Laurent series for $f(z)= \frac{1}{ (z-i)(z+2i)}$

Question

I'm struggling with this question. I tried to break $f(z)$ using partial fractions and modify each equation so it looks like $\dfrac{1}{1-z}$ series but that's where I get stuck. Any help would be greatly appreciated.

Expand $f(z)= \dfrac{1}{ (z-i)(z+2i)}$ as a series throughout each of the domains:

1. $|z|< 1$

2. $1 <|z|< 2$

3. $2 <|z|< \infty$

This is the partial fractions that I got. (1i/3)/(z+2i)-(1i/3)/(z-i)

Answer 1

Let me only give you the secret to doing this; you can go from there.

No matter what, you can express $\frac1{az+b}$ either as $\frac1 b\cdot\frac1{1+az/b}$ or as $\frac1{az}\cdot\frac1{1+b/(az)}$.

It’ll be your choice as to which form you take, to make the part after the “$1+$” smaller than $1$ in absolute value.