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Residue (Complex Analysis) — Definition, Formula & Examples

A residue is the coefficient of the (z−z0)−1(z - z_0)^{-1} term in the Laurent series expansion of a complex function around an isolated singularity z0z_0. It captures the essential information needed to evaluate contour integrals via the residue theorem.

Let ff be analytic in a punctured neighborhood of z0z_0. The residue of ff at z0z_0, denoted Res⁡(f,z0)\operatorname{Res}(f, z_0), is the unique complex number such that ∮Cf(z) dz=2πi Res⁡(f,z0)\oint_C f(z)\,dz = 2\pi i\,\operatorname{Res}(f, z_0) for any positively oriented simple closed contour CC enclosing z0z_0 and no other singularity. Equivalently, it is the coefficient a−1a_{-1} in the Laurent expansion f(z)=∑n=−∞∞an(z−z0)nf(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n.

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 z₀
  • z0z_0 = The isolated singularity (pole of order m)
  • mm = Order of the pole (m = 1 for a simple pole)

How It Works

To find a residue, you expand the function in a Laurent series around the singularity and read off the a−1a_{-1} coefficient. For a simple pole at z0z_0, a shortcut exists: 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, apply the general formula involving the (m−1)(m-1)-th derivative. Once you have all residues inside a contour, the residue theorem gives the contour integral 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) = \dfrac{z}{(z-1)(z-2)} at z=1z = 1.
Identify the pole: The factor (z−1)(z - 1) in the denominator gives a simple pole at z=1z = 1 (order m=1m = 1).
Apply the simple-pole formula: Multiply f(z)f(z) by (z−1)(z - 1) and take the limit as z→1z \to 1.
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: sum the residues and multiply by 2πi2\pi i. This technique appears throughout physics and engineering — for instance, in evaluating real improper integrals, inverse Laplace transforms, and quantum field theory propagator calculations.

Common Mistakes

Mistake: Using the simple-pole formula at a higher-order pole
Correction: First determine the order mm of the pole. If m>1m > 1, you must use the general formula involving the (m−1)(m-1)-th derivative; the simple-pole shortcut will give a wrong answer.