Compute the value of the complex integration If $a\in \mathbb C$ with $|a|<1$ then find the value of the integration: $$\dfrac{1-|a|^{2}}{\pi}\int_{|z|=1} \dfrac{|dz|}{|z+a|^{2}}.$$
I can't proceed anyway, my main difficulty is for $|dz|.$
Thanks in advance..........

$$z=e^{i \phi} \implies dz = i e^{i \phi} d\phi \implies |dz| = d\phi - -i dz/z$$
$$|z+a|^2 = (z+a) (\bar{z} + \bar{a})  = |z|^2 + \bar{a} z+a \bar{z} + |a|^2 = 1+|a|^2+  \bar{a} z + a z^{-1}$$
Then the integral is
$$-i \frac{1-|a|^2}{\pi} \oint_{|z|=1} \frac{dz}{\bar{a} z^2 + (1+|a|^2) z + a} $$
Now find the poles 
$$z_{\pm} = \frac{-(1+|a|^2) \pm (1-|a|^2)}{2 \bar{a}} = \begin{cases}-a & + \\ -1/\bar{a} & - \end{cases}$$
We only need the residue at the pole $z_+$ (since $|a| \lt 1$), which is $1/(1-|a|^2)$.  Thus, by the Residue Theorem, the integral is
$$-i \frac{1-|a|^2}{\pi} i 2 \pi \frac1{1-|a|^2} = 2$$
