A <pole> of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ <holomorphic> near zero and $g(0)\ne0$. Equivalently, the <principal part> is finite and starts with $\boxed{a_{-n}z^{-n},\ a_{-n}\ne0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.

Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Analysis / 5 / b / ii / Solution

 Solution (source code)