# Compute $\int_{|z| = r} \frac{|dz|}{|z-a|^2}$ where $|a| \neq r$

Compute $$\int_{|z| = r} \frac{|dz|}{|z-a|^2}$$ where $$|a| \neq r$$

I know that $$|dz| = -ir\frac{dz}z$$ and that $$z\overline{z} = r^2$$. So I get

$$\int_{|z| = r} \frac{|dz|}{|z-a|^2} = -i\int_{|z| = r} \frac{rdz}{(z-a)(\overline{z} - \overline{a})}$$

But I don't know how to proceed from here

The denominator of your integrand is missing a factor of $$z$$. The denominator should be $$z(z-\alpha)(\bar{z}-\bar{\alpha})$$. The idea now is to multiply this out and replace $$z\bar{z}$$ with $$r^2$$. After some multiplication and factoring, the denominator of your integrand becomes $$(r^2-\bar{\alpha}z)(\alpha-z)$$.
Thus, your integral takes the form: $$\int_{|z|=r}\frac{-ir}{(r^2-\bar{\alpha}z)(\alpha-z)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)(z-\alpha)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)}\cdot\frac{1}{(z-\alpha)}$$
The answer now depends on whether $$|\alpha| > r$$ or $$|\alpha| < r$$. In the latter case, use Cauchy's integral formula.