Not only did you teach this very clearly, your enthusiasm was undeniable and contagious!

This is going on my "1-sided cheatsheet" for my final next week ✨ 🙌🏼

A holomorphic function can have an absolute minimum inside the region iff that minimum is equal 0. See for example f(x)=x on the unit disk has min |f(x)|=0 at x0 = 0 not on the boundary

@drpeyam

2 жыл бұрын

That’s a good point, I never realized that!

@Zxv975

2 жыл бұрын

Is this really an "if and only if" statement? The converse would be along the lines of "if a function f has a minimum of 0, then f is holomorphic" which I can't imagine would be true, right?

@ibrahimmassy2753

2 жыл бұрын

@@Zxv975 Yes, is if and only if. Suppose f is holomorphic and doesn't have zeros, thus 1/f is holomorphic too, thus the maximum modulus of 1/f is in the boundary but this corresponds to the minimum modulus of f. In contrast, if f has zeros then 1/f is not holomorphic in the zeros of f thus maximum modulus doesn't apply to 1/f

@Zxv975

2 жыл бұрын

@@ibrahimmassy2753 Makes perfect sense. Thanks.

Best channel

A proof to this would make a nice video

I have a HUGE gap in something very basic that I never ask ... Why are the boundrys of a set k denoted \ partial k?

That's really something I have trouble understanding even after spending some time on it. Same with Liouville theorem. It would mean that there is no such thing as a "3D bell curve" entire complex function. Something that has a finite max at 0 and then tends to 0 as the |.| increases. And that's a huge disconnect for me who always somehow thought of complex numbers as |R x |R...

@benjaminbrat3922

2 жыл бұрын

So I guess my question is: how to think about the difference between |C and |R x |R

@MrYourcraft1

2 жыл бұрын

@@benjaminbrat3922 The difference lies in the definition of a differentiable function on R^2 as compared to a a holomorphic function in C. A function f: C -> C is holomorphic if the corresponding function f':R^2->R^2 is differentiable (so far so good) AND obeys the Cauchy-Riemann equations. So your idea of "3D bell curve" certainly exists and can easily be written in terms of g':R^2->R^2, but g' wont obey the Cauchy-Riemann equations, which means that g wont be holomorphic.

@benjaminbrat3922

2 жыл бұрын

@@MrYourcraft1 Thanks!! I'll reread that section. Would you remember where it is in the Stewart or the Boas?

@drpeyam

2 жыл бұрын

It’s not covered in Stewart, I’d recommend looking in brown and churchill

@pierreabbat6157

2 жыл бұрын

The bell curve goes to 0 as z goes to ∞, as long as z is more real than imaginary. If z is more imaginary than real, the bell curve goes to ∞. If you integrate it along any line through the origin (i.e. exp(-(xz)²) where |z|=1) that is not more imaginary than real, you get something involving √π, including when z=√i, which gives an Euler spiral.

Hi Peyam Jan. Thank you for the great lecture. I was wondering why for the maximum g(z) is exp(f), while for the minimum h(z) is exp(-if)? Why you did not use exp(-f) for h(z)? #Dr_Payam

But how do you proove this/where does it come from?

In my textbook by Saff & Snider their notation defines the "bar" that you wrote, ¯k ("k bar"), as the conjugate of that complex function. Is that what is being defined here as well: the conjugate of k? EDIT: "No, Kbar is K with its boundary" -Dr. Peyam

@drpeyam

2 жыл бұрын

No, Kbar is K with its boundary

@isaackay5887

2 жыл бұрын

@@drpeyam ​ Thank you! Needed to make sure...sometimes these things are a bit... _complex_ for me ...I'll see myself out 😅

I'm guessing this is a similar principle to the one they use in Operations Research, where the optimal business solution will always be on one of the vertices.

@drpeyam

2 жыл бұрын

Yep, well except it’s at the boundary, not the vertices

Green theorem on the general way, i dont remember the name but is F(∆D)=∆f(D)

Where is Chen Lu😒

0:16 scary

Now why is this theorem true? And more importantly. What's the motivation behind the proof.

As someone who has just come across your channel what is the 'Chen Lu'? I have not heard this function before and seems kind of racist if you are slurring words...

@benjaminbrat3922

2 жыл бұрын

That comes from BlackPenRedPen, I think. Him and piM are friend. I think it's one of their prof who pronounced "Chen lu" and "prada lu" for "chain rule" and "product rule" and they loved the teacher. Not sure of the exact detail.

@mertaliyigit3288

2 жыл бұрын

@@benjaminbrat3922 they thought those are special names of the rules, and after they finished calc I they finally understood its just badly pronounced "chen lu"

@roxashikari3725

2 жыл бұрын

It's a running joke between Dr. Peyam and blackpenredpen.

@gasun1274

2 жыл бұрын

racist is when when you say something with the accent you grew up with