<s>[INST] Finding the Taylor series (complex numbers) 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?
 [/INST] 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}
</s>