lpetrich
Contributor
"Galois-Free Guarantee! | The Insolubility of the Quintic" -- avoiding Galois theory and using only some properties of complex numbers, crediting mathematician Vladimir Arnold. But it does have some Galois-theory connections.
The quadratic formula is well-known: the solution of a general degree-2 polynomial requires a square root. Less well-known are solutions for cubic and quartic polynomials, degrees 3 and 4. The cubic one requires a cube root of a quadratic solution, and the quartic one a square root of a cubic solution.
But no further with addition, subtraction, multiplication, division, and nth roots.
At 04:11 the presenter has a proof of the Fundamental Theorem of Algebra, as it's called, that a degree-n polynomial in complex numbers has at n complex roots. The proof has two parts: showing that at least one complex root always exists, and using it to show that the remaining (n-1) complex roots also exist.
Consider polynomial
\( \displaystyle{ P(z) = \sum_{k=0}^{n} a_k z^k } \)
Make the independent variable z make a circle in the complex plane: \( z = z_0 + r e^{i\theta} \) for radius r and angle θ from 0 to 2π. For large r, \( P(z) \simeq a_n z^n \simeq a_n r^n e^{n i\theta} \) and it makes n loops around the origin. Shrinking r to 0, one arrives at the coordinate origin, the curve has to go through the origin, or both. That means that the polynomial has at least one root, which I will call z1. P(z) must be divisible by (z - z1):
\( P(z) = (z - z_1) P'(z) \)
where P'(z) has degree (n-1). Continue further, and one finds that
\( \displaystyle{ P(z) = a_n \prod_{k=1}^{n} (z - z_k) } \)
This is an existence proof as opposed to a proof by construction: it shows that the roots exist without showing how to find them. It also does not say whether or not any roots coincide, or how many sets of coincident roots that there are.