Calculating the contour integral: $\int_{|z-z_0|=r}\frac{1}{\bar{z}}dz$
Solution 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 $-\frac {r^{2}}{ \overline {z_0}^{2}}$].