Cauchy's Integral Formula and Proof 

Thank you! Glad you liked it!

Agreed

Jesus was the best mathematician.

Not only in youtube

ive watched like 5 videos and still don't get anything. Im an idiot.

I appreciate the kind feedback! Glad you found my work useful!

Watch from 2:55 of this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qTDDFMAt7j4.html If 'a' lies outside the contour, then f(z)/(z-a) is fully defined and holomorphic inside the contour. Therefore, by Cauchy's Theorem (in the video above), its contour integral would be zero. Hope that helps!

Why girls?????????

What has this gotta do with GIRLS??? If you love maths, love maths, and not girls. You cannot serve two masters at a time. From what you have said you cannot really become an expert in maths, because you have a secondary motive. And are the girls gonna give you good grades?

How to get laid, level Sciences. PD: Yeah, you can't be Math Master with two masters. I mean, you can be in the council, but not being a PhD in Maths Master.

The integral around a closed curve of a complex function is zero only if the function is analytic inside the curve (see: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qTDDFMAt7j4.html). When I took the limit, the integrals I end up with individually are not zero because their corresponding contours enclose a pole of the function g(z) being integrated.

Thank you for the kind feedback Joseph!

Glad I could clear up the question, and thank you for the kind feedback!

I'm wondering this, too

Thank you! We do need to show that the integrand is uniformly continuous, but in this case, the point a is not contained in the boundary of the integral (curve C), meaning the integrand is continuous on C. That means that we can apply differentiation inside the integral!

No, it's only zero when f(z)/(z-a), the function you're integrating, is differentiable (holomorphic) inside the region of the contour. Thus, when a is OUTSIDE the contour, f(z)/(z-a) is still defined INSIDE the contour, because there's no 'a' inside the contour to make the denominator zero. However, if 'a' is inside the contour, then f(z)/(z-a) has an undefined derivative (is not differentiable) inside the contour, at z = a, to be precise. That's why the integral wouldn't be zero in this case. For more details on the conditions for a zero integral, look at this video that I made: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qTDDFMAt7j4.htmlm13s Hope that helps!

Faculty of Khan Thank you!! So the integral wouldnt be zero if "a" its inside because the function wouldnt be holomorphic anymore in all the inside of the contour(because of the denominatorz-a) I have just discovered this channel and has great content, keep it up!

Yes, exactly, and thank you for the kind feedback!

No problem! Glad I could help!

I have a very basic doubt. Cauchys theorem says that closed Integral of f(z)dz is zero. Where z lies on the closed curve C. Then the point 'a' inside/outside the curve should have a value for cauchy intergral, and should only be zero when it's on the curve as that time the Cauchy intergral would be zero (by cauchys theorem). But he says that all a outside the curve would be zero. I don't get it please explain :/

"Ah I see its probably to do with a limit process", yeah pretty much. We're taking the limit so technically the curves never overlap.

it clearly isnt at z =a

no. The function has to be analytic EVERYWHERE on the interior (and boundary). Is f(z)/(z-a) analytic on the region which includes a? Clearly not, because it has a singularity at z=a... just like how 1/x is not differentiable at x=0. If it isn't even defined, it has no hope of being analytic!

No problem! Glad you liked it!

Thank you!

LOL

Yeah, I'm so glad he didn't get into analytic continuation here like so many others. Refreshing.

It is undefined then I think

Someone answer

Well, complex analysis comes up in a lot of ways in Math/Physics. One way that immediately comes to mind is the computation of various trigonometric integrals using the residue theorem (e.g. in ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-nfNzT2SpN4g.html); a problem that would be far more difficult to solve in real analysis. There's also applications of complex numbers in control theory, quantum mechanics, etc.

Thank you!

Glad you liked it! As for the blackboard, I took pictures and put them in here: imgur.com/a/bKjHV The link is also in the description.

Paul's online notes refers to this (Ctrl+F = cancel): tutorial.math.lamar.edu/Classes/CalcIII/GreensTheorem.aspx I haven't found a justification, but I can come up with an argument using regular integration. In typical integration, when you switch the limit of Integral from a->b of f(x) to Integral from b->a of f(x), a negative sign comes out (i.e. Integral a->b = -Integral b->a). So you can imagine that when you switch the direction of line integration, a negative sign would come out there as well, so the result up the line + down the line = 0. Hope that helps!

That doesn't help much. I see how switching the bounds on normal integrals negate the output. Even with Greens thrm integrals the orientation chances the result. But those are different then complex integrals. I did some work here imgur.com/89RhqM5 showing that the integrals will not cancel out. I'm going to go ahead and ask my prof and possibly get back to the thread with an answer. Thanks for your help!

Interesting, and no problem! I think with the work you did, you're on the right track to understanding the answer to your own question. However, I'll just point out that the substitution you did from t -> gamma-t is incomplete: You have to change both the limits and the sign of the differential; so in the end, the sum of those integrals turns out to be zero instead of twice the integral along one direction. Here's some work that I did to show what I mean: i.imgur.com/nr3Ia5t.png

Oh!!!! of course! I knew it was a simple error. it always is! thanks so much. sometimes its helpful to have another set of eyes. haha! I can finally move on now.

Yes I agree Phyi6er, he already proved line integral of closed curve, but this is also just the fundamental theorem of calculus which applies to complex functions as well, integral from a to b along path c is F(c(a))-F(c(b)) and integral from b to a along path c is just opposite F(c(b))-F(c(a)) , so adding those two cancel out and the direction of the line integral along curve c does change the sign of the result .

Thanks! I might add future videos where I do that, but for now, I'm going to continue with other topics. I appreciate the suggestion though!

Faculty of Khan Thanks. I absolutely don't mind. Being in high school it feels great to know about such things we aren't taught in school. You videos are easy to understand. :)

Hi Leo, Are you referring to what I did around 3:10? If so, then what's happening there is that I'm just reducing that gap (the gaps between points B and E and A and D) to zero. I can make that gap as small as I want because even with a really really small gap, I still have a closed contour (path W) inside which g(z) is holomorphic (so the integral of g(z) dz = 0 on W). Eventually, when those gaps become zero, the segments BA and DE overlap. At this point, two of those integrals cancel out (the ones along BA and DE) leaving us with the integrals along paths C and C'. Now if I'm interpreting your question correctly, then I did NOT (sorry for the uppercase, but that's just for emphasis) say that integral from A->D along curve C = integral along C (ACW) or that integral from E->B along curve C' = integral along C' (CW) when the gap isn't zero yet (i.e. the contours are still open) HOWEVER, when that gap becomes zero, the contours automatically become closed, since A now overlaps with D and E now overlaps with B. Of course, the open contour integrals when A and D don't overlap are going to be different from the corresponding closed contour integrals. Hope that clears up the confusion!

I use Camtasia for recording.

1) C is a curve, so it's not analytic. The *function* f(z) is analytic inside C. However, f(z)/(z-a) is not analytic at z = a, simply because it's undefined at z = a (denominator becomes 0). 2) You are right that 'a' is not a part of W, but as we decrease the gap between AB and DE, our equation ends up reducing to: anticlockwise integral of g(z) around C = anticlockwise integral of g(z) around C', and since a is inside the curve C', we can calculate f(a) (see the video after 4:09). 3) As z -> a, f(z) -> f(a). The reason z -> a is that we're making the circle C' progressively smaller in our final step (5:42), so within C', the only value that z can take on is 'a', or some complex number very close to 'a'. That's why z -> a, and by extension, f(z) -> f(a) Hope that helps!

Faculty of Khan Nice, I like the prof easy to folow. That part when everything on W=0, proved that C=C' when W is everywhere analytic. And then C' is easy to parameterise Think the teacher not got it. Thanks I like the proof but I have to try the one in the book nexttime insted.

@@FacultyofKhan I would like to add something important: your reasoning for (3) I think is slightly incorrect. You don't know that f(z) -> f(a), unless f is continuous. But of course, f is continuous (it is analytic after all!) so it works out. Awesome channel by the way! :)

Glad it helped!

Theta is the counterclockwise angle we've travelled along the circle C'. It equals zero at the point on the circle directly east from the center. It's analogous to the angle theta in polar coordinates, except our coordinate system is now centered at point 'a' instead of at the origin O. Hope that helps!

theeeeeeeeeeeeeta

Thank you for the kind feedback!

lol do you know me IRL that you could tell the difference? And to answer your question, my voice sounds different because I typically edit the audio on Audacity which changes pitch and some frequency components of my voice (editing out background noise, changing audio speed etc). I do that just to make my voice clearer and the audio sound better. Hopefully you found my video/audio of sufficient quality!

We've never met however; I watch a great deal of Khan Academy videos. Thanks for sharing!!

​@@ericklascaibar9900 Khan Academy is different from Faculty of Khan; incidentally, he does mention that he chose to name this Faculty of Khan after Khan Academy.

We're different, yes, but I didn't name this channel after Khan Academy lol. 'Khan' is my own last name - so it just made sense to include it. The 'Faculty of' part is because my videos teach university-level material, and universities typically divide their subject areas into 'faculties' (e.g. Faculty of Science, Faculty of Engineering, Faculty of Medicine).

>Is it because we students could get angry, or is it something inside you telling you it's not okay to state a theorem without proving it? Haha it's both. I tend to prove theorems when I think they're really important to the subject matter (e.g. this theorem and Residue Theorem are like the biggest theorems in Complex Variables). Also, I've actually stated theorems without proof in other videos, which, of course, resulted in some resentment.

@@FacultyofKhan I'm sorry that you get resentment either way, and I'm sorry I am part of that. Now that I have the chance, I want to thank you for making these videos that are helping me immensely to preparate for my exams!

No worries! If anything, your feedback and the feedback I get from others (resentment's probably not even the right word) will help me in future videos, where I'll have more worked examples. Good luck on your exams!

I am a non-Mathematician, an Engineer who uses these practically. I personally love the fact that he includes proofs, and feel that it's an absolute must for actual understanding. Because otherwise it's just a random formula that you don't know where it comes from. Without knowing how it's derived and all the assumptions that are made in obtaining it, it can be very easy for a practitioner to come to a very wrong (and costly) conclusion based on the accidental omission of certain assumptions which comes from not fully understanding how it was derived.

@@kabascoolr I agree, and as a matter of principle, I don’t incorporate any maths I don’t understand into my research (compact field theories, electromagnetism, computational algorithms). Any serious academic should appreciate a good proof, not even for the sake of reproducing it, but just to say, “I know from past experience that it makes sense and such and such are the main assumptions”. Edit: I dropped out of electrical engineering a few years ago, and I still remember one of the problems was how overwhelming it felt to be given equations, theorems, and entire physical models (like the lumped-element model of a circuit, for instance) without careful justification. It drove me nuts, and got so bad my scholarship program required me to see a therapist (by then, the damage was done). Proofs are like water to me 💧