Finding Laurent series of the following function

Question

How does one find the Laurent Series expansion of $f(z) = \frac{-1}{(z-1)(z-2)}$ in the annulus $\{ z \text{ such that } 1<|z|<2 \}$?

Answer 1

We have that $$\frac{-1}{(z-1)(z-2)}=\frac{1}{z-1}-\frac{1}{z-2}.$$

Note that, if $|1/z|<1$, that is, if $|z|>1$: $$\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-(1/z)}=\frac{1}{z}(1+z^{-1}+z^{-2}+\dots)=z^{-1}+z^{-2}+\dots$$

and, if $|z|<2$: $$-\frac{1}{z-2}=\frac{1}{2}\frac{1}{1-(z/2)}=\frac{1}{2}(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\dots)$$

In the annulus, both series converges, so you can sum them.