<s>[INST] Show that $f(z) $ equals power series representation when $0 <|z-1| < 2.$ I need to show that when $0<|z-1|<2$, the following equality holds:
$$\frac{z}{(z-1)(z-3)} = -3\sum_{n=0}^{\infty}\dfrac{(z-1)^n}{2^{n+2}} - \dfrac{1}{2(z-1)}$$
I wrote 
$\frac{z}{(z-1)(z-3)}$ as the sum $-\frac{1}{2}\left(\frac{1}{z-1}\right) + \frac{3}{2}\left(\frac{1}{z-3}\right)$. 
So given that the equality holds, I only need to show that: $$-3\sum_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}}  = \frac{3}{2}\left(\frac{1}{z-3}\right)$$
I've written $\frac{3}{2}\left(\frac{1}{z-3}\right)$ as $-\frac{1}{2}\sum_{n=0}^{\infty}\frac{z^n}{3^n}$, but I can't find out how to proceed from here!
Q: How do I show that: 
when $0<|z-1|<2$, the following equality holds:
$$\frac{z}{(z-1)(z-3)} = -3\sum_{n=0}^{\infty}\dfrac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$$
 [/INST] HINT:
$$\frac{1}{z-3}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\,\frac{1}{1-\frac{z-1}{2}}$$
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