then the value of $ \frac{1-\vert a \vert^2}{\pi} \int_{\gamma} \frac{\vert dz \vert}{\vert z+a \vert^2} $.

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If $a\in \mathbb{C}$ with $\vert a \vert <1$ , then the value of

$$ \frac{1-\vert a \vert^2}{\pi} \int_{\gamma} \frac{\vert dz \vert}{\vert z+a \vert^2} $$

where $\gamma$ is the simple closed curve with $\vert z \vert=1$, taken with the positive orientation.

I tried but I cant manipulate the denominator, I used $z \bar z =\vert z \vert^2$, Please help.