How to find the Laurent series for $$\frac{1}{(z-1)(z-2)}$$ on $|z|<2$
Here is what I did so far:
$$\frac{1}{(z-1)(z-2)}=\frac{1}{(z-2)}-\frac{1}{(z-1)}$$ now $$\frac{1}{(z-1)}=\frac{-1}{(1-z)}=-\sum_{n=0}^{\infty}z^n$$ as long as $|z|<1$ which is where we are looking for the series so this is fine.
Now to deal with the second part is where I cannot figure out how to proceed $$\frac{1}{(z-2)}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^n$$ but this is only valid on $|z|<2$ but this is not what is needed.
Alternatively we can factor out $z$ to get $$\frac{1}{(z-2)}=\frac{1}{z}\sum_{n=0}^{\infty}\left(\frac{2}{z}\right)^n$$ which is valid for $|2/z|<1 \implies |z|>2$ again this is not what we want.
Where am I going wrong?