I'm trying to expand $\frac{1}{(z-1)^2(z-2)}$ with $z$ complex on the annulus $2<|z|<3$. I try rewriting it in partial fractions as $ \frac{1}{(z-1)^2(z-2)}=-\frac{1}{z-1}-\frac{1}{(z-1)^2}+\frac{1}{z-2}. $ I know I can make the last summand above converge on $|z|>2$, by writing it as $\frac{1}{z}\cdot\frac{1}{1-2/z}$. However, I don't know how to deal with the other terms. I can make the first term converge on $|z|>1$, by rewriting it as $-\frac{1}{z}\frac{1}{1-1/z}$, but I don't see how to deal with the middle term to get it to converge on the desired annulus. What's the right thing to do?
So I rewrite $\frac{1}{z-1}$ as $\frac{1}{4-1+(z-4)}=\frac{1}{3}\frac{1}{1+(z-4)/3}$ which converges for $|z-4|<3$. But then I get a total series of form $ -\frac{1}{3}\frac{1}{1+(z-4)/3}-\frac{1}{9}\frac{1}{(1+(z-4)/3)^2}+\frac{1}{z}\frac{1}{1-z/2} $ where the regions of convergence of the first two terms are $|z-4|<3$ and that of the last term is $|z|>2$. How can I get convergence on the annulus?