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.
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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}.
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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.
The residue is one number, not the whole principal part. It is precisely the
coefficient of the (z - z_0)^{-1} term — not the sum of all the
negative-power coefficients, and certainly not the function's value at
z_0 (which does not even exist there). Why only that single term?
Because \oint_\gamma (z - z_0)^n\,dz = 0 for every integer
n except n = -1, so as you loop once
around the singularity every other coefficient integrates away to nothing. That lone survivor
is the deep reason residues alone pin down contour integrals — the punchline waiting
for us in the residue theorem.