Finding the Taylor series (complex numbers)

Question

I have $$ {\frac{1}{(z+1)(z-2)}} $$ I did $$ {\frac{1}{(z+1)(z-2)}} = {\frac{A}{z+1}}+{\frac{B}{z-2}} $$

and found $A=-1/3, B=1/3$

So now I have $$ -{\frac{1}{3}}\times {\frac{1}{z+1}}+{\frac{1}{3}}\times{\frac{1}{z-2}} = -{\frac{1}{3}} \sum(-1)^k z^k + {\frac{1}{3}} \sum(-1)^k (-{\frac{z}{2}})^k $$

But the answer in the book is different. What am I doing wrong?

Answer 1

We have if $|r|<1$, then $$\frac{1}{1-r}=\sum_{i=0}^\infty r^i$$

\begin{align} -{\frac{1}{3}}\times {\frac{1}{z+1}}+{\frac{1}{3}}\times{\frac{1}{z-2}} &=-{\frac{1}{3}} \sum(-1)^k z^k - \frac16 \times \frac{1}{1-\left( \frac{z}2\right)} \\&= -{\frac{1}{3}} \sum(-1)^k z^k - {\frac{1}{6}} \sum ({\frac{z}{2}})^k \\ &= \frac13 \left[\sum(-1)^{k+1} z^k - {\frac{1}{2}} \sum ({\frac{z}{2}})^k \right]\\ &= \frac13 \sum\left[(-1)^{k+1} - {2}^{-k-1} \right]z^k\\ \end{align}

Answer 2

It is indeed true that$$\frac1{(z+1)(z-2)}=-\frac13\times\frac1{z+1}+\frac13\times\frac1{z-2}$$and that$$\frac1{z+1}=\sum_{k=0}^\infty(-1)^kz^k.$$However,$$\frac1{z-2}=-\frac1{2-z}=-\sum_{k=0}^\infty\frac{z^k}{2^{k+1}}.$$