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Residue — Definition, Formula & Examples

A residue is the coefficient of the 1z−z0\frac{1}{z-z_0} term in the Laurent series expansion of a complex function around an isolated singularity z0z_0. It captures the essential behavior of the function near that singularity and is the key ingredient in evaluating contour integrals.

Let ff be analytic in a punctured disk 0<∣z−z0∣<R0 < |z - z_0| < R with an isolated singularity at z0z_0. The residue of ff at z0z_0, denoted Res⁡(f,z0)\operatorname{Res}(f, z_0), is the unique coefficient b1b_1 in the Laurent expansion f(z)=∑n=−∞∞an(z−z0)nf(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n corresponding to n=−1n = -1. Equivalently, Res⁡(f,z0)=12πi∮γf(z) dz\operatorname{Res}(f, z_0) = \frac{1}{2\pi i} \oint_\gamma f(z)\, dz, where γ\gamma is any positively oriented simple closed curve enclosing z0z_0 and no other singularity.

Key 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]
Where:
  • ff = A complex function with an isolated singularity at $z_0$
  • z0z_0 = The isolated singularity
  • mm = The order of the pole (for a simple pole, $m = 1$)

How It Works

To find a residue, you first classify the singularity. For a simple pole at z0z_0, multiply f(z)f(z) by (z−z0)(z - z_0) and take the limit as z→z0z \to z_0. For a pole of order mm, use the formula involving the (m−1)(m-1)-th derivative. Once you know the residues inside a contour, the Residue Theorem gives you the contour integral directly as 2πi2\pi i times the sum of those residues.

Worked Example

Problem: Find the residue of f(z)=z(z−1)(z−2)f(z) = \frac{z}{(z-1)(z-2)} at z=1z = 1.
Classify the singularity: The factor (z−1)(z - 1) in the denominator vanishes at z=1z = 1, giving a simple pole (m=1m = 1).
Apply the simple-pole formula: For a simple pole, the residue is the limit of (z−z0)f(z)(z - z_0) f(z) as z→z0z \to z_0.
Res⁡(f,1)=lim⁡z→1(z−1)⋅z(z−1)(z−2)=lim⁡z→1zz−2\operatorname{Res}(f, 1) = \lim_{z \to 1} (z - 1) \cdot \frac{z}{(z-1)(z-2)} = \lim_{z \to 1} \frac{z}{z-2}
Evaluate the limit: Substitute z=1z = 1 into the simplified expression.
11−2=−1\frac{1}{1-2} = -1
Answer: Res⁡(f,1)=−1\operatorname{Res}(f, 1) = -1

Why It Matters

Residues turn difficult contour integrals into simple algebra via the Residue Theorem, making them indispensable in complex analysis courses. They also appear in physics and engineering — for example, in evaluating inverse Laplace transforms and computing real improper integrals that resist standard calculus techniques.

Common Mistakes

Mistake: Using the simple-pole formula at a higher-order pole.
Correction: Always determine the pole order mm first. If m>1m > 1, you must use the general formula with the (m−1)(m-1)-th derivative, not just the limit of (z−z0)f(z)(z - z_0)f(z).

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