Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Analysis / 6 / a / iii

A pole of order $m\ge 1$ at $a$ has finitely many negative powers in its Laurent series, the lowest being exactly $-m$:Equivalently,Thus $(z-a{)}^{m}f(z)$ has a holomorphic extension with nonzero value at $a$, whereas multiplying by any smaller power does not remove the singularity.