Woohoo!!!! 😄

Marvel: Infinity War is the most ambitious crossover event! Bprp and Dr. Peyam: Hold my markers

Please make a video on complex analysis :)

Best explanation of contour integral I have seen. Hopefully, y'all continue with a series of more complex integrals.

This is the first time I know that a lot of Peyam’s videos were taken in his office. Always liked the picture beside the board.

This video summarizes my calc 4 exam for next month on residue theorem. Thank you very much!

Hokusai's Great Wave and Van Gogh's Starry Night : Peyam is definitely the Artist of mathematics ! Well, about contour integral, I have to work a bit more :-(

Is it me or is this a lot more interesting than the usual videos? It gives an insight to how you'd work in university.

Complex analysis is *really* cool but for people to understand it from the ground and then people don’t have to assume it’s magic :) Basically the reason why the residue is important can be explained by the Taylor series of the (analytic, meaning it has a complex derivative) function, breaking it up into a sum of different powers of z. Every term has a nice primitive function [remember integral of x^n=(x^(n+1))/(n+1) ] *except* the 1/z term! (What’s kind of interesting in real analysis is really important in complex analysis here.) It’s the derivative of the complex logarithm which is multi valued. If we integrate on a closed curve, are start and endpoint is the same and so those terms with a primitive function integrate to zero! The exception is 1/z and famously the (complex) logarithm increases by 2*pi*i in one lap around the origin. So the only thing we care about when integrating on a closed curve is where the function has any singularities (else the 1/z term coefficient is necessarily 0 as the function is “nice”) and in those singularities, what the coefficient of 1/z is, and we call that the residue! (Note: some singularities have residue 0, like 1/z^2 in z=0, and do not contribute to the integral. It also does not have to be a simple “pole”, you can also integrate e^(1/z) around z=0 for example. This is an essential singularity where the function approaches infinity and 0 in the same point but it does not matter!)

Finally I found a worth watching video about contour integrals

To compute the residue just notice that 1/(z^2+1) = 1/(z-i) * 1/(z+i) and the second factor has no poles at z=i, so its Laurent expansion coincides to its Taylor expansion. The -1-th coefficient of 1/(z^2+1) now must be 1 (i.e., 1/(z-i)'s -1-th coefficient) times the 0-th coefficient of 1/(z+i), which is its value at z=i, namely 1/2i. So the residue is 1/2i (or -i/2).

Love dr. P. Very approachable person and loves his subject.

Wow this is amazing, he made some amazing points throughout the video

Honestly this is the most intuitive and clear explanation I've found of the process of finding bounds with triangle inequality etc - thanks Dr. Peyam, this was very useful!

This is so wholesome haha I love complex analysis as well!

Hello ,Dr. πm) I would like to say thank you that you invited blackpenredpen to your lecture because they asked very useful questions for novices. As for me, I was graduated from the engineering faculty of the russian university and surprisingly I had all these stuff with proofs (surprisingly because even mathematicians have less mathematics than us- engineers ). Anyway , thank you that you reminded me of the best subject - Complex Analysis) . By the way ,Residue in Russian means "Вычет" 😁

This video brought back some memories from graduate school... I really miss those times...

Dr Peyam is genius dude

Peyam's enthusiasm for math is infectious

This was cool.....Dr. P. rocks!

Ohhh ... Dr Peyam is left-handed!!! :D

Worth every minute

Really enjoyed this one!

wow!! Dr. Peyam. Brilliant try. Thank you. BlackPenRedPen!!

MIND BLOWN.

One of your best videos in all the time, I watched it for several times! Thank you so much dear *bP🖋️rP🖍️*

Nice paintings man .... Van Gogh, Hokusai , and something modernist with many colours resembling the Inca Empire flag ....

Loved it....!!!!

A question about determined integrals that are equal to pi**N As this video shows Integral of 1/(x^2+1) from -inf to inf= pi ?Is there any function integrated on an interval that is equal to pi**2 or pi**N (where N is an integer N=2, 3.....). Obviously the limit of the integral do not contain explicit pi nor the integrated function. Nor the function contains trigonometric (or inverse of) functions.

Nasty flashbacks to undergrad ... while complex analysis is amazing, I found it very challenging.

Man, how beautiful is it to see a culture that appreciates learning. So wonderful.

Calcualtion of the residuum is much easier than demonstrated. As the pole is of first order, you can use the following identity for residuum calculation: \textstyle \operatorname {Res}_{a}f=\lim _{{z ightarrow a}}(z-a)f(z) The term (z-a) cancels out and the remaining function value is equal to the residuum. I have integrated some even more complex functions. This method is pretty easy.

Awesome Video 😄

Happy Pride Peyam!!

Great video !

Where was this channel when I was in college?!?! 😊

Uhh Dr.Peyam rocking the Rolex, okay, I see you xD

Dr. Peyam is stylin!! Look at the sweet watch!! Steady pimpin' these lil Math tricks! Get Σ playa!! (I don't really talk like this; its just fun and I'm just very enthused about your attire.) But yo shirt be lookin' ill af; is dat silk???

Wonderful 🌷

Pls can you make a video about macclaurin series thx.

Wooooowwwwwwwwwwwwwwwwwwwww I really appriciate the teacher , well done sir

You guys are fun. I want to see a Very Special Episode of "The Big Bang Theory" where you give them all swirlies.

Do the integral of y=x^^x

Oke, it’s very interesting

I would think that someone who could do 1000 integrals, would be a complex analysis jock.

Great to see this again. We used the limit rule. The Laurent series was for me an eye opener, or should I say an i opener.

Excelent video dr peyam. I wonder what would have happened ...if when we were about to integrate 1/(z^2+1) wich goes to zero....instead we had put Arctan(z) what would be the result? Again 0? From z=R+0i to z=-R+0i the integrand satisfies Cauchy Riemann so...why not....

I really liked the host

Man... I love your smile 😍😅😛

Bprp: because "i" don't like to be on the bottom. Dr. P.: Yes, Nice. Bprp: ... Someone don't watch my videos Hahaha hahaha

Chester has a great shirt😄

it flew , just over the cereberum

Nice Big Classroom He Has There

I enjoyed math for the first time

Sometimes you do something just to see that it agrees with what you already know. Other times you do it because it is actually easier.

Okay

"ok" - blackpenredpen 5/1/19

Super u and Dr πr

please can help me to solve let a and b are matrices, if a*b=b*a prove that a*(b^-1)=(b^-1)*a

When Dr.Peyam was parameterizing the line segment on the semi circle from (-R,R) he wrote that ( γ(t)=t) but what is the meaning of (γ(t)).

Hi Blackpenredpen, I have a question for you: When you write(for example) sin^-1(x), you mean 1/sin(x) or arc sin (x)?

Why didn't you use the evaluation of arctan(x) between inf and -inf? It was much simpler... Did you just want to show an alternative method?

I just always forgot why the semi circle is sufficient. Why not both poles have to be included. Because you can integrate using any curve? but you have to include at least 1 pole, clearly.. can't rememberrr

Isn't it easier to use Cauchy's Integral Formula? with f(z)=(z+i)^-1 and w=i? you get the exact same result. Also this only works for R>1 right?

BPRP : "OK" Me : master saitama?

Why in that case the absolute value of z its iqual to R?

What course would you learn this in?

8:40

Can't we use Euler's form for z??..

Can't this integral (i.e. from minus infinity to plus infinity of 1/(1+x^2) ) also be solved like this? 1: L=int(-inf,+inf, 1/(x^2+1) ) 1/(x^2+1) is even, so L can also be derived as 2 times the integral from 0 to infinity => 2: L=2*int(0,+inf, 1/(x^2+1) )=2*int(0,+inf, (1/(x+i))*(1/(x-i)) ) 3: f(x)=1/(x^2+1)=1/((x+i)*(x-i))=A/(x+i)+B/(x-i) => A=-B and -2Ai=1 => A=i/2=-B => 4: L=2*int(0,+inf, i/(2(x+i))-i/(2(x-i)) )=i*int(0,+inf, 1/(x+i)-1/(x-i) ) 5: F'(x)=2f(x)/i=1/(x+i)-1/(x-i) and so F(x)=ln|x+i|-ln|x-i|+C L=i*(F(+inf)-F(0) )=i*(lim( x->+inf, ln|(x+i)/(x-i)| ) - (ln(i)-ln(-i)) )=i*(0-i*pi/2-i*pi/2)=-i^2*pi=pi I wonder: how much am I doing (in)correct at step (4 to) 5 or maybe somewhere else?

3:10 shouldn't it be absolute value abs(γ'(t)) in general?

I'm the like number 1000........yeeeeeeeeeeeeeeeeeeeeee ;-)

I get it or should I say 2i get it.

Subtitules in spanish please!!!! Tranks.

8:11

Are you also a doctor? , so do you have phd?

First!

No entendí una pija pero vi todo el video 📹 por que se cagaban de risa 😂.

arctan(inf) - arctan (-inf) = π ?? I'd like an explanation please. You just glossed over that as if it's trivial

i stan the pride flag in the background

Did you make a mistake? The integral was 1/z^2 + 1 and you used the triangle inequality with 1/z^2-1?

Third

Wow that starry night picture(wrong focus sorry)

why is it x in the title, not z?

I umm... I have no clue what just happened.

We stan with an LGBTQ+ math professors!

🏳️‍🌈🏳️‍🌈🏳️‍🌈

অভাবের দিনে গান্জা পাইলো কৈ?😮

Second

In physics we learn this in 2nd semester lol.

Wait wait... A rainbow flag 🤔🤔

I get so lonely lonely lonely lonely lonely

Show the teacher some respect

周知院学院　周知のじーじつ

Anyone watch while playing minecraft?

Why is there this rainbow flag?

I was just wondering... Is Dr Peyam gay?

disappointed

LGBTQ flag ? 😡

this was awful.