Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 3 / i

The denominator factors as $(z+1-i)(z+1+i)$ and neither factor cancels with the numerator. Thus there are simple poles at $z=-1\pm i$. At infinity the quotient tends to one and has a regular expansion in $1/z$, so it has a removable singularity at infinity, with extended value one.

The numerator is nonzero at $1$ and $-2/3$, so the finite singularities are a pole of order three at $1$ and a pole of order two at $-2/3$. The numerator degree exceeds the denominator degree by three, with leading behaviour $z^3/9$. Hence infinity is a pole of order three.

There are logarithmic branch points at $z=-3i$ and infinity. At zero, every local branch has $\log (3i)=\ln 3+i(\pi /2+2\pi k)\ne 0$, so division by $z^2$ gives a pole of order two at zero. Decay at infinity does not remove its branching: with $w=1/z$, the function is $w^2[\log (1+3iw)-\log w]$, which changes branch when $w$ encircles zero.