Order of a zero

Definition

Let $U \subseteq \mathbb{C}$ be an open set, let $f : U \to \mathbb{C}$ be holomorphic on $U$, and let $z_0 \in U$. Assume that $f$ is not identically zero on any open disc with center $z_0$. The order of $f$ at $z_0$ is the smallest integer $m \ge 0$ with $f^{(m)}(z_0) \neq 0$, where $f^{(m)}$ is the $m$-th complex derivative of $f$. If $f(z_0) = 0$, this order $m \ge 1$ is the order of the zero $z_0$ of $f$. A zero of order $1$ is a simple zero. If $f$ is identically zero on an open disc with center $z_0$, the order of $f$ at $z_0$ is $\infty$.

Remarks

The order is $m$ if and only if $f(z) = (z - z_0)^m g(z)$ near $z_0$, where $g$ is holomorphic near $z_0$ and $g(z_0) \neq 0$. The order exists because holomorphic functions are analytic: if all the derivatives at $z_0$ are $0$, then the Taylor series is $0$ and $f = 0$ near $z_0$. For example, $z^3 (z - 1)$ has a zero of order $3$ at $0$ and a simple zero at $1$. The function $\sin z - z$ has a zero of order $3$ at $0$.

Explanations

L1 human

The order of a zero says how fast the function goes to $0$ at that point. Near a zero of order $m$, the function behaves like a constant times $(z - z_0)^m$.