Applying Cauchy Integral Formula

Complex Analysis · Axiom Academy

EXAMPLE Applying Cauchy's Integral Formula Evaluate a contour integral with a double pole — no parametrization, just the formula and one derivative. Let C be the circle |z| = 2 traced counterclockwise . Evaluate The squared factor (z - 1)^2 means the singularity is a double pole, so we will reach for the generalized Cauchy formula — the version with a derivative. The double pole at z = 1 sits inside the circle C , since |1| = 1 < 2. Everywhere else inside, the integrand is analytic. Nice work. A double pole turned into a one-line answer — the only "calculus" was a single derivative. Spot the order first: the power on (z - a) in the denominator is the pole order n . Here (z - 1)^2 gives n = 2 . Generalized Cauchy formula: for an order- n pole inside C , , where g is the analytic part. For n = 2 that is . Result: with g(z) = e^z , a = 1 , and g'(1) = e , the integral is . Inside is everything: the formula only fires when the pole is enclosed. A pole outside C leaves the integrand analytic throughout the interior, so Cauchy's theorem makes the integral 0 . Read off the order, peel off g , differentiate n - 1 times — this is the engine that powers residue calculus.

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