Evaluate $\int_{|z| = r} \frac{1}{a-\overline{z}}dz$

Question

Evaluate $\int_{|z| = r} \frac{1}{a-\overline{z}} dz$ where $|a|\neq r$.

I was trying to find a way to connect this integral to $\int_{|z|=r} \frac{1}{a-z}dz$. However this method does not work, and the only thing I got is $\overline{\int_\gamma f(z) dz} = \int_{\overline{\gamma}}\overline{f(\overline{z})}dz$ which is not so useful here.

Answer 1

I would rewrite it as $$\int_{|z|=r}\frac{dz}{a-r^2/z}=\int_{|z|=r}\frac{z\,dz}{za-r^2}$$ since $z\overline z=r^2$ if $|z|=r$.

Answer 2

Noting that $ \vert \bar{a} \vert = \vert a \vert \ne r$ we have that since $\frac{1}{z-\bar{a}}$ is holomorphic in $\mathbb{C}-\left\lbrace \bar{a}\right\rbrace$ that $\int_{\vert z \vert = r}\frac{1}{z-\bar{a}}dz = 2\pi i I(\gamma,\bar{a})$, where $I(\gamma,\bar{a})$ denotes the winding number of the curve $\gamma$ around the point $\bar{a}$, so it should be $2\pi i$ if $\vert \bar{a} \vert < r$ or $0$ if $\vert \bar{a} \vert > r$.