<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] 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,$$\frac1{z-2}=-\frac1{2-z}=-\sum_{k=0}^\infty\frac{z^k}{2^{k+1}}.$$
</s>