I am fully satisfied that you did not only mention the name of the theorems but instead used them to show how the integral becomes 0. Very straightforward proof and well-prepared video using two other theorems. Thank you, sir!
Thank you very much Sir, your explanation are always crystal clear. I was really delighted hearing about the Cauchy- Goursat theorem which has been taught to us (many years ago) as the Cauchy theorem. This is a way to pay tribute to forgotten names of people who greatly contributed to our nowadays understanding of Maths and physics. I have also really appreciated you mentioning the name of Green. Would you be so kind, if your spare time permits it, to prepare a lecture on the use of Green's functions, the delta function and convolution in order to solve in an elegant manner the Poisson equation in electromagnetics ? We surely would appreciate. Michel from Grenoble (France ).
Also the Green's theorem says that the integration over a boundary is equal to the integral over the domain, given some conditions. So should we not make the assumption that the contour can be continously deformed so its boundary perfectly fits with the boundary of the analytic domain, D. Since the original integration is done over the boundary of the domain C which is obviously a subset of D. So C would have to be made equal to D in order for Greens theorem to apply for the domain D.
Yes. Its an analytic function and its a closed curve, so the theorem applies. You could also double-check it using polar coordinates, using x = R*cos(u) with R = 1, and integrating with respect to u from 0 ti 2*pi which also would give 0
@@johndufek7492 f(z) is analytic only implies f'(z) exists everywhere in D, but not continuous. Without continuity, the proof is much much longer. In the book "Complex variables and applications" by James Ward Brown and Ruel V. Churchill on page 150 gives the detailed proof without continuity.
@@qinjackie2115 To be analytic, f''(z) must also exist. If f'(z) is not continuous everywhere, then how can it be differentiable in D? The same argument also applies to greater degrees of the derivative by induction. Hope it helps
