I want to evaluate the integral
$$\int_{\partial\Bbb D}\frac{e^z\sin z}{(1-e^z)^2}\, dz$$
where $\partial\Bbb D$ is the boundary of the unit disk. Then its clear that the unique singularity of the integrand is at $z=0$, thus I checked if this singularity is a simple pole evaluating
$$\lim_{z\to 0}z\cdot\frac{e^z\sin z}{(1-e^z)^2}=\lim_{z\to 0}\frac12\cdot\frac{z^2}{\cosh z -1}=1$$
Thus using the residue theorem I found that the value of the integral is $2\pi i$. However checking the answer with WolframMathematica it seems to give a value of zero.
Im not sure if my solution is correct. Can someone confirm it? Thank you.