Integrate a Laurent series term by term around a small circle centered at an isolated singularity $a$: every power $(z - a)^n$ gives $0$ except $n = -1$, which gives $2\pi i$. So only one coefficient matters for integrals, and it gets a name.
Idea clave
The residue of $f$ at an isolated singularity $a$ is the coefficient $c_{-1}$ of $(z - a)^{-1}$ in the Laurent series about $a$; then $\oint_{|z - a| = \varepsilon} f\,dz = 2\pi i\operatorname{Res}(f, a)$. At a simple pole:
$$\operatorname{Res}(f, a) = \lim_{z \to a}(z - a)f(z), \qquad\text{and if } f = \frac{g}{h} \text{ with } g(a) \ne 0,\ h(a) = 0,\ h'(a) \ne 0: \quad \operatorname{Res}(f, a) = \frac{g(a)}{h'(a)}.$$
Por qué funciona
At a simple pole, $f = \frac{c_{-1}}{z - a} + c_0 + c_1(z - a) + \cdots$, so $(z - a)f(z) = c_{-1} + c_0(z - a) + \cdots \to c_{-1}$. For $f = \frac{g}{h}$: $(z - a)\frac{g(z)}{h(z)} = \frac{g(z)}{\frac{h(z) - h(a)}{z - a}} \to \frac{g(a)}{h'(a)}$, using $h(a) = 0$ and the definition of the derivative. The condition $h'(a) \ne 0$ is what makes the zero of $h$ (and so the pole) simple.
Worked examples
Ejemplo resuelto: Cover-up
Find $\operatorname{Res}\left(\dfrac{1}{(z - 1)(z + 2)}, 1\right)$.
- $z = 1$ is a simple pole. Multiply by $z - 1$: $\frac{1}{z + 2}$.
- Evaluate at $1$: $\frac{1}{3}$.
Respuesta: $\frac{1}{3}$
Ejemplo resuelto: Using $g/h'$
Find the residue of $\dfrac{z + 2}{z^2 + 1}$ at $z = i$.
- $g(z) = z + 2$, $h(z) = z^2 + 1$, $h'(z) = 2z$; $h'(i) = 2i \ne 0$, so the pole is simple.
- $\operatorname{Res} = \frac{i + 2}{2i} = \frac{(2 + i)(-i)}{2} = \frac{1 - 2i}{2} = \frac{1}{2} - i$.
Respuesta: $\frac{1}{2} - i$
Ejemplo resuelto: A trigonometric quotient
Find the residue of $\cot z$ at $z = \pi$, and of $\dfrac{1}{e^z - 1}$ at $z = 2\pi i$.
- $\cot z = \frac{\cos z}{\sin z}$, and $\sin$ has a simple zero at $\pi$ with $\sin'(\pi) = \cos\pi$: residue $\frac{\cos\pi}{\cos\pi} = 1$.
- $h(z) = e^z - 1$ has $h'(2\pi i) = e^{2\pi i} = 1$: residue $\frac{1}{1} = 1$.
Respuesta: $1$ and $1$
- Simple pole, factored denominator: cover up the factor $(z - a)$ and evaluate the rest at $a$.
- Simple pole of $\frac{g}{h}$ with $h(a) = 0$, $h'(a) \ne 0$: use $\frac{g(a)}{h'(a)}$. This is the quickest for $\tan$, $\cot$, $\frac{1}{e^z - 1}$ and denominators like $z^4 - 1$.
- Any isolated singularity: read $c_{-1}$ from the Laurent series (the only option at an essential singularity).
A removable singularity has residue $0$, and so does any point where $f$ is analytic. The residue at a pole can be $0$ too: $\frac{1}{z^2}$ has a double pole at $0$ with residue $0$, which is why $\oint\frac{dz}{z^2} = 0$.
Why "residue"? It is what is left over: after integrating around a small circle, every term of the Laurent series disappears except $c_{-1}$.
Error común
Using $\frac{g(a)}{h'(a)}$ at a pole that is not simple. For $\frac{1}{z^2(z - 1)}$ at $0$, $h(z) = z^3 - z^2$ has $h'(0) = 0$: the formula would divide by zero, and the pole is double. Check the order first (does $h'(a) \ne 0$? is the factor $(z - a)$ to the first power?). For higher-order poles, use the derivative formula of the next lesson.