Complex series: $\frac{z}{(z-1)(z-3)} = -3 \sum\limits_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$ for  $0 < |z-1| < 2$ 
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.

$$\dfrac{z}{(z-1)(z-3)} = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}$$
$$\dfrac1{z-3} = \dfrac1{(z-1) -2} = -\dfrac12 \left( \dfrac1{1-(z-1)/2}\right) = -\dfrac12 \left( \sum_{k=0}^{\infty} \left( \dfrac{z-1}2\right)^k\right)$$
Hence, $$\dfrac3{(z-1)(z-3)} = - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)$$
Hence,
\begin{align}
\dfrac{z}{(z-1)(z-3)} & = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}\\
& = \dfrac1{z-1} - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)\\
& = - \dfrac12 \dfrac1{z-1}- 3 \left( \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+2}}\right)\right)
\end{align}
