Complex Analysis L09: Complex Residues

[@parsahamidi]

Incredible explanation, the most intuitive description of Laurent series and the residue theorem that I've seen. Don't know how I'd be understanding this stuff without this playlist!

[@abhiramsatyavolu8078]

Totally agree with you on this, it’s amazing that we get this for free!

[@juniorcyans2988]

I'm taking complex analysis this semester. I watched your videos again and again. I love your channel so much!

[@aftermath4096]

This is honestly the first time I see such a clear intuitive explanation for the residue theorem. It seems almost trivial now.

[@idolgin776]

I am currently trying to learn the patterns and intuition for complex analysis, and your explanations are awesome. Thanks!

[@xinglinli9874]

Thank you, Steve

[@aviaser]

Thank you Steve. Great explanation! Greets from Argentina!

[@JoseAlves-ww3pp]

Thanks Steve, from Cape Town!

[@eig_himanshu]

Thanks Steve from India

[@rachappachimirela3462]

Thanks Dr.Rachappa

[@Syndicalism]

Thank you for such a clear explanation Professor Steve, this video helped make things finally click for me. I wish I had found your channel and playlist last semester when I was still taking Complex Analysis and drowning trying to wrap my head around Residue Theorem. Fortunately I'm able to get your help for Dynamical Systems this semester and in the future when I take Vector Analysis as well as PDE. Keep up the great work!

[@jimlbeaver]

That’s a remarkable result and it makes me strangely uncomfortable. Really interesting…thanks!

[@petervanderwaart1138]

I think that if you are going to use the 3d plot of the upward spiral, you should be more thorough. For example, what points in the domain are mapped to the axis of the spiral? Log(0) is undefined, but what about points arbitrarily close to the origin? Does the usual 3d depiction show all the surfaces of the range? When you get to the residue theorem, its very easy to think going up the spiral is going around the pole, but it's not. The geometry of the range does not recapitulate the geometry of the domain. It is completely different.

[@solomonapuu542]

Thank you!

[@eig_himanshu]

Could you please discuss Complex analysis with Potential Flow

[@cliffknoll6226]

I think there might be a (serious?) problem in the presentation (I’m an engineer not a mathematician, and truly delighted by the best courses hands down on a variety of subjects). So, the use of the Log depends on the derivation of the fundamental theorem, which you showed on a disk. I looked the fundamental theorem up and it applies on an annulus but requires (i think) splitting the annulus into 2 deformable half annuli. But your last lecture doesn’t mention that, moreover it relies on the Cauchy-Goursat, which specifically cannot tolerate a singularity in the middle of the disk or the closed path isn’t 0, which it can’t be anyway (at this juncture) because it’s i*2pi! I love these lectures but i’m struggling a little with this. I assume the 2 half annuli can be introduced in the last lecture, and toss the (beloved) Cauchy-Goursat out the window? Also, i think this renders all applications of fundamental theorem void (ie all n in Z) without introducing some trick (half annuli, or maybe slit in the domain to exclude the singularity?). I derived 1/z in Cartesian coordinates from scratch, which made me feel a little more secure with this (you did polar in this lecture and i wasn’t sure about separation into two real functions as done with Cartesian in the last lecture (i need to work through polar in more detail later)). If I’ve missed something, or completely whack, please let me know, somebody

[@strippins]

Lecture 8 now includes a slit in between the two curves to create path not including a singularity?

[@AZZEDDINE2801]

Thank you

[@soccerbrain75]

Will part 8 of this series come back online? Is it currently been reworked?

[@hoseinzahedifar1562]

kzbin.info/www/bejne/o2KuoIKnq9qrhJo

[@exec9292]

What's the point of this though?

[@AABB-px8lc]

In short, it simplify very hard to grasp formula to almost 1 grade level constant.

[@Andres186000]

The point is that when you want to calculate an integral around a closed loop, you only need to calculate the residue, which is a matter of just evaluating a limit.

[@ralphhebgen7067]

One point is it helps to solve certain real-valued integrals that would be hard/impossible to evaluate.