$f(z) =\frac{z}{(z+1)(z-2)}$
given $1 < |z+1| <3$
So I have to write it as the power of $(z+1)$, but using partial fractions I get $-\frac{z}{3(z+1)} + \frac{z}{3(z-2)}$ and I don't know how to deal with this $z$ on the nominator. How can I write it as power of $(z+1)$?
(it is probably something very simple, but I have no ideia)
One may write, for $1<|z+1|<3$, $$ \begin{align} \frac{z}{z-2}&=\frac{(z-2)+2}{z-2} \\&=1+\frac{2}{z-2} \\&=1-\frac23\cdot\frac{1}{1-\dfrac{z+1}3} \\&=1-\frac23\cdot \sum_{n=1}^\infty \left(\dfrac{z+1}3\right)^n \end{align} $$ finally, $$ \frac{z}{(z+1)(z-2)}=\frac{1}{3(z+1)}-\frac{2}{9}-\frac23\cdot \sum_{n=1}^\infty \left(\dfrac{z+1}3\right)^n. $$