Complex Analysis: Conformal Mappings

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[shardulkakade]

Your videos are the rare type of videos where one can digest the information, however complex, on the screen at any point without pausing. The topic just... flows. Your Complex analysis videos are some of the clearest I've ever seen and I just want to thank you.

Since your videos are uploaded at almost the exact time as I take a break, I have the privilege of being one of the first people in the world to watch your videos

[DKAIN_]

Man i am in highschool and I always wondered how does a graph of a complex function looks like, can you do a video on it👍

[algorithminc.]

Great channel ... thanks. I look forward to scoping your other videos. Subscribed. Cheers ...

[stevekass]

At 5:47, you say that curves C'_1 and C'_2 are the images of C_1 and C_2 under the mapping w = f(z) = z^2, but the images of C_1 and C_2 under that map are half-lines, not the lines you show; they shouldn’t extend left of the vertical axis in the w-plane. The images f(C_1) and f(C_2) do intersect at the origin, but is the "angle between f(C_1) and f(C_2) at the origin" well-defined if the curves only extend in one direction from the origin?

[manstuckinabox]

EDIT: oh wait these are general conformal mappings.... I ain't changing my comment LOL, however there is a theorem if not mistaken that states the only conformal mappings from C to itself are mobious transformations.

what I always found interesting about them is two things:

first, they're highly similar to transformation matrices in linear algebra. And second we can create from a specific subset of conformal mappings a group with the usual composition operator on functions, it's not obvious at first glance why it's important, but I'd like to take you back to the video you uploaded on the integral of a specific function satisfying f(x)+f(1/x) = pi/2.

the reason why you can't immediately jump and say, "Uh-yuk that's arctan!" is because you're a- killing the fun b- not answering the question, is arc tan the only function satisfying this property?


anyways, if we notice that we have a finite group of mobious transformations namely: x; 1/x, 1+x/1-x, x/x-1, 1-x, 1/1-x, we can construct from these 6 linearly independent equations and solve for the function.

the idea of the group is not prevelent, but if you think in this direction it becomes quite easy to generalize a method to solving for a more general case of problems.

and this idea also links back to my first point on why they're interesting, since they're so similar to transformations, in fact I'd argue we can show some morph ism from linear transformation to mobious transformations, I don't mean to the set of complex numbers, but to the set of complex functions satisfying this property.

Lastly their names is really worth mobin time.

[GeoffryGifari]

Wait... so the function f(z)=z^2 can work as a conformal map? is this because its holomorphic?
What kind of holomorphic functions can be a conformal map?

[Calcprof]

Recommended: Z. Nahari's book on conformal mapping -- and its Dover so it's inexpensive. I had graduate complex analysis from Nehari, and his intuition is incredible.

[membermember]

Are we gonna do a video on Schwarz-Christoffel (rlly nice stuff) ?

[Daniel-ycur]

Love this topic ❤ recently did a lecture on it