\ &=2\frac{1-|a|^2}\pi\int_0^\pi\frac{\mathrm{d}\theta}{|a|^2+1+2|a|\cos(\theta)}\tag{3}\\ &=2\frac{1-|a|^2}\pi\int_0^\infty\frac{\frac{2\,\mathrm{d}u}{1+u^2}}{|a|^2+1+2|a|\frac{1-u^2}{1+u^2}}\tag{4}\\ &=4\frac{1-|a|^2}\pi\int_0^\infty\frac{\mathrm{d}u}{(1+|a|)^2+(1-|a|)^2u^2}\tag{5}\\[6pt] &=\left.\frac4\pi\tan^{-1}\left(\frac{1-|a|}{1+|a|}u\right)\right]_0^\infty\tag{6}\\[12pt] &=2\tag{7} \end{align} $$ Explanation:
$(1)$: $z=\frac{a}{|a|}e^{i\theta}$
$(2)$: $\cos^2(\theta)+\sin^2(\theta)=1$
$(3)$: use symmetry
$(4)$: Weierstrass substitution
$(5)$: algebra
$(6)$: arctangent integral
$(7)$: evaluate