Solving $\oint_C \frac{3z-2}{z^2 - 2iz} \,dz $ using Cauchy's Integral Formula I want to solve the following: $$\oint_C \frac{3z-2}{z^2 - 2iz} \,dz$$ where $C$ is the circle of radius $2$ centered at $z = i$ with a counterclockwise orientation. I used Cauchy's Integral Formula: $$\oint_C \frac{f(z)}{z-z_0} \,dz = 2\pi if(z_0)$$ My solution was as follows: $$\oint_C \frac{3z-2}{z^2 - 2iz} \,dz = \oint_C \frac{\frac{1}{z}(3z-2)}{\frac{1}{z}(z^2 - 2iz)} \,dz = \oint_C \frac{3-\frac{2}{z}}{z - 2i} \,dz $$ Hence, $$\oint_C \frac{3z-2}{z^2 - 2iz} \,dz = 2\pi i(3-\frac{2}{2i})=6\pi i - 2\pi$$ However, I plugged this problem in to MATLAB and Wolfram Alpha and they both tell me that the answer is just $6\pi i \approx 18.850i$. MATLAB's answer to my problem Can someone point out what I did wrong?

> see that, $\color{blue}{\frac{3z-2}{z^2 - 2iz} = -\frac{i}{z}+\frac{3+i}{z - 2i}}$ By Cauchy formula we get $$\oint_C \frac{3z-2}{z^2 - 2iz} \,dz=-i\oint_C \frac{dz}{z} \,dz +(3+i)\oint_C \frac{dz}{z - 2i} =\color{red}{2i\pi(-i+3+i)=6i\pi}$$