The Cauchy-Goursat theorem is really non-intuitive and is very astounding. Can someone geometrically explain to me why its true?
I'm specifically talking about this version of the theorem:
For some complex function, the closed contour integral over a simply connected region is $0$ if the function is analytic in the region and $2\pi$ if the region contains a single non-analytic point, [where the function is continuous].
Best Answer
It sounds like you want a kind of "visual" proof, or at least intuition. The go-to source for that is Needham's Visual Complex Analysis. Check out page 435 (of the pdf) of the linked book, which offers a few different explanations. Personally, I find the geometric intuition to be the following: if you can shrink the contour to a single point without crossing a singularity of the function, then the integral is 0. Then the idea follows immediately because you have function $f$ bounded by some $M$, (it has no singularities inside the contour!), and you're integrating it on an arbitrarly short loop around a point, of length $\epsilon$, so your integral is bounded above by $M\epsilon$, and $\epsilon$ can be made arbitrarily small.
So in some sense you can think of a contour as a stretched rubber band, and each singularity as a peg (an imperfect analogy). In other words, it really helps to accept the fact that the choice of contour for contour integration is quite arbitrary (as long as you respect singularities). The precise formulation of this analogy is that what really matters is the winding number of the contour around each singularity. If the winding is 0, then the singularity does not contribute.