It's unbelievable an amazing course like this available completely free on KZread. The guy is really good!

I really appreciate you making these lectures public. It's dense material, but that's exactly what I'm looking for. Thank you

What a ride in the complex analysis world! Thank you so much for putting it together! What a ride it is

This isn't mere mathematics. It is a work of community service, a work of kindness, and a work of charity.

Bravo! What an excellent set of lectures on complex numbers! Really well taught by Dr Brunton.

thanks a lot, completed the whole complex analysis 10 hours before my finals, You're a brilliant teacher!!!

Excelent course, greetings and congratulations.

I believe those "tricks" for showing that parts of integral in complex plane are 0 etc...are called Jordans lemmas (theorems)....unfortunately I do not have my textbook with me and it has been over 20 years...

@quantum4everyone

Жыл бұрын

Yes, that is correct. Jordan’s lemma holds for a broader set of integrands, and his formulation is slightly odd because everything is rotated by 90 degrees because his exponential in the integrand has no i in it. But the essence of the argument would be very similar.

Thanks for a great series. I was very well taught.

superb sir!!!!!! you way of explanation is fantabulous

How many hours do you have in one day? 70? More than 70? I just scrolled xN speed (with N huge) this series about complex analysis. Very well done. Lots of students in Engineering dealing with dynamical systems and control (so, almost every student in Engineering) curious about some detail about the math behind them and coming across these lectures should be so thankful to you. Obviously they're not enough without personal effort and study, but they're a good point to start for sure. Anyone who wants a concise and quite precise introduction to complex analysis and many other mathematical topics useful in engineering, could have on Schaum's Outlines, Advanced Mathematics for Engineers and Scientists: 10-15 pages of theory for every topic, and proofs left as an exercise to the reader.

@Eigensteve

Жыл бұрын

Awesome, thanks for the kind words -- glad you like them!

Great great great lecture...Thank you so much.

Thank You! I just finished a dynamics homework with no reference to a Laplace transform table. Un-necessary for sure, but I feel like a boss lol. Anyway the only bit I had to dig for myself was finding residues of higher order poles, but without your introduction I'd have struggled to make sense of the literature.

Your lectures are really great, thanks a lot! Is it possible to follow a similar apporach to obtain Fourier transforms?

Great class!

Thank you for the course, I appreciate the conceive form. You've been mentioning, that in good ol' days there would have been a whole semester course on complex analysis. Could you maybe recommend any sources to dive deeper into the topic?

At 34:50, the way theta and the contour are defined requires integration from pi/2 to -pi/2; but the limit is still zero. Jordan's lemma proves this in general...

@chrislubs1341

Жыл бұрын

Saw this, but noted he redefines theta to (PI - theta) to get the corect integral, which is important to keep inequalities from reversing due to a sign error. This video might serve to suggest useful FOURIER TAUBERIAN THEOREMS.

ATHF reference in last video of a Complex Analysis lecture... the future is now!

Hey Steve, I really like your videos, and I'm curious - are you writing in reverse or did you flip the image? Either way, it's a cool effect!🤔

Question: why would ML bound work? At 31:48, he assumes that exp(gamm*t)*gamma does not go to infinity, but it may, if gamma >1 and t -> infinity?

42:30 Is all the following gymnastics necessary? Since -Rcos(theta)t is always negative between -pi and pi, as we tend R to infinity, the integrand goes to 0, so the integral goes to 0

Would a semicircular contour be harder? Like if it was just a vertical line and a semicircle without C+ and C-?

About 38:20 or thereabouts am I right in thinking: Given R² + 2Racosθ + a² then holding R and a fixed while theta varies satisfies (R- a)² ⩽ R² + 2Racosθ + a² ⩽ (R+a)² since -1 ⩽ cosθ ⩽ 1 hence |R-a| ⩽ √(R² + 2Racosθ + a²) ⩽ |R+a|

at 44:00 I think this inequality can be deduced using Taylor series expansion up to 2 terms for cosine

what a ride..

Does someone now what marker pen this guys use ?

minute 41:00 I think the trick is |R-a| = sqrt((R-a)^2) wich leads to sqrt(R^2 - 2aR + a^2) If theta = pi, the inequality becomes equal, so true, but if theta is different from pi, the term 2aR cos(theta) < 2aR which is also true. Then you're right

Hello Steve, it will be nice if you make some videos related to statistics and probability theory

so, this 50 mins to prove C+, C-, Cr are 0 - is this proof just for the simplest f(s)=1/(s-a) ? ... do we need to re-do this math proof for each other possible f(s) ?

@chrislubs1341

Жыл бұрын

Be aware different inverse transforms f(t) corrispond to F[s] as distinquished by domain of F[s], so pick a desired f(t) by considering the Bromwich integral.

In time 25:22: I think ds = -dx. isn't it correct?

@byronwatkins2565

Жыл бұрын

Yes. But, the limit is still zero.

The (reverse) triangle inequality directly states |s-a| >= ||s|-|a|| = |R-a|.

*stops to e^at @ t=021.140*

Lack of consistency. Before talking about inverse Laplace transform, it is wise to first define the Laplace transform and how it is a generalization of the Fourier transform? Time to frequency domain? People with no background of signal theory will be confused.

complex step finite difference

Why is the default inverse definition the infinite integral and not the Cauchy integral formula?

You call this an integral, you don't have bacon on the curve?

At 27:04, I wonder how the norm of a complex variable is just the real part. Should we regard Cauchy inequality, || e^x+e^iR ||