Finding Laurent series in annulus $1<|z|<2$ I want to find the Laurent series for $f(z)=\dfrac{1}{(z-1)(z-2)}$ inside $1<|z|<2$. 
I can apply the formula here to get $$f(z)=\sum_{k=0}^\infty a_kz^k+\sum_{k=1}^\infty b_kz^k$$ where $$a_k=\dfrac{1}{2\pi i}\int_{|z|=2}\dfrac{f(z)}{z^{k+1}}dz$$ and $$b_k=\dfrac{1}{2\pi i}\int_{|z|=1}f(z){z^{k-1}}dz$$ But how can I calculate those values?
 A: Far more simply, observe that $$\frac{1}{(z-1)(z-2)}=\frac{(z-1)-(z-2)}{(z-1)(z-2)}=\frac1{z-2}-\frac1{z-1}.\tag{$\star$}$$
Note that for $z\neq0,2,$ we can write $$\frac1{z-2}=-\frac1{2-z}=-\frac12\cdot\cfrac1{1-\left(\frac{z}{2}\right)}$$ and $$\frac1{z-2}=\frac1{z}\cdot\cfrac1{1-\left(\frac{2}{z}\right)}.$$
Now, one of $\frac1{1-\left(\frac{z}{2}\right)}$ and $\frac1{1-\left(\frac{2}{z}\right)}$ can be expanded as a geometric series in the disk $|z|<2,$ and the other can be expanded as a geometric series in the annulus $|z|>2$. That is, we will use the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $|w|<1$. You should figure out which one works in the disk $|z|<2,$ as that is the relevant one.
Likewise, we can rewrite $\frac1{z-1}$ in two similar forms, one of which is expandable in $|z|<1$ and one of which is expandable in $|z|>1.$ You should figure out which one works in the annulus $|z|>1,$ as that is the relevant one.
Using $(\star)$ with the expansions you found above will give you the desired Laurent series.
