Cauchy's Integral Theorem

A holomorphic function's loop integral is zero.

The idea

Theorem (Cauchy's Integral Theorem).

Let $\gamma$ be a closed contour and suppose $f$ is holomorphic on an open set containing both $\gamma$ and the region it encloses. Then $\oint_{\gamma} f(z)\,dz = 0.$

Equivalently, if $f$ is holomorphic on a simply connected open set $D$, meaning one with no holes, so that every closed curve in $D$ can be shrunk to a point without leaving $D$, then $\oint_{\gamma} f(z)\,dz = 0$ for every closed contour $\gamma$ lying in $D$. The figure's left panel shows such a contour shrinking to a point; its right panel shows the one thing that can block the shrinking.

Holomorphic on the contour alone is not enough: $f$ must be holomorphic on everything the contour encloses, and the theorem claims nothing when even one interior point is left out. This is why a nonzero loop integral is informative: it proves that $f$ has a singularity — a point where it fails to be holomorphic — somewhere inside.

Two consequences follow. Path independence: two contours in $D$ with the same endpoints give the same integral, because traversing one forward and the other backward makes a closed contour. Deformation: if $f$ is holomorphic in the region between two closed contours, one lying inside the other, their integrals are equal — so a contour may be slid and reshaped freely as long as it never crosses a point where $f$ fails to be holomorphic.

Ways to work on it

• Walkthrough. The model integral, the hypotheses, and a direct application.
• Practice. Decide a closed-contour integral over the unit circle.
• Hardest. Evaluate a closed-contour integral when a singularity is enclosed.

Not sure where to start? Take the ten-question placement test.