Computing Residues at Simple Poles

Complex Analysis · Axiom Academy

LESSON Computing Residues at Simple Poles The single most useful residue in all of complex analysis — and two clean ways to read it off without hunting for the whole Laurent series. 1. Multiply Away the Pole, Then Take the Limit Near a simple pole the Laurent series has exactly one negative-power term. So if we multiply f by the factor (z-z_0) that is causing the blow-up, that troublesome term turns into a plain constant — the residue — and every other term still carries a positive power of (z-z_0) that dies as . Multiply by (z-z_0) : the blow-up becomes the constant a_ -1 For at z_0=2 , multiplying gives (z-2)f(z)=1 for every z , so the limit is . Here the function was already a pure residue term — no constant or higher-order part to throw away. 2. The Quotient Shortcut: No Limit Needed Most functions you meet are quotients of analytic functions. When h has a simple zero at z_0 (so h(z_0)=0 but ) and , the limit formula collapses into a one-line evaluation. z_0 is a zero of the denominator — that's what creates the pole of f . A simple zero: the tangent has nonzero slope, so it's a pole of order exactly one. If the numerator also vanished, the factor would cancel and the pole might disappear. h'(z_0) — the tilt of the tangent at the root — is exactly the number you divide by. For at z_0=i , take g(z)=z^2 and h(z)=z^2+1 , so h'(z)=2z . Then g(i)=-1 , h'(i)=2i , and No limit, no Laurent series — just two evaluations and a divide.

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