Residues and the Residue Theorem: Complex Integrals

The residue of f at an isolated singularity a, written Res(f, a), is the coefficient a₋₁ of the term 1/(z − a) in the Laurent expansion of f around a. It earns its name because when you integrate f around a closed loop encircling a, every Laurent term integrates to zero except a₋₁/(z − a), whose integral is 2πi·a₋₁ — it is the only part that "remains".

This single number is extraordinarily powerful. The residue theorem states that the contour integral of f around a closed curve equals 2πi times the sum of the residues at the singularities inside the curve, converting difficult integrals into short algebraic computations.

Key Concepts

• Residue: Res(f, a) = a₋₁, the coefficient of (z − a)⁻¹ in the Laurent series at a.
• Simple pole residue: if a is a simple pole, Res(f, a) = lim (z → a) (z − a)f(z).
• Residue at a removable singularity: always 0, since the principal part is empty.
• Residue theorem: ∮_C f(z) dz = 2πi × (sum of Res(f, aₖ) for all singularities aₖ inside C), for C a positively oriented simple closed contour.
• Residue at infinity: defined via the substitution w = 1/z; the sum of all residues including infinity is zero.

Formulas

• Simple pole: Res(f, a) = lim (z → a) (z − a) f(z).
• Pole of order m: Res(f, a) = 1/(m − 1)! · lim (z → a) dᵐ⁻¹/dzᵐ⁻¹ [(z − a)ᵐ f(z)].
• Quotient shortcut: if f = g/h with g(a) ≠ 0, h(a) = 0 and h′(a) ≠ 0, then Res(f, a) = g(a)/h′(a).
• Assumptions: f must be analytic on and inside C except for finitely many isolated singularities; C must not pass through any singularity.

Worked Example

Evaluate ∮_{|z|=1} eᶻ/z² dz, where the contour is the unit circle traversed counterclockwise.

1. The integrand f(z) = eᶻ/z² has one singularity inside |z| = 1: a pole of order 2 at z = 0.
2. Use the order-2 formula with m = 2: Res(f, 0) = 1/1! · lim (z → 0) d/dz [z² · eᶻ/z²] = lim (z → 0) d/dz (eᶻ).
3. d/dz (eᶻ) = eᶻ, and at z = 0 this equals 1. So Res(f, 0) = 1.
4. By the residue theorem, the integral equals 2πi × 1 = 2πi.
5. Sanity check via series: eᶻ/z² = 1/z² + 1/z + 1/2! + …, and indeed the 1/z coefficient is 1, confirming the residue.

Common Misconceptions

• "The residue is the value of the function at the pole." No — f is not even defined at a pole. The residue is the 1/(z − a) coefficient of the Laurent series.
• "Only poles have residues." Essential singularities have residues too (e^(1/z) has residue 1 at 0); removable singularities have residue 0.
• "The residue theorem needs the function analytic everywhere inside." It needs analyticity except at isolated singularities — handling those singularities is precisely the theorem's job.

Real-World Applications

• Real integrals: integrals like ∫₀^∞ dx/(1 + x⁴) are evaluated by closing the contour and summing residues.
• Inverse Laplace transforms: recovering time-domain signals from transfer functions is a residue computation.
• Series summation: sums like Σ 1/n² can be evaluated by integrating πcot(πz)/z² around large contours and collecting residues.

The residue is the one Laurent coefficient that survives contour integration, and the residue theorem turns ∮_C f(z) dz into 2πi times a sum of such coefficients. Learn the simple-pole and order-m formulas, and a large class of intimidating integrals becomes routine algebra.