Let \(f\) be holomorphic on a punctured disc \(D_r(a)\setminus\{a\}\) and let \(m\ge1\) be an integer. The function \(f\) has a pole of order \(m\) at \(a\) if \(\lim_{z\to a}(z-a)^{m}f(z)\) exists in \(\CC\setminus\{0\}\).
If \(f\) has a pole of order \(m\) at \(a\), then
\(\lim_{z\to a}(z-a)^{k}f(z)=0\) for every integer \(k>m\);
\(\abs{(z-a)^{k}f(z)}\to\infty\) as \(z\to a\) for every integer \(k<m\).
In particular \(m\) is the smallest integer \(k\) for which \(\lim_{z\to a}(z-a)^kf(z)\) exists in \(\CC\), and the definition agrees with D-AUD6K.
Let \(c\coloneqq\lim_{z\to a}(z-a)^mf(z)\neq0\). For \(k>m\), \((z-a)^kf(z)=(z-a)^{k-m}\cdot(z-a)^mf(z)\to0\cdot c=0\). For \(k<m\), \(\abs{(z-a)^kf(z)}=\abs{z-a}^{k-m}\abs{(z-a)^mf(z)}\), where the first factor tends to \(\infty\) and the second to \(\abs{c}>0\). Finally, \(h(z)\coloneqq(z-a)^mf(z)\) is bounded near \(a\), so by Riemann's removable singularity theorem it extends holomorphically with \(h(a)=c\neq0\), and \(f(z)=(z-a)^{-m}h(z)\) is a pole of order \(m\) in the sense of D-AUD6K; conversely such a factorization gives \((z-a)^mf(z)=h(z)\to h(a)\neq0\).
Backlinks
FD-EKGLWOrder of a pole as the least $m$ with $(z-a)^{m+1}f(z)\to0$variant-of