$$\frac{1}{(z-2)(z+1)}=\frac{A}{z-2}+\frac{B}{z+1}$$
\begin{cases}A+B=0 \\ A-2B=1 \end{cases}
$$A=\frac{1}{3} \\ B=-\frac{1}{3}$$
$$\frac{1}{(z-2)(z+1)}=\frac{\frac{1}{3}}{z-2}+\frac{-\frac{1}{3}}{z+1}$$
Laurent Series $$\rvert z \rvert <1$$
$$\frac{1}{(z-2)(z+1)}=-\frac{1}{6} \ \frac{1}{-\frac{z}{2}+1}-\frac{1}{3} \ \frac{1}{1-(-z)}= \\=-\frac{1}{6} \sum_{n=0}^{+\infty} (\frac{z}{2})^n-\frac{1}{3} \sum_{n=0}^{+\infty} (-z)^n $$
$$1 < \rvert z \rvert <2$$
$$\frac{1}{(z-2)(z+1)}=-\frac{1}{6} \ \frac{1}{-\frac{z}{2}+1}-\frac{1}{3} \ \frac{1}{z(1-(-\frac{1}{z}) )}= \\ \\ = -\frac{1}{6} \sum_{n=0}^{+\infty} (\frac{z}{2})^n-\frac{1}{3} \sum_{n=0}^{+\infty} (-\frac{1}{z})^{n+1} $$
$$ \rvert z \rvert >2$$
$$\frac{1}{(z-2)(z+1)}=\frac{1}{3} \ \frac{1}{z(1-\frac{2}{z})}-\frac{1}{3} \ \frac{1}{z(1-(-\frac{1}{z}) )}= \\ \\ = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{2}{z})^{n+1}-\frac{1}{3} \sum_{n=0}^{+\infty} (-\frac{1}{z})^{n+1} $$
Is it correct?
Thanks!