Show that when $0 < |z-1| < 2$, $$\frac{z}{(z-1)(z-3)} = -3 \sum_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$$
I thought to attack this using a partial fraction decomposition and then breaking the partial fractions into Maclaurin series. I got the $$-\frac{1}{2(z-1)}$$ from that, but the other part has me a bit stumped.
I have: $$ \frac{3}{2(z-3)}$$
for the other partial fraction. And I factor out the 3/2 and then have:
$$\frac{1}{z-3}$$
So I get: $$\frac{1}{1 - (-(z-2))}$$
but this does not give me what I need.