Find laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}. Find the Laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}.
So far I have the following; I'm not 100% sure if it is right.
$\frac{z-1}{(z-2)(z-3)}$ = $\frac{2}{(z-3)}$-$\frac{1}{(z-2)}$
For $\frac{1}{(z-2)}$=$\frac{1}{z}$ $\frac{1}{1-(\frac{2}{z})}$=$\frac{1}{z}$$\sum_{k=1}^n\frac{2^k}{z^k}$=$\sum_{k=1}^n\frac{2^k}{z^{k+1}}$ for $|z|<1$.
I am having trouble with the other fraction.  I have seen similar questions asked, but I cannot seem to get the information I need.  Any input would be much appreciated!

Related problems: (I), (II). Use partial fraction to write your function as
$$\frac{z-1}{(z-2)(z-3)}= - \frac{1}{z-2}  + \frac{2}{z-3} = -\frac{1}{z(1-\frac{2}{z})}-\frac{2}{3(1-\frac{z}{3})} $$
$$ = -\frac{1}{z}\sum_{k=0}^{\infty}\frac{2^k}{z^{k}}-\frac{2}{3}\sum_{k=0}^{\infty}\frac{z^k}{3^k} \,. $$
Now, the first series converges in $|z|>2$ and the second series converges in $|z|<3$ which implies the common region is $ 2< |z| < 3 $ 
