Fundamental Theorem of Algebra: Any univariate polynomial with complex coefficients has at least one complex root. That is to say, the field of complex numbers is algebraically closed.
Although it bears this name, it is not actually the most fundamental theorem of algebra; because at that time, algebra was primarily concerned with solving polynomial equations with real or complex coefficients, it was named the Fundamental Theorem of Algebra.
Based on this premise, we can deduce:
A univariate n-th degree algebraic equation with complex coefficients has n roots within the complex range (which may include conjugate complex roots).
This is proven using mathematical induction and polynomial “division” as follows:
Given that a first-degree equation has 1 root, and an n-th degree equation has at least 1 root. Assume that an (n-1)-th degree equation has (n-1) roots; we aim to prove that an n-th degree equation has n roots.
Let the function be f(x) = a_0 + a_1 x^1 + \dots + a_n x^n.
We calculate \frac{f(x)}{x-x_1}, where x_1 is a pre-specified constant. The process of calculation is similar to how we perform numerical long division:
Ultimately, our result is: \begin{aligned} f(x) =& \, a_0 + a_1 x^1 + \dots + a_n x^n \\ =& \, (x-x_1)[a_n x^{n-1} + (a_{n-1} + a_n x_1)x^{n-2} + \dots + (a_1 + a_2 x_1 + \dots + a_n x_1^{n-1})] \\ & + a_0 + a_1 x_1^1 + \dots + a_n x_1^n \end{aligned}
Let x_1 be a root of the equation a_0 + a_1 x^1 + \dots + a_n x^n = 0, therefore: \begin{aligned} & \, a_0 + a_1 x^1 + \dots + a_n x^n \\ =& \, (x-x_1)[a_n x^{n-1} + (a_{n-1} + a_n x_1)x^{n-2} + \dots + (a_1 + a_2 x_1 + \dots + a_n x_1^{n-1})] \end{aligned}
Then the solutions that satisfy a_n x^{n-1} + (a_{n-1} + a_n x_1)x^{n-2} + \dots + (a_1 + a_2 x_1 + \dots + a_n x_1^{n-1}) = 0 are also roots of the equation f(x) = 0. This equation has (n-1) roots, and adding x = x_1, the equation f(x) = 0 has a total of n roots.
The proof is complete.
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