the complex derivative is strange...

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[Phlosioneer]

It should be noted that the limits shown are a necessary but NOT sufficient condition. It’s not enough to only consider linear paths towards 0 in a complex limit. There are functions that have a defined and equal limit for all lines through the origin, but on one specific curve (e.g. a parabola) they have a different limit. With some work, you can show that all “pathological” functions like that will also be discontinuous elsewhere, sidestepping the need to take a limit and completing the proof. It’s still a huge landmine for anything else related to complex limits, though.

[raptor]

Any complex differentiable function is analytic, so any real differentiable function which is not analytic will not be complex differentiable

[dang-xntct]

when you're so early there's no sponsorblock segment yet.

[_IQ_EQ]

One of the very few videos I have understood completely.

Michael Penn speaks slowly and explains immediately the possible doubt points like he does when he explains why u(x, y) instead of just u(x).

A great teacher. If only others took so much pain and emphasized on students' understanding and not completing syllabus.

[goodplacetostop]

25:12 Homework
26:00 Good Place To Stop

[markolson]

Huh. So *that's* where the Cauchy-Riemann relations come from. I've wondered for almost fifty years. (My complex analysis prof simply stated them and dove in, and I never subsequently made use of complex analysis to have a need to a deeper understanding.) Thanks!

[danieleferretti]

An analogous video about quaterionic derivation would be great!

[pierreabbat]

e^(-1/x²) is real differentiable at 0, but not complex differentiable at 0 (it goes to infinity on the imaginary path).
e^(-1/x⁴) is not complex differentiable at 0 either. The derivative goes to 0 on both real and imaginary paths, but to ∞ on the diagonals.

[numbers]

saw on stackexchange: f(x) = |x|^2 is differentiable (everywhere), but f(z) = |z|^2 is not holomorphic -- the complex derivative is only well-defined at z = 0. You can pretty much get any value for the complex derivative anywhere else by choosing your path wisely.

[GreenMeansGOF]

The complex conjugate function and the real part function are just the identity on R. So the R version is f(x)=x which IS differentiable.

[kilianklaiber]

This is a really good explanation. I had to learn this way back in university and used it so seldom that I had to look it up everytime it was necessary.

[ianmathwiz]

2:19 Wouldn't the complex conjugate function be an example of a function that's R-differentiable but not C-differentiable? Restricted to R, the complex conjugate is just the identity function, which is obviously differentiable, but it's the canonical example of a non-differentiable function on C.

[hyperduality]

The complex derivative is conformal as it is independent of angle at a point -- Cauchy Riemann equations.
Space/time symmetries are dual to Mobius maps -- stereographic projection.
Real is dual to imaginary -- complex numbers are dual.
"Always two there are" -- Yoda.

[thierrypauwels]

What I remember of it, is that you can combine both equations at the end to get an equation for f, without having to use its real and imaginary parts:
f_x = u_x + i v_x = v_y - i u_y = - i (u_y + i v_y) = -i f_y.
And you get finally the single condition f_x + i f_y = 0.

[talastra]

I couldn't do this, but I followed the explanation better than usual.

[seitanarchist]

I definitely have to take issue with your "everything* diff. in R is diff. in C" statement. The counterexamples aren't exactly sparse. You need to start with a real analytic function to be able to extend it to a complex differentiable function.

[iabervon]

There's a function with a real derivative everywhere except one point, but no complex derivative anywhere, at least in the most obvious extension (the absolute value, although extending it differently gives some fun functions not differentiable at paths or with weird branch cuts). It's interesting that the point at 0 for 1/x is no big deal with complex numbers, despite that.

[jamesmcadory]

This was so well explained thanks so much!

[Dravignor]

Is this the start of a complex analysis series?

[e.d.]

Such a *sad* thing! This video ignores something that almost every textbook also ignores:
The hidden assumption in this video is that if the derivative of a complex function along the Real Path is identical to the derivative of the same complex function along the Imaginary Path, then they are both identical to the derivative along any other path!
*But* is that assumption correct? Let's think of a *Real* function of two variable x and y i.e. Q(x, y): If one proves that the derivative along the x-axis is identical to the derivative along the y-axis, does that means that they are both identical to the derivative along any other direction/path?