Find Laurent series, in powers of $z$, of $$f(z)=\frac{\sin(2z)}{z}$$ valid in the region $|z|>0$.
The singularity is $0$ but $0$ isn't inside the region of the domain so what do you exactly expand?
Do you just expand $\sin(2z)$ and then divide it by $z$?
The function's analytic in your region (and almost analytic at $\;z=0\;$ ...), so using the power series for $\;\sin z\;$ which has infinite convergence radius:
$$\frac1z\sin2x=\frac1z\sum_{n=1}^\infty(-1)^{n-1}\frac{(2z)^{2n-1}}{(2n-1)!}=\sum_{n=1}^\infty(-1)^{n-1}\frac{2^{2n-1}z^{2n-2}}{(2n-1)!}$$
and we get in fact a power series, as expected.