Complex Analysis 33 | Residue for Poles

Residues at isolated singularities can be computed cleanly once the singularity is identified as a pole—and if the singularity is not “explosive,” the residue must be zero. The core takeaway is a dichotomy: a bounded holomorphic function near an isolated singularity contributes no residue, while a pole of order n has a residue determined by a specific differentiation formula.

Start with an entire function f (holomorphic on all of ℂ) and “remove” a point z0 from its domain, creating an isolated singularity at z0. Because the contour integral around z0 depends only on values on the loop, the integral that defines the residue stays the same as for the original entire function. But for a holomorphic function with no singularity inside the loop, Cauchy’s theorem forces every closed contour integral to vanish. That means the residue at the artificially created singularity is 0. More generally, if f remains bounded near an isolated singularity at 0, then the residue must vanish: the residue is proportional to a contour integral, and the standard estimate bounds that integral by (maximum of |f| on the circle) × (length of the circle). Since the circle radius ε can be made arbitrarily small, the bound shrinks to 0, forcing the integral—and thus the residue—to be 0.

When the residue does not vanish, the singularity must be unbounded in a controlled way, and poles provide the standard mechanism. A pole at 0 of a holomorphic function f is defined by the existence of a holomorphic function h on a punctured disk such that h = 1/f, with h extending holomorphically at 0. Intuitively, poles are “reciprocal zeros”: f blows up exactly where 1/f has a removable singularity.

Poles can be characterized by power-series behavior. If 0 is a pole of order n, then f can be written on a small disk as f(z) = (z − z0)^(−n) · G(z), where G is holomorphic and non-vanishing near z0. Expanding G into a power series yields a Laurent expansion for f with a nonzero coefficient a_{−1} on the (z − z0)^(−1) term; that coefficient is the residue.

This structure leads to a practical residue calculation rule. If z0 is a pole of order n, then multiplying by (z − z0)^n removes the singular part, producing a holomorphic function. The residue then comes from differentiating n−1 times and evaluating at z0: Res(f, z0) = (1/(n−1)!) · lim_{z→z0} d^{n−1}/dz^{n−1} [(z − z0)^n f(z)]. All other Laurent terms vanish under the differentiation/evaluation process, leaving exactly the (z − z0)^(−1) coefficient.

An example makes the rule concrete. For f(z) = 1/(z^2(1+z)), the point 0 is a pole of order 2. Using n = 2, the residue at 0 becomes the first derivative of z^2 f(z) evaluated at 0. Since z^2 f(z) = 1/(1+z), differentiating gives −1/(1+z)^2, and evaluating at z = 0 yields −1. This residue value is what later powers contour integrals via the residue theorem.

In short: bounded isolated singularities yield residue 0; poles of order n yield residues via an (n−1)-derivative limit formula, with the residue equal to the coefficient of the (z − z0)^(−1) term in the Laurent expansion.