Complex Analysis 33 | Residue for Poles [dark version]

Residues at isolated singularities can be computed cleanly once the singularity is identified as a pole—and if the function stays bounded near the point, the residue must be zero. That boundedness principle is the first major takeaway: when a holomorphic function has an isolated singularity at z0 but does not “explode” near z0, the contour integral around a small circle shrinks to nothing, forcing the residue to vanish.

To justify that, the discussion starts from the residue definition via a small closed contour integral. For a holomorphic function f on a punctured disk around z0, the residue is tied to the value of ∮ f(z) dz around a circle of radius ε. If f remains bounded on that circle, then the integral is bounded by (maximum |f| on the circle) × (length of the circle), i.e., a constant times 2π ε. Since ε can be made arbitrarily small, the integral must be 0, so the residue is 0. This generalizes the earlier observation that if a function is entire and one artificially removes a point (creating an isolated singularity), the residue at that point still comes out zero.

The next step is to focus on poles, the main kind of isolated singularity that produces nonzero residues. A pole at z0 is characterized by the existence of a holomorphic function h on a small disk such that h(z) = 1/f(z) and h extends holomorphically at z0. Equivalently, a pole behaves like the inverse of a zero: the function blows up like a negative power of (z − z0). A canonical example is f(z) = 1/(z − z0), where the reciprocal is z − z0, holomorphic everywhere, confirming z0 as a pole.

Poles also come with an order n. The transcript gives an equivalent structural form: if z0 is a pole of order n, then f can be written as f(z) = (z − z0)^(−n) G(z), where G is holomorphic near z0 and does not vanish there. Expanding G into a power series yields a Laurent expansion for f with a nonzero coefficient at (z − z0)^(−n). The residue is identified as the coefficient of (z − z0)^(−1), often denoted a_(−1).

The practical payoff is a residue calculation rule for poles. If z0 is a pole of order n, then multiplying by (z − z0)^n produces a holomorphic function. Differentiating (n − 1) times and evaluating at z0 isolates the (z − z0)^(−1) term. The resulting formula is: Res(f, z0) = 1/(n − 1)! · lim_{z→z0} d^(n−1)/dz^(n−1) [(z − z0)^n f(z)].

An example seals the method: for f(z) = 1/(z^2(1+z)), the point z0 = 0 is a pole of order 2. Using n = 2, the rule requires only one derivative of (z − 0)^2 f(z) = 1/(1+z). Differentiation gives −1/(1+z)^2, and evaluating at z = 0 yields −1. That value is the residue at 0, setting up how contour integrals will later be evaluated via the residue theorem.