Mod-02 Lec-03 Calculus of residues (Part I)

[@henrywang6931]

Prof. Balakrishnan's lecture is full of mathematical insight, his explanation captures the essence of the problem and is very intuitive. This is a real treat for someone who already knew the topic a little. He is truly a wonderful lecturer!

[@brotherstech3901]

No doubt, Professor Balkrishnan is the best of IIT and on NPTEL channel. (Love from students).

[@ozzyfromspace]

I’ve always been scared to learn about this so called Residue Theorem, but I took a chance on this professor and he did not disappoint. I’m amazing that I can see the big picture idea of why I’d want to find the Residue of a function when attempting to compute its clockwise closed path integral in a region where it is entire. I’m only h-school educated so the fact that this lecture got through to me is really quite astounding, and now I feel a bit more confident about picking up a book when I’m done with his lecture series.

[@krishnareddynandipati3248]

I Heard his all of CM and QM lectures....became addicted......and now MP....Thank you sir.

[@vishalrao7010]

Same here, big fan of his teaching and him.

[@aravindhansivakumar2740]

Same bro. I am also. Even I didn't see him. But if somebody ask where did you get this knowlt on CM and MMP. Without any doubts, I can say. I have learned this all from Dr. balki sir. He is my teacher. 😍

[@saicharanmarrivada5077]

Clearly the best lecture on Complex analysis ever

[@brotherstech3901]

Finally I watched the whole lecture after 3 failures. I complete d my basics then came back. Everything is now Crystal clear. *😊

[@pramod120895]

Haha.... Atlast I found a guy with same mindset of mine..

[@mayankpandey4663]

Which level is this for Bsc or Msc?

[@mayankpandey4663]

Can you please tell me what should I do before going through these lecture so that i don't have any problem while watching this??

[@akhils350]

@@mayankpandey4663 i think phd

[@elamvaluthis7268]

@@mayankpandey4663 BSc level pure mathematics completely and attend this lecture it will be clear to you.

[@IftakHussain]

what happened in the middle of the lecture at 21:35.......sure technical fault...the technical staff should know that everybody doesn't have the chance to learn from this great man so please try to extract the whole lecture and then upload it...otherwise it becomes quite confusing as a learner...

[@SuryaGhosh7]

page 15 of the lecture notes: nptel.ac.in/courses/115106086/download/mathphys_nptel.pdf

[@anandapatmanabhansu]

Can you explain how essential singularity. ( that part was missing during 21:35

[@asrithperuri629]

best maths teacher in engineering life

[@dhimanroy1671]

Prof Balakrishnan is All square! Whatever the subject it it, the professor is always conducting the course with crystal clear concepts.

[@stephendickman9020]

Absolutely beautiful spokesman for an elagant subject!!!!!!!!!!

[@bhavyaagrawalla6319]

His explanation is so good even a high school student can understand

[@shifagoyal8221]

Fortunate are those who could attend your classes in IIT . I suppose such lectures should be preserved for the students to refer.

[@TheScientificPhilosophy]

at 21:00 essential singularity is missing yet very well explained it all , thanks !

[@anandapatmanabhansu]

Thankyou sir♥

[@MrAzizisse]

Indian Institute of Technology, must invest in moving black boards which the prof. can pull up or down to his eye level in order to write comfortably without bending his back which tires him. US schools like MIT, Sandford, Yale, all use such moving black boards. such black boards don't require the prof to bend his back or over extend his arm to reach the line he wants to write at. its very inexpensive. more over, i must thank you for the vast amount of lecture series available on this channel. however more recent lecture series from this prof. on more broader topics would be great. cosmology and astrophysics is not available here .

[@elamvaluthis7268]

Very nice explanation.

[@karabomothupi9759]

Revolutionary

[@samyakkshinde6864]

Pure gold🏅

[@ephbranch3496]

This man is God himself _/\_ please visit Roorkee for a guest lecture sometime - IITR students/fans

[@dalitshiv834]

No bro

[@AbhayKumar-og9ep]

Awesome lecture bro !!

[@debopamghosh959]

Please make sure that there is no skip between the video.... there is a skip at 21:18.... it's very difficult and annoying.... please be careful further, it's my request....

[@martinjones6202]

well sir is a math teacher i think...is this physics?????

[@dalitshiv834]

This is Mathematical physics chapter 1

[@surerakashtara]

Great lecture

[@esayasasfaw4405]

it is best learning

[@rajthebes3991]

Even a dead mind can learn concepts from your lectures.

[@shanecarlson7488]

Uhh. Its broken at 21:00 or so. It just skips forward it seems

[@ghasshee]

indeed. and differentiate the equation (m-1) times and you will get residue .

[@seanki98]

That awkward moment when one of the best lecturer's video skips a very essential part >:(

[@sishirjana565]

Sir ke amar pranam

[@asadullah337]

You are great Sir. From: Pakistan

[@Myrslokstok]

Beutifull!

[@Vercongent]

@10:09 it should be 1/z - z/3! + ---

[@Vercongent]

video guy screwed up @21:20!!!! >:O

[@Vercongent]

Vercongent I will save you the trouble by noting that the residue of a function which has a pole of order m at z=a is lim_{z \to a} 1/(n-1)! * d^(n-1)/dz^(n-1) [(z-a)^m *f(z)] . After this you naturally consider what happens when m \to \infty and in this case z=a is called an essential singularity and this is where the video picks back up

[@sijojosephdr]

Vercongent Your formula is wrong lim_{z \to a} 1/(n-1)! * d^(n-1)/dz^(n-1) [(z-a)^n *f(z)] is the right one !!! He teaches brilliantly man !!!!

[@hubomba]

He used the right formula, he just did an algebra mistake. Happens to everyone.

[@ephbranch3496]

Wow your so smart finding silly calculation mistakes

[@rituyadav5808]

nice

[@Abhishek-hy8xe]

Churchill and Brown is very dry.

[@pradeepshekhawat1296]

yessss

[@easylearningwithlabour9030]

how to calculate the residue at 21.00 ?what will be the formula?

[@adhimta3306]

Easy Learning with Labour after multiplying, differenciate till you reach the targeted coefficient, then apply the limite so all the higher order terms go to zero and finally solve for your coefficient (just multiply by a constant).

[@mateenkhan4243]

Which textbook is referred? ???

[@elamvaluthis7268]

Refer to Mathematical physics by Irwing and Mathematics physics by Arfken.

[@rituyadav5808]

why 1/sin nπz has simple pole ??..plz tell anybody.

[@seanki98]

So, we are clear that a function f has a simple pole if f(z)*(z-a) is both analytic and non zero at z=a, yes? You can check that the definition he gave is equivalent to this. The expression I wrote down clearly is analytic, since the denominator in the singular part of f(z) cancels through, and the rest of f(z) is regular. f(z)*(z-a) must be non-zero at z=a as well, because the only term left over when you substitute is c_(-1), which obviously must be non-zero. Otherwise, f(z) would not have a pole in z=a. Now, f is said to have a simple zero (a zero of order one) if f(z)/(z-a) is both analytic and non-zero at z=a. Just accept this as the defintion- you can check that this is a reasonable definition because it relates to multiplicity of zeros of a function- say a polynomial. Ok, I claim now that if f(z) has a simple zero at z=a, then 1/f(z) must have a simple pole at z=a. This is very easy to check. Suppose f(z) has a simple zero at z=a. Then, f(z)/(z-a) is non-zero and analytic at z=a. Therefore, (z-a)/f(z) is non-zero and analytic at z=a [I just took the reciprocal...]. Notice that this must mean that 1/f(z) must have a simple pole at z=a, because (1/f(z))*(z-a) is non-zero and analytic at z=a. Now, consider f(z)=sin(z). This has the power series f(z)=z-z^3/3!+... Clearly, it has a simple zero at z=0. You can check that... a=0 here. What is a little less obvious is that all the zeros of sin(x) is simple. Well, it is actually kinda obvious- anywhere sin(z) is zero, the derivative, cos(z) cannot be zero. A function that has a simple zero at z=a must have a non-zero derivative at z=a, you can check that. I make a very simple substitution and arrive at f(z)=sin(nπz). These clearly have simple poles for z=0, z=-1, z=1,... z any integer. But that must mean, as we saw in the paragraph before, that 1/f(z) must have a simple pole at the same values of z. Hence, 1/sin(nπz) has simple poles for any z equal to an integer. Does that answer your question?

[@feeras19864]

nice mistake at 10:00