$$\int_c \frac{dz}{z-3i}$$
$$f(z) = \frac{1}{z-3i}$$ where C the circle |z|=π counterclockwise.
So since f(z) is not analytic when z = 3i, we have to integrate another way since we cannot apply cauchy theorem of integrals.
$$\int_c \frac{1}{z - 3i}dz = \int_c (z - 3i)^{-1}$$
Where do I go from here?
Using residue theorem, as $z=3i$ is internal to the circle $|z|=\pi$ $$\oint_{|z|=\pi}\frac{dz}{z-3i}=2\pi i\text{Res}\left[\frac{1}{z-3 i},\{z,3 i\}\right]=2\pi i$$ because $$\text{Res}\left[\frac{1}{z-3 i},\{z,3 i\}\right]=\underset{z\to 3 i}{\text{lim}}\frac{z-3 i}{z-3 i}=1$$