The Fundamental Theorem of Algebra

The fact that polynomials are complex analytic in the entire complex plane lies at the heart of a beautiful and simple proof of the fundamental theorem of algebra.

Although it might appear like we are hiding the complexity of the proof by building up so much machinery that it's possible dispose of this mighty theorem in a few lines, in fact it is truly direct. For the most powerful ideas are necessarily the simplest; if they were not simple it would not be possible for them to explain so much.

[1] Complex Analysis, Elias M. Stein, Rami Shakarchi (Chapters 1 & 2)

[2] A Course of Modern Analysis, E. T. Whittaker, G. N. Watson (Chapter 5)

[3] Geometry and the Imagination, David Hilbert, Stephan Cohn-Vossen (Chapter 4, section 38)

[YellowBunny]

This was a great video. To me the first 5 or so minutes with mostly animations were a bit too slow while all those integrals near the end of the video were a bit too quick to follow along without pausing the video. Maybe talking a bit more about each of the steps might help there.

[TheOneThreeSeven]

Let me know how this video works for you, I know it is at a more advanced level than most youtube math videos but I think it's about time to kick it up a notch around here. Were you able to follow everything, or do you have any questions? Let me know!

[mathsfermattest]

I watched your video from your comment on mathslogger

[vitalysarmaev]

That's fascinating and just wonderful! Great job!

[hassanakhtar]

Superb animation and I loved this video. I think it would be nice if there was some light music in the intro and the ending. Im not an expert though.

[cameronspalding]

@14:36 for the middle one the integrals should be f(gamma)*dz/d(gamma) and it is being integrated wrt gamma

[gggg]

Great video, very helpful. Thanks a lot

[douglas]

at 2:31, why 1/|p(z)| is upper bounded?
p(z) can be as close to zero as we want, so 1/|p(z)| can be as big as we want, so it is not bounded, correct?

[sereadavid]

but if p(z) tends towards 0 then 1/abs(p(z)) just gets bigger and bigger, hence you can't set an upper limit. So how come 1/abs(p(z))<M, where M is finite?

[ARBB]

Clean animation and visual design, very nice work on that front. However, the pacing is a little off, and you sometimes trip yourself off too quickly when presenting concepts, while taking lots of time in certain unnecessary and calmer parts.

Regardless, very nice.

[paradoxicallyexcellent]

A much nicer proof is to use the integral definition of winding number. Let gamma be a large circle and consider the winding number of the curve f circ gamma about zero. For a large enough radius circle, this winding number is nonzero. It follows that if we continuously shrink this circle, we encounter a zero of the polynomial.

Your proof requires changes of variables and proof by contradiction which I believe obfuscates why FTA is actually true.

[cameronspalding]

The Cauchy Reimann equations are linear