Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 5D (Complex Methods)

Take the circle with the usual positive, counterclockwise orientation. The only singularity is at zero, and the Laurent series givesThe coefficient of $z^{-1}$ is $1/2$, so the residue theorem givesThe essential singularity at zero still has a well-defined residue; the theorem does not require it to be a pole.

The integrand has simple poles at $z=\pm 1/2$, both inside the positively oriented unit circle. At either pole $z_0$, the residue is $z_0/(-8z_0)=-1/8$. Their sum is $-1/4$, so the residue theorem gives