<s>[INST] Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$. Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$.
I tried breaking it up using partial fractions but it did not seem to get me any closer to what I want.
 [/INST] $$\frac z{(z-1)(z-3)}=-\frac1{2(z-1)}+\frac3{2(z-3)}=-\frac1{2(z-1)}-\frac34\frac1{1-\frac{z-1}2}\;\;(**)$$
Now, we have that
$$|z-1|<2\implies \frac{|z-1|}2<1\;,\;\;\text{so}$$
$$(**)=-\frac1{2(z-1)}-\frac34\sum_{n=1}^\infty\frac{(z-1)^n}{2^n}=-3\sum_{n=0}^\infty\frac{(z-1)^n}{2^{n+2}}-\frac1{2(z-1)}$$
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