I have to find the Laurent series, around $z=0$ of $$\frac {sin(z)}{z^2(z-\pi)}$$
when $|z|>\pi$.
I've already calculated the series separately, but I'm having trouble putting the whole thing together. This is what I've done so far:
$$sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$$
$$\frac{sin(z)}{z^2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}z^{2n-1}$$
$$\frac{1}{z-\pi}=\frac{1}{z(1-\frac {\pi}{z})}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac {\pi}{z})^n$$
Now, I would normally do the Cauchy product, but in this case the result does not look at all like a power series. I've also tried fraction descomposition, and it the result is pretty much the same. Any ideas?
PD: Please, if somebody have any criticism, tell me. I apologize for the possible spelling mistakes. I'm not an English speaker, and sometimes the language gets a bit messy.
Cauchy product is fine, but it's not the Cauchy product of power series (since the series you got for $\frac{1}{z-\pi}$ is not a power series) but the Cauchy product of Laurent series. If $$f(z) = \sum_{k=-\infty}^\infty a_k z^k,\qquad g(z) = \sum_{k=-\infty}^\infty b_k z^k$$ in some annulus $r < \lvert z\rvert < R$, then $$f(z)g(z) = \sum_{n=-\infty}^\infty \left(\sum_{k=-\infty}^\infty a_k b_{n-k}\right)z^n$$
in at least that annulus. I don't think you get a nice short closed form for the coefficients here, but that is the normal case, only sometimes you get nice formulae for the coefficients.