The Power of Residues

Complex Analysis · Axiom Academy

A contour integral, with no integrating You already know how to read a function's singularities and pull out its residues . The payoff is enormous: the integral of a function around a closed loop doesn't depend on the loop's shape or any parameterization at all — it depends only on the residues of the poles trapped inside . Sum over the poles z_k that lie inside C . No parameterizing. Find the poles, add their residues, multiply by 2 i . Watch the closed contour sweep all the way around. Each pole it has enclosed contributes its residue, and the running total climbs by 2 i per unit of residue — here two simple poles of residue 1 each, so it settles on = 4 i . The integral around the whole loop is just 2 i times the residues swept up inside it. Drag the radius to grow the circular contour outward from the origin. Each time the loop crosses a pole, that pole switches from outside to inside and the integral jumps by 2 i times its residue. A pole left outside contributes exactly nothing. Move a pole in or out and the answer changes by exactly one residue's worth — the shape of the loop never matters. Closing the loop to crack a real integral Here's the move that feels like magic. To attack the real integral _ - ^ x^2+1 , join the real axis to a big semicircular arc in the upper half-plane, enclosing the pole at z=i . Crank the radius R : the arc's contribution shrinks toward 0 , while the straight real-axis piece grows to capture the whole integral — leaving 2 i\, (f,i)= .

This is the written version of the interactive lesson above. See the full Complex Analysis course.