Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 5A (Complex Methods)

Factor the denominator as $z^2(1+z^2)$. Near zero,so zero is a double pole with residue zero. At $z=\pm i$ the poles are simple and the denominator derivative is $2z+4z^3=-2z$. ThereforeThese are all the poles.

The only pole is the simple one at $z=1$, with $\boxed{{\mathrm{Res}}_1=e}$. The point $z=0$ is an essential singularity, not another pole: multiplying $e^{1/z^2}$ by a nonvanishing analytic factor cannot remove its infinite principal part. For completeness its residue can also be read fromThe coefficient of $z^{-1}$ is $-{\sum }_{k\ge 1}1/k!=1-e$. This extra residue belongs to the essential singularity; it does not change the list of poles.

The zeros of the denominator satisfy $e^z=\ell \pi$ with nonzero integer $\ell$, since an exponential never vanishes. Equivalently the poles areAt such a point $e^{z_{m,k}}=(-1{)}^{k}m\pi$, and the derivative of the denominator is $e^z\cos (e^z)=(-1{)}^{m+k}m\pi \ne 0$. All these poles are simple, with