Cauchy-Goursat Theorem Proof - Complex Analysis by a Physicist

Рет қаралды 17,348

3 жыл бұрын

In this video we do a proof of the Cauchy- Goursat Theorem in complex analysis.
The Cauchy-Goursat Theorem is of particular importance because if the required conditions for the theorem are present then we can easily determine that come complex integrals are zero.
Music Credit: Mirror Mind By: Bobby Richards

Пікірлер: 27

@yarenkaya7872 Жыл бұрын

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!

@MichaelMarteens 8 ай бұрын

Amazing that these videos are so straight the point and yet understandable. Also cool that you often include many examples.

@michelthiam9895 9 ай бұрын

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 ).

@yaojoelkoko8600 Жыл бұрын

Clear in my mind Thanks

@sahanasastha846 6 ай бұрын

It's very useful and a satisfied video. Thank u sir

@abhinavm3808 2 жыл бұрын

Thank you !

@hodece-2056 2 жыл бұрын

Thank you sir

@elgatito00 3 жыл бұрын

Awesome!

@nspace-cowboy 3 жыл бұрын

Thanks!

@saeedyari5715 7 ай бұрын

great

@ranukumar1353 2 жыл бұрын

Thanks sir...it was good and concise too...

@nspace-cowboy 2 жыл бұрын

Glad you liked it

@loretafreckles5754 Жыл бұрын

Nicely explained and also thanks for the dark mode hahah I've trying to find complex analysis like this

@nspace-cowboy Жыл бұрын

Happy to help!

@hanaalmira2749 2 жыл бұрын

if i have integral of x dz with |z|=1, can i use this theorem so that i have the solution is 0?

@zlatanbrekke6538 Жыл бұрын

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

@patrickadejoh8203 5 ай бұрын

But this is the same proof as the Cauchy's Integral theorem. That means they're the same right?

@murtazaabasskhan1103 Жыл бұрын

Good work!

@nspace-cowboy Жыл бұрын

Thanks!

@murtazaabasskhan1103 Жыл бұрын

@@nspace-cowboy Anytime! :)

@Channel-zb1fi 6 ай бұрын

Is it a bad idea to let the curve C be a function of the form r = a(t) + i * b(t), a

@Channel-zb1fi 6 ай бұрын

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.

@goto6799 2 жыл бұрын

Is continuity a necesaary condition for the integral to be zero

@johndufek7492 2 жыл бұрын

I believe so since f(z) is analytic within the domain D. That being said, analytic implies continuous.

@qinjackie2115 2 жыл бұрын

@@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.

@yarenkaya7872 Жыл бұрын

@@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