Residues

Of all the infinitely many coefficients in a Laurent series, exactly one survives integration around the singularity. Recall that \oint_\gamma (z - z_0)^n\,dz = 0 for every integer n except n = -1, where it equals 2\pi i. So integrating the Laurent series term by term wipes out everything but the 1/(z - z_0) term.

The residue of f at z_0 is the coefficient a_{-1} of 1/(z - z_0) in the Laurent series: \operatorname{Res}(f, z_0) = a_{-1}.
It is the single coefficient that survives: \oint_\gamma f(z)\,dz = 2\pi i\,\operatorname{Res}(f, z_0).

That turns a contour integral into the hunt for one number. The headline is that you usually do not need the whole Laurent series to find it.

Residues at poles, without the full series

Simple pole (order 1). If f has a simple pole at z_0, multiply by (z - z_0) to clear it, then take the limit:

\operatorname{Res}(f, z_0) = \lim_{z\to z_0} (z - z_0)\,f(z).

Pole of order m. Clear the pole with (z - z_0)^m, then differentiate m - 1 times to dig out the a_{-1} coefficient:

\operatorname{Res}(f, z_0) = \frac{1}{(m - 1)!}\lim_{z\to z_0} \frac{d^{\,m-1}}{dz^{\,m-1}}\Bigl[(z - z_0)^m f(z)\Bigr].

The quotient shortcut. When f = g/h with a simple zero of h at z_0 (and g(z_0) \ne 0), the limit collapses to a derivative of the denominator:

\operatorname{Res}\!\left(\frac{g}{h},\, z_0\right) = \frac{g(z_0)}{h'(z_0)}.

Worked residues

Step 1 — a trivial simple pole. For f(z) = \dfrac{1}{z - 3} at z_0 = 3:

\operatorname{Res}(f, 3) = \lim_{z\to 3}(z - 3)\cdot\frac{1}{z - 3} = 1.

Step 2 — the quotient rule in action. For f(z) = \dfrac{1}{z^2 + 1} at z_0 = i, take g = 1, h = z^2 + 1, so h'(z) = 2z:

\operatorname{Res}(f, i) = \frac{g(i)}{h'(i)} = \frac{1}{2i} = -\frac{i}{2}.

Step 3 — confirm with the limit form. Factor z^2 + 1 = (z - i)(z + i) and clear the pole:

\operatorname{Res}(f, i) = \lim_{z\to i}(z - i)\frac{1}{(z - i)(z + i)} = \frac{1}{2i}.

Both routes agree. With the residue in hand, \oint_\gamma \frac{dz}{z^2 + 1} = 2\pi i \cdot \frac{1}{2i} = \pi for a contour enclosing only i — a real integral, evaluated by a single coefficient.

The residue is the punchline of this whole stage. The disk of convergence told us where a series lives; the Laurent series gave us a language for singularities; classifying them told us what kind of trouble each one is. Now a single number from that series — a_{-1} — collapses a contour integral to one multiplication by 2\pi i. Sum the residues inside a contour and you can evaluate integrals that have no elementary antiderivative at all, including stubborn real integrals that yield to no other method. That is the residue theorem, and the residue is its atom.

Compute a residue live

Pick a simple-pole function with the control: \dfrac{1}{z - a} (residue at z_0 = a) or \dfrac{1}{z^2 + 1} (residue at z_0 = i). The pole is marked, and the readout computes the residue via \lim_{z\to z_0}(z - z_0)f(z) — for the first function this is always 1; for the second it is 1/(2i) = -i/2. Move the a slider to slide the first function's pole along the real axis and watch its residue stay pinned at 1.