Let \(f\) be holomorphic on a punctured disc \(D_r(a)\setminus\{a\}\) with a pole at \(a\). The order of the pole is the smallest integer \(m\ge0\) such that \[ \lim_{z\to a}(z-a)^{m+1}f(z)=0. \]
If \(f(z)=(z-a)^{-n}h(z)\) near \(a\) with \(n\ge1\), \(h\) holomorphic near \(a\), and \(h(a)\neq0\), then the order defined here is \(n\).
We have \((z-a)^{m+1}f(z)=(z-a)^{m+1-n}h(z)\). If \(m\ge n\), this tends to \(0\). If \(m=n-1\), it tends to \(h(a)\neq0\); if \(m<n-1\), its modulus tends to \(\infty\). So the smallest \(m\) with limit \(0\) is \(n\).
What is the order of a pole of $f$ at $a$?