Cauchy Riemann equations

In this video I derive the famous Cauchy-Riemann equations for a differentiable function of one complex variable. Those are equations that determine whether a complex function is differentiable or not, in terms of its real and imaginary parts. Zot zot :)

[blackpenredpen]

It's the Oreo-Cauchy-Riemann-Peyam Equations!!

[ibonitog]

Awesome video! Love your style of teaching, I'd be so grateful if I had a Prof like you!

[Kingradek]

"X not"
"Why not?"
:DDD
I always think that he is having a weird conversation with himself :'DDD

[Bbb]

Brilliant, and hillarious. I haven't ever laughed more than in your lecture, Dr. Peyam! As another comment said, your enthusiasm is amazing. Glad we made it through the marathon! SUCH A good video. Thank you Dr. Peyam.

[tomatrix]

I’m on a marathon watching your videos, Really enjoying them.

[abhishekchakraborty]

Your enthusiasm is contagious. I love it.

[dalitas]

This kind of maths is exactly what I enjoy, I always though numberphile did a too broad a perspective, but your channel has all the complex analysis I could ever need

[Fematika]

Yes!! Now prove it in the other direction!

[dashbaljinbishuubazar]

Thank you very much for the lecture. love the efforts you are putting in. Appreciate it.

[cedricp.]

Hi ! Nice video :)
I have two questions:
- Is the converse of this proposition true ?
- You can only split the limit if you know that the two limits you get exist. Here, how do you prove that Ux and Vx exist ?
Thank you !

[iilugs]

Hey doctor Peyam! great video as always, I love them! Well, I have a question for you. It´s a little bit paradoxical, and I wolud love if you would tell me where's the mathematical flaw.
So, the taylor Serie for e^x = 1/0! + x/1! + x^2/2! + x^3/3! ... and so on.Based on that, substituting x by 1, we can say that e is equal to the sum of the reciprocals of the factorials of the natural numbers, including 0.
Thus (i feel very eloquent when I use this expression), we can say that:
e = 1 + 1 + 1/2! + 1/3! + 1/4!.... <=> (adding -2 in both sides and, on the right side, multiplying and dividing by 1/2)
<=> e - 2 = 1/2(1+1/3+ 1/(4*3) + 1/(5*4*3) . . . ) <=>(multiplying and dividing by 1/3)
<=> e - 2 = 1/2(1+1/3(1+1/4 + 1/(5*4) + 1/(6*5*4) . . .)
if we kept dividing and multiplying by each reciprocal of natural number, as n factorial includes every natural number up to n, we can assure that said reciprocal will be one of the factors.
With this in mind, we can say that :
e-2 =
Multiplying a number times a fraction is the same as dividing the number by the reciprocal of the fraction, so, e can say that
e-2 = ...)+1)/4+1)/3+1)/2 + 2 (sorry, I couldn't manage to upload a pic of the fraction to the commet.

Ok, now that we've established that e = 2 + infinite fraction lets save that knowledge for later.

Now, let me talk about a particularly adorable piece of math, that is ramanujans funny square roots.
Srinivasa Ramanjan (an indian mathematician from the 20th century, as you probably know) once wrote that:

= 3


Well, 3 = 2 +1 = 2+ 2/2 = 2 + (1+1)/2 = 2 + (1+3/3)/2 = 2 + (1+(1+2)/3)/2 = 2 + (1+(1+8/4)/3)/2 = 2 + (1+(1+(1+7)/4)/3)/2 ... and so on.

If we repeated this process to infinity, we would get the same fraction that equals e !!! That means that 3 = e, and we could do this "fraction expansion" with any other natural number bigger that 3 and get the exact same fraction! that means e = 4, and e=5, and that is impossible, and I can't find the error in my reasoning.
I would love if you could point out the error.
Anyways, thanks a lot for reading, and please continue doing your amazing work, you are a super likeable person, and I love your demonstrations. Also, the video about the half derivative absolutely blew my mind! Great Job!

Ps: I would lovve if you could prove L'Hôpital's Rule, I cant get the demonstration that's on wikipedia
PPs: I would also love to see you proof Euler's solution for the Basil Problem ((pi^2)/6 = sum of the reciptocals of the perfect squares)

[Channel-zbfi]

Why is it sufficient to only check the vertical and horisontal direction in order to guarantee the differentiability of a function. Should it not be necessary to check all possible paths to z0?

[deeptochatterjee]

Wow, I just finished Calc III and I'm amazed that I understood this

[camilomuianga]

Hey Dr peyam, in first hand i'm congratulation for this interesting vidie, however, i Just came out with a doubt and i'd be glad with you clarify it!
Well, cauchy-Riemann equations we can related it with analytic functions right?
With we succed to see that the Cauchy-Riemann equations was sutisfied is that enough to we say is analytic functions?

[francribaj]

Thank you!Lesson very much appreciated subbed.:)

[leonine]

Thanks you so much, the explanation is very helpful 🙏

[BarkanUgurlu]

Hi Dr. Peyam, great explanation. Could you please make a video on Lipschitz condition? Greetings from Istanbul.

[michaelpurtell]

Is that Black pen filming and if so do you both teach at same institution. By the way excellent presentation (must be cause I understood it and I am not so good at math)

[isaaconyach]

I like your enthusiasm, now i feel like scoring some coke

[plaustrarius]

Little late to the party, I'm trying to learn a bit about holomorphic functions. I'm trying to convince myself of the polar form of the CRE, would love to see a video like this that proves it from more basic principals! (likely you have one already haha!)

Blown away that z^i is holomorphic and so is a^z! (only did the example for a being positive and real so far.)