So I need to integrate
$$\oint_\gamma \frac{e^z}{z^2-1}dz$$
where $\gamma$ is the circle of radius $2$ centered at $0$. So by the residue theorem I have that
$$\oint_\gamma \frac{e^z}{z^2-1}dz = 2 \pi i \bigg( \text{Res}(\frac{e^z}{z^2-1};1)+ \text{Res} (\frac{e^z}{z^2-1};-1) \bigg)= \pi i (e-e^{-1})$$
So I calculated the residues at the two points of singularity. Namely, $z=\pm 1$. Did I do it correctly?
Best Answer
Your final answer is right, but you should write $$\text{Res}\left(\frac{e^z}{z^2-1}, \pm 1\right)$$ instead of those functions. The functions you put inside does not even have singularity there.