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Contour Integrals, Laurent Terms & Residues

Useful preparation: Analyticity & Cauchy–Riemann

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Meaning & notation

Residues are the reciprocal-term coefficients that survive integration around singularities. The residue theorem combines these local contributions; contour orientation determines their signs.

Contour integral: For a piecewise continuously differentiable path $\displaystyle \gamma : \left[a, b\right] \to C$ and a continuous integrand along it, integrate $\displaystyle f\, \left(\gamma\, \left(t\right)\right)\, \gamma'\, \left(t\right)$ over the parameter interval. Reversing the path reverses the sign.
Laurent series and residue: A Laurent expansion around a includes integer powers, possibly negative, of $\displaystyle z - a$ and converges on an annulus. Its coefficient of $\displaystyle \left(z - a\right)^{-1}$ is the residue at an isolated singularity a.
Pole: A pole of order m has a finite negative-power part with highest reciprocal power $\displaystyle \left(z - a\right)^{-m}$ and a nonzero coefficient there. A simple pole has $\displaystyle m = 1$ ; a removable singularity is not a pole.

Conditions before calculation

For the version used here, C is a piecewise smooth simple closed contour, traversed once counterclockwise. The function is holomorphic on an open neighborhood of the contour and its interior except at finitely many isolated poles inside.
No pole may lie on C. Only enclosed poles contribute. Clockwise traversal multiplies the result by −1; more general closed curves require winding numbers rather than a simple inside/outside count.
The simple-pole formula $\displaystyle \frac{p\, \left(a\right)}{q'}\, \left(a\right)$ applies when p,q are holomorphic near a, $\displaystyle q\, \left(a\right) = 0$ , $\displaystyle q'\, \left(a\right) \ne 0$ , and $\displaystyle p\, \left(a\right) \ne 0$ . If the numerator also vanishes, examine cancellation first.

Why the method works

1. Locate singularities before computing

Factor the denominator and check numerator cancellation. Compare each remaining pole with the actual contour; a correct residue at an exterior pole contributes nothing.

2. Extract the reciprocal coefficient

For a simple pole, multiplying by $\displaystyle z - a$ removes its singular factor. Evaluating the remaining holomorphic expression gives the residue.

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

3. Account for the orientation

One counterclockwise turn contributes $\displaystyle 2\, \pi\, i$ times each enclosed residue. This comes from integrating the reciprocal term; other Laurent powers integrate to zero around a small circle.

\oint_C f(z)\,dz=2\pi i\sum_{a\text{ inside }C}\operatorname{Res}(f,a)

Worked examples

EXAMPLE 1

Integrate $\displaystyle f\, \left(z\right) = \frac{z + 1}{\left(z - 2\right)\, \left(z + 3\right)}$ around $\displaystyle \left|z\right| = \frac{5}{2}$ counterclockwise.

The simple poles are 2 and −3. Neither is on the circle; only 2 has modulus less than $\displaystyle \frac{5}{2}$ .
Remove the factor $\displaystyle z - 2$ and substitute 2.
$\operatorname{Res}(f,2)=\left.\frac{z+1}{z+3}\right|_{z=2}=\frac35$
Multiply the single enclosed residue by $\displaystyle 2\, \pi\, i$ . The exterior pole is excluded.
$\oint_C f(z)\,dz=\frac{6\pi i}{5}$

Conclusion: The integral is $\displaystyle \frac{6\, \pi\, i}{5}$ ; clockwise traversal would give $\displaystyle \frac{-6\, \pi\, i}{5}$ .

EXAMPLE 2

Integrate $\displaystyle g\, \left(z\right) = \frac{z^{2} + 1}{\left(z - 1\right)\, \left(z + 2\right)}$ around $\displaystyle \left|z\right| = 3$ clockwise.

Both simple poles, 1 and −2, lie strictly inside the circle, and there are no poles on it.
Compute each residue after canceling its own simple-pole factor.
$\operatorname{Res}(g,1)=\frac23,\qquad\operatorname{Res}(g,-2)=-\frac53$
Their sum is −1. Clockwise orientation uses $\displaystyle -2\, \pi\, i$ times that sum.
$\oint_C g(z)\,dz=-2\pi i(-1)=2\pi i$

Conclusion: The integral is $\displaystyle 2\, \pi\, i$ ; reversing orientation changes its sign.

Pitfalls & counterexamples

Tempting claim: Any denominator zero on or inside the contour can simply be added to the residue sum.

Correction: Poles on the contour invalidate the stated theorem and may make the ordinary contour integral undefined.

Test the claim: For $\displaystyle f\, \left(z\right) = \frac{1}{z - 1}$ on the unit circle, the path passes through $\displaystyle z = 1$ , where the integrand is undefined.

Check your understanding

What is the residue at 0 of $\displaystyle \frac{4}{z^{3}} - \frac{7}{z} + 2$ , and its integral around a counterclockwise unit circle?

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Compare your reasoning

The residue is −7 and the integral is $\displaystyle -14\, \pi\, i$ .

Only the coefficient of $\displaystyle z^{-1}$ contributes, although the singularity is a pole of order 3.

Connect this idea

Laurent expansions separate local singular behavior. Residues combine that local information with contour geometry and orientation to evaluate otherwise difficult integrals.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.