Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$.

Question

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.

Answer 1

Hint: the series on the right-hand side is a geometric series. The value of a geometric series is $\frac{\mbox{First Term in Series}}{1-\mbox{Common Ratio}}$. The first term in the series is $1/4$ and the common ratio is $(z-1)/2$. Of course, the geometric series converges if and only if $|(z-1)/2|<1$. I haven't checked to see whether the equation is actually true.

Answer 2

$$\frac z{(z-1)(z-3)}=-\frac1{2(z-1)}+\frac3{2(z-3)}=-\frac1{2(z-1)}-\frac34\frac1{1-\frac{z-1}2}\;\;(**)$$

Now, we have that

$$|z-1|<2\implies \frac{|z-1|}2<1\;,\;\;\text{so}$$

$$(**)=-\frac1{2(z-1)}-\frac34\sum_{n=1}^\infty\frac{(z-1)^n}{2^n}=-3\sum_{n=0}^\infty\frac{(z-1)^n}{2^{n+2}}-\frac1{2(z-1)}$$