Computing ∮(1/z)dz Around Circle

Complex Analysis · Axiom Academy

EXAMPLE Computing a Circular Contour Integral Parametrize the circle, then integrate a non-analytic function directly Evaluate the contour integral , where C is the circle |z| = 2 traversed counterclockwise. Because the integrand is not analytic , Cauchy's theorem does not apply — we must parametrize C and integrate directly. The contour C : the circle |z| = 2, traced counterclockwise. Every point is z = 2e iθ for θ from 0 to 2π. Nicely done. You evaluated a contour integral from scratch by parametrizing the path and integrating directly. Parametrize the path: for a circle |z| = R , use with from 0 to , so . The conjugate trick: on |z| = R we have , hence — this rewrites a conjugate in terms of alone. Non-analytic means no shortcuts: fails the Cauchy–Riemann equations, so Cauchy's theorem can't force the integral to 0 . Direct computation gives . Whenever the integrand involves , |z| , or , reach for parametrization — the analytic shortcuts simply do not apply.

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