Mathematics.

singularities

The Residue Theorem

Complex Analysis40 minDifficulty7 out of 10

You should know: cauchy integral formula, poles and singularities

Overview

The residue theorem generalizes Cauchy's integral formula to functions with several isolated singularities inside a closed contour: if f is analytic inside and on a simple closed contour C except for finitely many isolated singularities z1, ..., zn inside C, then ∮_C f(z) dz = 2πi times the sum of the residues of f at those singularities. This turns the calculus problem of evaluating a contour integral into the algebra problem of computing residues (Laurent coefficients a_{-1}), and for a pole of order m, the residue can be computed by an explicit derivative formula without expanding the full Laurent series. The theorem is the single most powerful computational tool in complex analysis, and it is routinely used to evaluate real definite integrals that are difficult or impossible by elementary real methods.

Intuition

Cauchy's integral formula already shows that a single pole of 1/(z-a) inside a loop contributes exactly 2πi to the integral (all other 'smooth' behavior of f gets absorbed and evaluated at a). The residue theorem simply says: if there are several isolated troublemakers (poles) inside the loop instead of just one, each contributes its own 2πi·(residue) independently, and you add them all up. This turns contour integration around a complicated boundary into a purely local computation — you never have to actually walk around the full contour, you just need to know what's happening in tiny neighborhoods of each singularity trapped inside.

Formal Definition

Definition

Let f be analytic on and inside a positively oriented simple closed contour C, except for isolated singularities z1, ..., zn inside C:

∮Cf(z) dz=2πi∑k=1nRes⁡(f,zk)\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, z_k)
The residue theorem
Res⁡(f,z0)=lim⁡z→z0(z−z0)f(z)\operatorname{Res}(f,z_0) = \lim_{z\to z_0}(z-z_0)f(z)

Residue at a simple pole (order 1)

Simple pole formula
Res⁡(f,z0)=1(m−1)!lim⁡z→z0dm−1dzm−1[(z−z0)mf(z)]\operatorname{Res}(f,z_0) = \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^m f(z)\right]

Residue at a pole of order m

General pole formula

Worked Examples

  1. f(z) = 1/(z(z-2)) has simple poles at z=0 and z=2. Only z=0 lies inside |z|=1 (since |2|=2>1).

    poles at z=0,2; only z=0 inside C\text{poles at } z=0, 2; \text{ only } z=0 \text{ inside } C

  2. Compute the residue at the simple pole z=0 using the limit formula.

    Res⁡(f,0)=lim⁡z→0z⋅1z(z−2)=lim⁡z→01z−2=1−2=−12\operatorname{Res}(f,0) = \lim_{z\to 0} z\cdot\frac{1}{z(z-2)} = \lim_{z\to 0}\frac{1}{z-2} = \frac{1}{-2} = -\frac{1}{2}

  3. Apply the residue theorem with only this one enclosed pole.

    ∮Cf(z) dz=2πi⋅(−12)=−πi\oint_C f(z)\,dz = 2\pi i \cdot \left(-\frac{1}{2}\right) = -\pi i

Answer: ∮_C dz/(z(z-2)) = -πi.

Practice Problems

Difficulty 6/10

Evaluate ∮_C dz/(z²+1) where C is the circle |z|=2 (counterclockwise), which encloses both poles z=i and z=-i.

Difficulty 6/10

Evaluate ∮_C dz/(z²+1) where C is the circle |z - i| = 1 (counterclockwise), which encloses only z=i.

Difficulty 7/10

Evaluate ∮_C e^z/z³ dz where C is |z|=1 (counterclockwise), using the pole-of-order-m residue formula.

Quiz

The residue theorem states that for f with isolated singularities z1,...,zn inside a closed contour C:
The residue at a simple pole z0 of f can be computed as:
∮_C dz/(z-i) where C is |z-i|=1 (counterclockwise) equals:

Summary

  • The residue theorem: ∮_C f(z) dz = 2πi times the sum of residues of f at the isolated singularities of f enclosed by C.
  • Residue at a simple pole: Res(f,z0) = lim_{z→z0}(z-z0)f(z); for a pole of order m, use the (m-1)-th derivative formula.
  • This turns contour integral evaluation into a local, algebraic computation at each enclosed singularity, and underlies many real-integral evaluation techniques.

References

  1. WebsiteWikipedia — Residue theorem