Residue at a simple pole

Statement

Let $c \in \mathbb{C}$, let $r > 0$, and let $f$ be holomorphic on $0 < |z - c| < r$. Assume that $f$ has a simple pole at $c$, that is, a pole of order $1$. Then $$\operatorname{Res}(f, c) = \lim_{z \to c} (z - c) f(z),$$ where $\operatorname{Res}(f, c)$ is the residue 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 $1$.