Calculating the contour integral: $\int_{|z-z_0|=r}\frac{1}{\bar{z}}dz$

Question

I'm trying to find out what the following contour integral equals to: $$\int_{|z-z_0|=r}\frac{1}{\bar{z}}dz$$ given that $z_0$ is a point in the complex plane, where $|z_0|\neq r >0$. Due to the fact that the function is not holomorphic, I'm not able to apply any theorems that require this condition. If I parametrize the contour, I get: $$i\int_0^{2\pi}\frac{re^{2it}}{\bar{z_0}e^{it}+r}dt$$ When $z_0=0$, the answer is trivial. However, I have difficulty calculating this integral when $z_0$ is any other point. I would appreciate some help.

Answer 1

Hint: $\frac 1 {\overline z}=\frac 1 {\overline {z-z_0}+\overline{z_0}}$ and $\overline {z-z_0}=\frac {r^{2}} {z-z_0}$ when $|z-z_0|=r$. Now apply Cauchy's Theorem to see that the given integral is $0$ if $|z_0| <r$. For $|z_0| >r$ use Residue Theorem.

[ The residue at $z=z_0-\frac {r^{2}} {\overline {z_0}}$ is $-\frac {r^{2}}{ \overline {z_0}^{2}}$].

Answer 2

$$I=\int_0^{2\pi}\frac{re^{2it}}{\bar{z_0}e^{it}+r}dt$$

Put $e^{it}=w\implies ie^{it}dt=dw$

$$I=\frac{1}{i}\oint_{|w|=1} \frac{wdw}{w\overline{z_o}+r}$$

There is a simple pole at $w=-r/\overline{z_0}$, at which the residue is $-r/\overline{z_0}^2$. Using Residue theorem, can you proceed from here?