find laurent series of $\frac{1-\cos(z)}{z^3(2z+3)}$ in $0<|z|<1$.
the singularities are at $z=0$ and $z=\frac{3}{2}$.
can I just do this?:
$\frac{1}{z^3(2z+3)}$ $(1-\cos(z))$ = $\frac{1}{z^3(2z+3)}$ $\left[1-\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}+...\right)\right]$
(because in $0<|z|<1$ ,$\frac{1}{z^3(2z+3)}$ well defined(?))
this is what i got
$\frac{1-\cos(z)}{z^3(2z+3)}$ = $\frac1{z^3(2z+3)}$ $z^2(\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\frac{z^6}{8!}+...)$
= $\frac1{z(2z+3)}$ $(\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\frac{z^6}{8!}+...)$
and using partial fraction decomposition
= ($\frac{1}{3z}-\frac{2}{3(2z+3)}$)$(\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\frac{z^6}{8!}+...)$
=($\frac{1}{3z} -\frac{2}{3^2}\sum_{n=0}^\infty (\frac{-2z}{3})^n$) [$\sum_{n=0}^\infty \frac{z^{2n}}{(2n+2)!}$]
from here I don't know what to do...any help?