View Single Post

Mentor

## Why is integral of 1/z over unit circle not zero?

 Quote by Poopsilon Clearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.
The theorem we need is the one that says that if B and C are two contours with the same endpoints, and B can be continuously deformed to C without going outside an open set on which f is analytic, then ##\int_B\, f(z)dz=\int_C\, f(z)dz##. That is a pretty deep theorem, but it's surprisingly easy to prove (see e.g. Saff & Snider). It follows almost immediately that if C is a closed contour, and f is analytic on an open set that contains C, then ##\int_C\, f(z)dz=0##.