I was faced with a problem finding the Laurent series of complex function in all possible areas at the point $(-2i)$.
So, my function is:
$$f(z)=\frac{2z-3+2i}{(z+2i)^2(z-3)^2}$$
It can be represented as the sum of partial fractions:
$$f(z)=\frac{\frac{3}{13}-\frac{2i}{13}}{(z-3)^2}-\frac{\frac{3}{13}-\frac{2i}{13}}{(z+2i)^2}$$
If I understand correctly, I need to find the Laurent series in the following areas:
$$D_1=\{0<|z+2i|<|3+2i|\}$$
$$D_2 = \{|3+2i|<|z+2i|<\infty\}$$
So, I got something like this:
$D_1$:
$$\frac{1}{(z-3)^2}=\sum_{n=1}^{\infty}n\frac{(z+2i)^{n-1}}{(3+2i)^{n+1}}$$
$$\frac{1}{(z+2i)^2}=\sum_{n=1}^{\infty}n\frac{(z+2i)^{n-1}}{(4i)^{n+1}}$$
$D_2$:
$$\frac{1}{(z-3)^2}=\sum_{n=1}^{\infty}n\frac{(3+2i)^{n-1}}{(z+2i)^{n+1}}$$
$$\frac{1}{(z+2i)^2}=\sum_{n=1}^{\infty}n\frac{(4i)^{n-1}}{(z+2i)^{n+1}}$$
What am I doing wrong?
EDIT: The answer is correct, thanks.
In the region $D_1$, what you wrote about $\frac1{(z-3)^2}$ is just fine. But, again in $D_1$, you simply have that the Laurent series of $\frac1{(z+2i)^2}$ is just $\frac1{(z+2i)^2}$ itself. There is nothing more to add here.