Laurent expansion of $f (z) = \frac1{z(z − 1)(z − 2)},$ (in powers of $z$) for $0 < |z| < 1,$ $1 < |z| < 2,$ and $|z| > 2.$ Find the Laurent expansion of $$f (z) = \frac{1}{z(z-1)(z-2)}$$
(in powers of $z$) for
a. $0 < |z| < 1$
b. $1 < |z| < 2$
c. $|z| > 2$
 A: The first thing you'll want to do, here, is what is known as a partial fraction decomposition. Write $$\frac{1}{z(z-1)(z-2)}=\frac{A}{z}+\frac{B}{z-1}+\frac{C}{z-2},$$ and solve for $A,B,C.$
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 these can be expanded as a multiple of a geometric series in the disk $|z|<2,$ and the other can be expanded as a multiple of 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 region works for which rewritten version, and find the respective expansions in both cases.
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 $|z|>1.$
Using the partial fraction decomposition with the expansions you found above will give you three different Laurent expansions of $f(z),$ one for each of the regions $0<|z|<1,$ $1<|z|<2,$ and $|z|>2$.
