Laurent series for $\sin z \sin(1/z)$? How can I find a laurent series for $\sin z \sin(1/z)$ for $z \neq 0$?
Is it possible to multiply two laurent series? I saw that in wikipedia, it's not generally possible.

Operate formally. Fix finite $z\not=0$. Then each expand each sine into Taylor series:
$$\begin{eqnarray}
   \sin(z) \sin\left(\frac{1}{z}\right) &=& \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{(2n+1)!} \sum_{m=0}^\infty (-1)^n \frac{z^{-2m-1}}{(2m+1)!} 
\\ &=& \sum_{n=0}^\infty \sum_{m=0}^\infty  \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} = \Big| \text{insert identity}\Big|
\\ &=& \sum_{k=-\infty}^\infty \sum_{n=0}^\infty \sum_{m=0}^\infty  \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} \delta_{k,n-m} 
\\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=\max(0,-k)}^\infty  \frac{1}{(2(k+m)+1)!(2m+1)!}
\\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=0}^\infty  \frac{1}{(2m+2 |k| +1)!(2m+1)!}
\end{eqnarray}
$$
The latter sum can be evaluated in terms of Bessel functions, giving:
$$
 \sin(z) \sin\left(\frac{1}{z}\right) = \frac{1}{2} \sum_{k=-\infty}^\infty (-z^2)^k \left(I_{2|k|}(2)- J_{2|k|}(2)\right)
$$
Indeed:
$$\begin{eqnarray}
  \sum_{m=0}^\infty  \frac{1}{(2m+2 |k| +1)!(2m+1)!} &=& \sum_{m=0}^\infty \frac{1+(-1)^m}{2} \frac{1}{(m+2 |k| +1)!(m+1)!} \\
  &=& \sum_{m=0}^\infty \frac{1-(-1)^m}{2} \frac{1}{(m+2 |k|)!(m)!} \\
 &=& \frac{1}{2} I_{2|k|}(2) - \frac{1}{2} J_{2|k|}(2)
\end{eqnarray}
$$
