Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 2 / 13D / i / Solution

The denominator vanishes when $z=(2j+1)i\pi /2$. For $R>0$, only $z_0=i\pi /2$ lies inside the rectangle. This is a simple pole, since the denominator derivative there is $2$. Its residue isThe positive orientation and the residue theorem consequently give