Around a positive circle centered at zero, the function 7/z³+2/z+5 has integral 4πi. The cubic pole term and constant term contribute zero. Only the coefficient two of z^(−1) survives.
That coefficient is the residue. It is a local quantity attached to an isolated singularity, and it can be zero even when the function has a severe pole. Residues convert many contour calculations into a finite sum of local coefficient calculations.
Predict
Predict the residue of 1/z². It is zero, although the function is unbounded near zero. Predict the residue of exp z/z² by expanding only far enough to find the z^(−1) term.
The needed term comes from the linear coefficient of the numerator. This is why a double pole often leads to one derivative rather than two in the residue formula.
Experiment
Compare three local calculations.
Function
Singularity order
Residue
3/(z−a)
Simple pole
3
1/(z−a)²
Double pole
0
exp(z)/(z−a)²
Double pole
exp(a)
The last entry comes from the first-order Taylor coefficient of exp z at a. Pole order and residue answer different questions.
Understand
For f holomorphic on a punctured disk about a, define Res_a f as the coefficient of (z−a)^(−1) in its Laurent expansion. Equivalently it is (1/(2πi)) times the integral around a sufficiently small positive circle. This definition includes essential singularities, whose residue is still a single coefficient.
For a positive simple contour enclosing finitely many isolated singularities, assume f is holomorphic on the intervening region and on a neighborhood of the contour. Remove disjoint small disks around the singularities. Cauchy’s theorem on the remaining region makes the outer integral equal the sum of counterclockwise small-circle integrals. Laurent integration then gives ∫f dz=2πi ΣRes f. For a more general cycle under the corresponding domain hypothesis, each residue is weighted by the contour’s winding number about its point.
At a simple pole of p/q with q(a)=0 and q′(a)≠0, factor q=(z−a)h with h(a)=q′(a). The residue is p(a)/q′(a). This value can be zero when a numerator zero cancels the supposed pole; the formula still correctly returns the coefficient.
More generally, if (z−a)ᵐf(z)=g(z) is holomorphic near a, then the residue is g^(m−1)(a)/(m−1)!. It extracts exactly the Taylor coefficient that becomes exponent minus one after division. Using a pole order smaller than the actual order invalidates that construction.
Apply
Compute the residue of exp(3z)/z³. The z² coefficient of the numerator is 9/2, so the residue is 9/2 and the positive-circle integral is 9πi. Check the factorial by differentiating twice as an independent method.
Repair “Add the residues of every pole in the formula.” Only poles contributing to the contour under its winding and domain conditions belong in the sum. A pole outside a simple positively oriented contour contributes zero, and a pole on the contour invalidates the ordinary statement.
Continue
The next lesson uses residues to calculate real integrals, but the contour’s added pieces must be estimated rather than discarded. Optional depth: compute the residue at an essential singularity by multiplying convergent Laurent series and carefully identifying all pairs of exponents that contribute to exponent minus one.
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Practice and reasoning
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Find the residue at z=2 of (z+1)/[(z−2)(z+3)].
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The residue of 1/z³ at zero is:
Derive the residue theorem for a simple positive contour enclosing finitely many isolated singularities and no singularity on the contour.
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If p,q are holomorphic, q(a)=0 and q′(a)≠0, derive Res_a(p/q)=p(a)/q′(a).
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Compute the residue at zero of exp(2z)/z² and its integral around a positive circle.
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A closed cycle has indices 2 and −1 about simple poles with residues 3 and 5, and satisfies the residue theorem’s domain hypotheses. Compute its integral.
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Reflect and continue
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