Let \(\Omega\subseteq\CC\) be a connected open set, let \(f\) be analytic on \(\Omega\) and not identically zero, and let \(z_0\in\Omega\) with \(f(z_0)=0\). Then there exist a unique integer \(n\ge1\), a neighborhood \(U\subseteq\Omega\) of \(z_0\), and a function \(g\) holomorphic and nonvanishing on \(U\) such that \[ f(z) = (z-z_0)^n g(z)\quad\text{for all } z\in U. \] This \(n\) is the order of the zero \(z_0\), and it is characterized by each of the following:
\(f^{(k)}(z_0)=0\) for \(0\le k\le n-1\) and \(f^{(n)}(z_0)\neq0\);
the Taylor expansion of \(f\) at \(z_0\) has the form \(f(z) = \sum_{k\geq n} c_k (z-z_0)^k\) with \(c_n\neq 0\).
Write \(f(z)=\sum_{k\ge0}c_k(z-z_0)^k\) on a disc \(U\) about \(z_0\), with \(c_k=f^{(k)}(z_0)/k!\). If every \(c_k\) vanished, \(f\) would vanish on \(U\), and by the identity theorem on the connected set \(\Omega\), \(f\) would be identically zero. So there is a least \(n\) with \(c_n\neq0\), and \(n\ge1\) because \(c_0=f(z_0)=0\); this \(n\) satisfies both characterizations. Put \(g(z)\coloneqq\sum_{k\ge n}c_k(z-z_0)^{k-n}\), which converges on \(U\), is holomorphic there, and satisfies \(f(z)=(z-z_0)^ng(z)\) and \(g(z_0)=c_n\neq0\); shrinking \(U\), continuity makes \(g\) nonvanishing on \(U\). If also \(f(z)=(z-z_0)^mh(z)\) near \(z_0\) with \(h(z_0)\neq0\) and, say, \(m>n\), then \(g(z)=(z-z_0)^{m-n}h(z)\) near \(z_0\), forcing \(g(z_0)=0\); hence \(n\) is unique.