The numerator is nonzero at $1$ and $-2/3$, so the finite singularities are \b[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 \b[infinity is a <pole> of order three].

Solution

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 / b / Solution

 Solution (source code)