At a simple pole, multiplying by $z - a$ exposes the residue. At a pole of order $m$, multiply by $(z - a)^m$ instead: the residue is then hidden one step deeper, as a Taylor coefficient that a few derivatives can reach.
Idea clave
If $f$ has a pole of order $m$ at $a$, then
$$\operatorname{Res}(f, a) = \frac{1}{(m - 1)!}\lim_{z \to a}\frac{d^{m - 1}}{dz^{m - 1}}\Big[(z - a)^mf(z)\Big].$$
Often a series expansion is faster: the residue is the coefficient of $(z - a)^{m - 1}$ in the Taylor series of $(z - a)^mf(z)$.
Por qué funciona
With $f = \frac{c_{-m}}{(z - a)^m} + \cdots + \frac{c_{-1}}{z - a} + c_0 + \cdots$, the function $g(z) = (z - a)^mf(z) = c_{-m} + \cdots + c_{-1}(z - a)^{m - 1} + c_0(z - a)^m + \cdots$ is analytic at $a$. Its Taylor coefficient of $(z - a)^{m - 1}$ is $c_{-1}$, and Taylor's formula says that coefficient equals $\frac{g^{(m - 1)}(a)}{(m - 1)!}$.
Worked examples
Ejemplo resuelto: An exponential over a power
Find $\operatorname{Res}\left(\dfrac{e^{2z}}{z^3}, 0\right)$.
- Pole of order $m = 3$, and $z^3f(z) = e^{2z}$.
- $\frac{1}{2!}\frac{d^2}{dz^2}e^{2z}\Big|_{0} = \frac{4}{2} = 2$.
- Series check: the $z^2$ coefficient of $e^{2z}$ is $\frac{2^2}{2!} = 2$.
Respuesta: $2$
Ejemplo resuelto: A double pole
Find $\operatorname{Res}\left(\dfrac{1}{(z - 1)^2(z + 1)}, 1\right)$.
- Order $2$: $(z - 1)^2f(z) = \frac{1}{z + 1}$.
- Differentiate once: $-\frac{1}{(z + 1)^2}$, and at $z = 1$ this is $-\frac{1}{4}$ (then divide by $1! = 1$).
Respuesta: $-\frac{1}{4}$
Ejemplo resuelto: A square in the denominator
Find $\operatorname{Res}\left(\dfrac{1}{(z^2 + 1)^2}, i\right)$.
- $(z^2 + 1)^2 = (z - i)^2(z + i)^2$: a double pole at $i$, and $(z - i)^2f = (z + i)^{-2}$.
- Differentiate: $-2(z + i)^{-3}$; at $i$: $\frac{-2}{(2i)^3} = \frac{-2}{-8i} = \frac{1}{4i} = -\frac{i}{4}$.
Respuesta: $-\frac{i}{4}$
Ejemplo resuelto: A series is quicker
Find $\operatorname{Res}\left(\dfrac{\sin z}{z^4}, 0\right)$.
- $\frac{\sin z}{z^4} = \frac{1}{z^4}\left(z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots\right) = \frac{1}{z^3} - \frac{1}{6z} + \cdots$.
- The residue is $-\frac{1}{6}$. (The pole has order 3, not 4, because $\sin z$ vanishes at $0$.)
Respuesta: $-\frac{1}{6}$
Which method to use? For a pole of order 2 the derivative formula is usually quickest: one derivative of $(z - a)^2f(z)$. For order 3 or more, or when the top is $\sin$, $\cos$ or $e^z$, expanding in a series and reading off one coefficient is often faster and less error-prone than taking several derivatives of a quotient.
Ejemplo resuelto: A pole of order 2 at a complex point
Find $\operatorname{Res}\left(\dfrac{z}{(z - i)^2(z + 1)}, i\right)$.
- $(z - i)^2f(z) = \frac{z}{z + 1}$. Its derivative is $\frac{(z + 1) - z}{(z + 1)^2} = \frac{1}{(z + 1)^2}$.
- At $z = i$: $\frac{1}{(1 + i)^2} = \frac{1}{2i} = -\frac{i}{2}$.
Respuesta: $-\frac{i}{2}$
A useful special case: if $f = \frac{g(z)}{(z - a)^m}$ with $g$ analytic at $a$, the residue is $\frac{g^{(m - 1)}(a)}{(m - 1)!}$, the coefficient of $(z - a)^{m - 1}$ in the Taylor series of $g$.
Error común
Forgetting $\frac{1}{(m - 1)!}$, or using the wrong order $m$. For $\frac{e^z}{z^4}$, $m = 4$ and the residue is $\frac{1}{3!}(e^z)'''(0) = \frac{1}{6}$, not $1$. And for $\frac{\sin z}{z^4}$ the order is $3$, since the top vanishes at $0$; using $m = 4$ still works (an overestimate of $m$ is harmless if you differentiate correctly), but using too small an $m$ gives nonsense. Check against the Laurent series when in doubt.