Complex integrals are ... different.

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[fordtimelord]

Thanks a million for starting with such a basic example and running it all the way thru, especially with the subtle suggestions your intuition may guide you. This is excellent teaching.

[DanielDH]

So you made integrals more *complex* …

[MegaICS]

i want a compilation of all the 3:32 moments 😂

[charleyhoward]

I ABSOLUTELY LOVE his reaction at 3:33; very Human !

[chrisdock]

Just want to say your videos are fantastic, particularly regarding format and clarity of exposition.

[goodplacetostop]

14:09 Getting a C or D on a test does not describe you as a person. A test score is simply a happening, not a person. Have a good day ☀️

[Vladimir_Pavlov]

In the theory of functions of a complex variable, analytical functions and non-analytical functions are distinguished.
The integral of an analytic function between two points does not depend on the path (trajectory) of integration, for a non-analytic function it does.
Conditions of analyticity of the function f(z)=u(x, y)+i v(x, y) in the sense of Cauchy-Riemann equations:
∂u/∂x=∂v/∂y and ∂u/∂y= - ∂v/∂x.
Function f(z)=z =x+iy is analytic.
The function f(z)=z* =x-iy is not analytic.
No miracles.

[indocesare]

Actually whenever integrating over a domain in which the function is holomorphic one can say that the result is path-independent, right?

[abdullahyousef]

Can you make a series about discrete mathematics, and thank you.

[MrWarlls]

For calculating the integral of /z following gamma1, you can notice that /z = z exp(-iπ/4). You just need multiply your first result. For the second integral, z = /z following gamma2 and /z = -z following gamma3 😉

[manucitomx]

Complex Analysis has always been fun to me because it IS complex.

Thank you, professor.

[mattcarnevali]

Reminds me of state functions Pchem where chemical properties are path-independent and depend only on the initial and final points of the system

[demenion]

just using the cauchy-riemann differential equations allows you to easily calculate whether a given function is path-independent or not which is always nice to remember

[PunmasterSTP]

Michael's videos are beta than just about anything else out there...

[dariosilva]

In the "Suggest a problem" section, there is no way to upload a picture, just plain text. So how would you recieve a geometry problem?

[ojasdeshpande]

Here we are told that in calculus course you need to forget what you learnt in complex numbers and say √(-1) is undefined 🤔

[jupytr]

Yes, but the integral of z(bar)dz(bar) is path independent

[andikusnadi]

At 2:46 why use 0 to 1, kindly give me some englintment.

[TheLethalDomain]

Having watched this after seeing that the complex conjugate of a baseline complex number does not satisfy the Cauchy-Riemann differential equations summarizes why I expected the complex conjugate to be path dependent, yielding a different answer over different paths. If it is not holomorphic in differential form, we shouldn't expect it to be holomorphic in integral form due to the fundamental theorem of calculus. This lack of complex analytic behavior necessarily implies that the complex conjugate term is not continuously differentiable. Therefore, there is a specific path dedicated to its integration instead of multiple paths.

The final part of this video is particularly nice, because it's essentially a proof of the fundamental theorem of complex variable calculus.

[anon]

Nice video as always
I guess there are more families of functions that have path independence?
And what about those that are path dependent? Are there special paths I can take that all yield the same value anyway?