Solving $\oint_C \frac{3z-2}{z^2 - 2iz} \,dz $ using Cauchy's Integral Formula

Question

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?

Answer 1

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

Answer 2

Let $$I =\oint_C \frac{3z-2}{z^2-2iz} dz$$ The integrand $\frac{3z-2}{z^2-2iz}$ is not analytic at the points $z=0$ and $z=2i$ both of which lie inside $C: |z-i|=2$. Writing it as: $$\frac{3z-2}{z^2-2iz} = \frac{3z-2}{-2i}[\frac{1}{z}-\frac{1}{z-2i}]$$ Therefore, $$I =\oint_C \frac{3z-2}{-2iz} dz - \oint_C \frac{3z-2}{(-2i)(z-2i)} dz $$ $$=\frac{2\pi i}{-2i}[[3z-2]_{z=0}-[3z-2]_{z=2i}]=-\pi[-2-6i+2]=6\pi i$$ using the Cauchy’s integral formula.

Answer 3

Hint:

Use following fraction decomposition and apply Cauchy's Integral Formula $$\frac{3+i}{z-2 i}-\frac{i}{z}$$