In Conway's Functions of One Complex Variable vol. 1, Chapter 4, Section 8 we see Goursat's theorem a little different. The proof in the video is the almost the same we see there, but we are doing it for another purpose: supposing f only differentiable (and not continuously differentiable) you want to show that the integral of f over any triangular path is zero to invoke Morera's theorem and conclude that f is holomorphic.
Why do we know in 9:26 that this part has an antiderivative? I understand this for f` but not for f(z_0). I thought we only know, that f has an Derivate…
@@brightsideofmaths Thank you very much. I get it now. :D BTW: This videos are very helpful particulary for my exams preparation. :) Thank you very much for your effort.
7:58 But if it approaches a point, then aren't we supposed to finish the proof? Since I thought that integrating on a point gives zero and the only absolute value that is less than or equal to zero is also zero. Maybe my argument is not rigorous enough.
One of my favourite theorems in complex analysis. Simpel, yet so elegant…
These videos are worth millions. Thank you so much for making them available to us for free
Glad you like them! The videos are for free but it's worth to support me such that I can create more :)
@@brightsideofmaths I definitely will support you when I have the means to 👊
In Conway's Functions of One Complex Variable vol. 1, Chapter 4, Section 8 we see Goursat's theorem a little different. The proof in the video is the almost the same we see there, but we are doing it for another purpose: supposing f only differentiable (and not continuously differentiable) you want to show that the integral of f over any triangular path is zero to invoke Morera's theorem and conclude that f is holomorphic.
Thank you for giving a proof without Green's Theorem!
Yes, I only wanted to use the stuff we already know. Otherwise, it just feels like magic.
Why do we know in 9:26 that this part has an antiderivative? I understand this for f` but not for f(z_0). I thought we only know, that f has an Derivate…
Yeah, thanks for the question. It's a linear function. And we know that each polynomial has an antiderivative.
@@brightsideofmaths Thank you very much. I get it now. :D BTW: This videos are very helpful particulary for my exams preparation. :) Thank you very much for your effort.
@@SC0PC I am very glad that my videos can help you :)
Very interesting, thanks !
7:58 But if it approaches a point, then aren't we supposed to finish the proof? Since I thought that integrating on a point gives zero and the only absolute value that is less than or equal to zero is also zero. Maybe my argument is not rigorous enough.
Note that we still integrate around the point.
If we already know that the integral of a closed curve is zero, what does it matter that the image is a triangle?
How do we know that? For example, the closed curved integral of 1/z along a circle around the origin is not zero.
that so good
For those interested, there's a nice account on Contour Integration and
Cauchy’s Theorem by Cosgrove (don't paste link bc YT erases it).
I don't know these videos but recommendations for other good resources to learn mathematics are always welcomed!
@@brightsideofmaths Most certainly...keep it up! Greetings from Rosario, Argentina.
what does Ran stand for in 10:29?
range, the set of points that \gamma^(n) can take