Let $F(z)= \frac{1}{(z-1)^2(z+2)}$.
I need to find a Laurent expansion for $F$ on the annulus $A = \{z: \sqrt{2}<|z-i|<\sqrt{5}$}.
So far I have managed to write $F$ as $F(z)=\frac{-1}{9(z-1)}+\frac{1}{9(z+2)}+\frac{1}{3(z-1)^2}$. Then writing $z-i = w$, I got the following expansions:
$\frac{-1}{9(z-1)}$ = $\frac{-1}{9w} \sum_{n=0}^{\infty} (-1)^{n}(\frac{i-1}{w})^n $
$\frac{1}{3(z-1)^2}$ = $\frac{1}{3w^2} \sum_{n=0}^{\infty} (-1)^{n}(1+n)(\frac{i-1}{w})^n$.
But how do we get the expansion for $\frac{1}{9(z+2)}$?
You are interested in the Laurent series centered at $i$, rather than $0$. So, use the fact that\begin{align}F(z)&=F\bigl((z-i)+i\bigr)\\&=\frac1{\bigl((z-i)+i-1\bigr)^2\bigl((z-i)+i+2\bigr)}\\&=\frac1{9\bigl((2+i)+(z-i)\bigr)}+\frac1{9\bigl((1-i)-(z-i)\bigr)}+\frac1{3\bigl((1-i)-(z-i)\bigr)^2}\\&=\frac1{9(2+i)}\frac1{1-\frac{z-i}{2+i}}+\frac1{9(1-i)}\frac1{1-\frac{z-i}{1-i}}+\frac1{3(1-i)^2}\frac1{\left(1-\frac{z-i}{1-i}\right)^2}\\&=\frac1{9(2+i)}\sum_{n=0}^\infty\left(\frac{z-i}{2+i}\right)^n-\frac1{9(1-i)}\sum_{n=-\infty}^{-1}\left(\frac{z-i}{1-i}\right)^n+\\&\phantom{=}\quad-\frac1{3(1-i)^2}\sum_{n=-\infty}^{-2}(n+1)\left(\frac{z-i}{1-i}\right)^n.\end{align}It's only at the last step that I've used the fact that $\sqrt2<|z-i|<\sqrt5$.