Complex Integration Using Branch Cuts

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René Gastelumendi

EDIT: Note that at 9:00, there should be a less than or equal to sign, since the modulus of the integral is less than or equal to the integral of the modulus. Thanks to some of the commenters below for pointing that out!

[FacultyofKhan]

If you made it all the way here from the beginning of my playlist, then congratulations! This marks the final video in my Complex Variables playlist, which should cover enough material for an entire first course in Complex Variables. If you'd like more videos on Complex Analysis, head on over to my advanced topics playlist or feel free to make a request!

[not_intelligent]

The fact that such lecture is free is ridiculous

[UnforsakenXII]

This is the one part that I never bothered learning in complex analysis but it's been kicking my ass in quantum field theory so I guess now's a good time before I head off to graduate school, lool.

[chronomo]

9:00 Isn't modulus of the integral is smaller or equal to the integral of the modulus of the integrand? This would be a mere technicality, as the ML argument still holds with this appraisal.

[Dairy_AK]

Thanks for adding two more vedios to playlist. I love all of them.
May you please upload the vedio on integrating the functions involving square roots or cube roots of complex numbers?

[rickmcn]

With the probable fact that I've missed something and I'll look slightly silly, I believe 8:53 you make a slight mistake. 'The modulus of the integral is the integral of the modulus', I think actually we can only say 'the modulus of the integral less than or equal to the integral of the modulus', and the the argument from there still holds.

[patipateeke]

Thank you so much for this video lecture series. i followed all of the videos and learned a lot. It is awesome that you made this course available to everyone! Just a question: which tools are you using to note down your lectures (software + notes taking tablet?)?

[mnada]

Thank you for these high quality lessons, I am an independent learner and it occurred to me to study complex analysis. I have watched the entire play list one pass for orientation but my understanding was not to the level that give satisfaction and planning to revisit the subject again, would you recommend any prerequisite specially in complex foundation prior studying these lessons again ?

[trisharoy]

I am soo very excited for the new video😁. Its a luxury to be seeing your lectures

[phyziks]

Sir love yours videos and explanations ❤

[Czeckie]

I find it interesting to note that the integrand taken from 1 to infinity instead is equal to the Catalan's constant, the alternate sum of squares of odd numbers inversed

[marcovillalobos]

Please please please do conformal mapping for differential equations next!!!

[weamah]

Thank you for this playlist, just finished it. Great quality! Is there any chance you can cite the source(s) you have used? I'm quite curious as to how you choose the specific topics to include in a playlist.

[gwen]

Great video! Do you have problem sheets or recommended books with problems covering these topics?

[UnforsakenXII]

The only thing weird about branch cut integrals is how would you know which branch cut to use when nobody explicitly gives it to you? Different branch cuts would give different values for the integral so in a scenario where you would applying this, how would you pick the right one?

[arkie]

Great video, maybe I'm missing something here but why doesnt he include all the singularities in his contour? How do we "see" that we only need to make a contour containing one of three singularities?

[tokkia]

Can you explain how you got the integral equation at 3:18? And why is the last integral the most important?

[neerajchauhan]

Could you please make a video on the contour integration used by Herbert Goldstein in Kepler problem (action angle variables).

[ZainAGhani]

Brilliant video once again! I have a question which is slightly off topic, however it is regarding contour integration to solve improper integrals. Let’s say we have simple pole on the real axis. Why can’t we just use a semi-circle, why do we have to be detour around the point. I thought if a singularity lies inside or ON a contour, then applying residue theorems yields the answer?

[enrienri]

Hi, first of all, i just love your channel! I'm having a super hard time with this kind of integrals.. I want to ask: would have been the same if I take, as a contour, a keyhole? and so put the branch cut on the real positive axes? Thanks for your time