Fundamental theorem of Algebra - A Simple but Beautiful Geometric proof

Proof that any polynomial with degree n has exactly n roots.

Time stamps :
0:00 Terms and Definitions
3:25 Complex Number Multiplication
8:34 Proof

[deeppurrple]

Math teachers, you can plagiarise.
Best explanation in this genre.

[FishSticker]

6:43 small correction: you got 2root(2) by rating the root of 8, but then when you square it, you say that it is equal to 4root(2) when you should have said it was equal to eight

[denysvlasenko]

You do not need to prove there are n roots. Existence of just one root R means that (if n>1) polynomial has a factor: (x-R)(polynomial_n-1). Then by induction it's trivial that polynomial_n-1 also has another root.

[pausesmaths]

10:22 " It might dance around this circle maybe a various a little bit up a little bit down but generally it will be something like this "
But at 11:11, we understand that a circle has to pass through the origin
But as you said, those are not perfect circles, maybe the origin 0 is somehow avoided from a circle to another one. For me, your proof suffers from a lack of an argument of continuity along the line joining C to O ... which is for me the key of the proof.
Tell me if I'm wrong and why ...
Good work though. It illimunates this proof.

[francescaerreia]

I think I understood everything you said because I think every individual sentence made sense to me as you stepped through, but I still don’t see where you proved this here. If I lost you, I don’t even know where.

[mgmartin]

Excuse me. What do you mean by “Gauss chad”?

[FishSticker]

Your proof technically doesn’t account for terms other than the leading one and the constant, it’s a slight modification to get there. besides that, great video

[annaclarafenyo]

Not quite the best proof along these lines, because it doesn't translate to a good algorithm. You are using an "inside-outside" argument in the proof to argue that 0 is "inside" the big circle, but "outside" the small circle which means that as you deform the big circle to the small one, zero must cross at some point. The rigorous version of this is called the "Jordan Curve Theorem", and if you look inside any proof of the Jordan Curve Theorem, you'll see it uses a winding-number argument. Points in the inside of the curve are those where the curve makes a winding number of one (or minus one, depending on orientation of the curve), while points on the outside have a winding number of zero.

You can use the winding-number argument directly to prove the theorem without going through the intermediate step of the Jordan curve theorem. Just notice that the function f(z)=z^n has a winding number of n around the biggest circles, and it can only have a nonzero winding number around a zero of the polynomial. Since winding number is additive when you chop a curve in two, one of the two halves has a nonzero winding. So you just keep chopping up the curve until you get a nonzero winding number around an arbitrarily small region, and that's converging to a zero of the polynomial. This is constructive up to minor Brouwer issues (you need to bound the topmost coefficient away from zero), and IMO this is the "book proof" of the theorem. It is essentially due to Gauss, although he didn't quite formulate it in terms of winding number, the ideas are analogous enough to give him credit.

[patato]

The FTA only states an every polynomial over the complex numbers has at least one root. This implies the statement you made.

[deeppurrple]

Some day you should come out with a better visual presentation.

[rainerausdemspring]

A polynomial is not a function. These are completely different things.