Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / 3 / ii

The denominator factors as $(z+2-i)(z+2+i)$. Neither root cancels against the numerator, and both are simple, so the finite singularities areThis is a rational function, so it has no branch points. At infinity, putting $w=1/z$ gives $w/(1+4w+5w^2)$, analytic at zero; infinity is therefore a removable singularity with value zero.