# Finding the Laurent series expansion of $f(z)$

Find the Laurent series expansion of $f(z)=\frac{z^2-1}{(z+2)(z+3)^2}$ at $0<|z+3|<1$

I have a couple of doubts in how to handle this problem: First of all, should I do it with partial fractions? Because now I have a polynomial in the numerator, so I don't know how to proceed if that's the case.

If not, can I use the general term for the Laurent series?

$$a_n=\frac{1}{2\pi i}\int \frac{f(\zeta)d\zeta}{(\zeta +3)^{n+1}}$$

But what happens with the $(z+3)^2$ in the numerator? Im pretty mixed up.

$$\frac{z^2-1}{(z+2)(z+3)^2}=\frac3{z+2}-\frac2{z+3}-\frac8{(z+3)^2}=$$
$$\frac{-3}{1-(z+3)}-\frac2{z+3}-\frac8{(z+3)^2}=-3\left(1+(z+3)+(z+3)^2+\ldots\right)-\frac2{z+3}-\frac8{(z+3)^2}$$
$$-\frac8{(z+3)^2}-\frac2{z+3}-3\sum_{n=0}^\infty(z+3)^n$$
Observe that the development of $\;\frac{-3}{1-(z+3)}\;$ in the above power series is valid since we re given $\;|z+3|<1\;$ . You get here $\;z=-3\;$ is a double pole with residue equal to $\;-2\;$ .