Laurent expansion of $\frac{2}{(z-1)(3-z)}$ The question asks me to find all the possible Laurent series expansions of $$f(z)=\frac{2}{(z-1)(3-z)}$$ about the origin so $$z_0 =0 $$ 
First I convert $f(z)$ into partial fractions to get $$f(z)=\frac{1}{z-1}+\frac{1}{3-z}$$
We can see there are three domains
$$D_1: |z|<1$$ $$D_2 :1<|z|<3$$  $$D_3 :|z|>3$$.
I first look at $\frac{1}{z-1}$ and I write it in the form $-(1-z)^{-1}$. By writing in this form I can see that it converges in $D_1$.
By applying the maclaurin series I get $$\sum^\infty_{n=0} z^n$$
How do I get the expression $\frac{1}{z-1}$ in the form so that it converges between $D_2$ and $D_3$ ?

Hint: $$\frac{1}{z-1}=-\frac{1}{1-z}=\frac{z^{-1}}{1-z^{-1}}$$ and $$\frac{1}{3-z}=\frac{1}{3}\frac{1}{1-\frac{z}{3}} = -\frac{z^{-1}}{1-3z^{-1}}$$
Can you see how to get a Laurent series for $\frac{z^{-1}}{1-z^{-1}}$ so that it converges when $|z|>1$ - that is, when $|z^{-1}|<1$?
