---
tag: 01991NCP7H
kind: statement
role: proposition
name: Residue at a simple pole
aliases: [residue-formula-for-simple-poles]
msc: ["30E20"]
uses: [0JFGKKWWFK, 0YK27TJ4WF]
sources:
  - url: https://en.wikipedia.org/wiki/Residue_(complex_analysis)
    locator: Simple poles
---

## Statement

Let $c \in \mathbb{C}$, let $r > 0$, and let $f$ be [[holomorphic|0DG0Z72CQD]] on $0 < |z - c| < r$.
Assume that $f$ has a simple pole at $c$, that is, a [[pole of order|0YK27TJ4WF]] $1$.
Then
$$\operatorname{Res}(f, c) = \lim_{z \to c} (z - c) f(z),$$
where $\operatorname{Res}(f, c)$ is the [[residue|0JFGKKWWFK]] of $f$ at $c$.
In particular, let $f = g / h$, where $g$ and $h$ are holomorphic near $c$, $g(c) \neq 0$, $h(c) = 0$ and $h'(c) \neq 0$. Then $\operatorname{Res}(f, c) = \frac{g(c)}{h'(c)}$.

## Remarks

Proof: near $c$, $f(z) = \frac{a_{-1}}{z - c} + a_0 + a_1 (z - c) + \cdots$, so $(z - c) f(z) \to a_{-1}$.
For example, $\operatorname{Res}\left(\frac{1}{z^2 + 1}, i\right) = \frac{1}{2i}$, with $g = 1$ and $h = z^2 + 1$.
At a pole of order $m$, the formula is $\operatorname{Res}(f, c) = \frac{1}{(m-1)!} \lim_{z \to c} \frac{d^{m-1}}{dz^{m-1}} \left( (z - c)^m f(z) \right)$.

## Explanations

### L1 human

At a simple pole, $f$ looks like $\frac{a}{z - c}$ plus something harmless. Multiplying by $z - c$ and letting $z \to c$ gives $a$, the residue.
For example, the residue of $\frac{1}{z}$ at $0$ is $1$, and the residue of $\frac{1}{z - 2}$ at $2$ is also $1$.