Let \(\Omega\subseteq\CC\) be open, let \(f\) be holomorphic on \(\Omega\), let \(z_0\in\Omega\), and let \(n\ge1\) be an integer. The point \(z_0\) is a zero of order \(n\) of \(f\) if there exist an open neighborhood \(U\subseteq\Omega\) of \(z_0\) and a holomorphic function \(g\colon U\to\CC\) with \(g(z_0)\neq0\) such that \[ f(z)=(z-z_0)^n g(z)\qquad\text{for all } z\in U. \] A zero of order \(1\) is a simple zero.
Let \(\Omega\subseteq\CC\) be a connected open set and let \(f\) be holomorphic on \(\Omega\) and not identically zero. Then every \(z_0\in\Omega\) with \(f(z_0)=0\) is a zero of order \(n\) of \(f\) for exactly one integer \(n\ge1\).
Existence. Let \(A\) be the set of \(w\in\Omega\) with \(f^{(k)}(w)=0\) for all \(k\ge0\). \(A\) is closed in \(\Omega\) because each \(f^{(k)}\) is continuous. \(A\) is open: if \(w\in A\), the Taylor series of \(f\) at \(w\) vanishes identically and equals \(f\) on a disc about \(w\), so that disc lies in \(A\). Since \(f\not\equiv0\) and \(\Omega\) is connected, \(A=\varnothing\). Hence there is a least \(n\) with \(f^{(n)}(z_0)\neq0\), and \(n\ge1\) because \(f(z_0)=0\). On a disc \(D\subseteq\Omega\) about \(z_0\), \(f(z)=\sum_{k\ge n}a_k(z-z_0)^k\) with \(a_n\neq0\); take \(U=D\) and \(g(z)\coloneqq\sum_{k\ge n}a_k(z-z_0)^{k-n}\), which converges on \(D\) and satisfies \(g(z_0)=a_n\neq0\).
Uniqueness. Suppose \((z-z_0)^n g(z)=(z-z_0)^m h(z)\) near \(z_0\) with \(g(z_0)\neq0\), \(h(z_0)\neq0\), and \(n<m\). Then \(g(z)=(z-z_0)^{m-n}h(z)\) for \(z\neq z_0\) near \(z_0\), and letting \(z\to z_0\) gives \(g(z_0)=0\), a contradiction.