Fundamental theorem of algebra

Statement

Let $n \ge 1$ be an integer and let $a_0, a_1, \dots, a_n$ be complex numbers with $a_n \neq 0$. Let $p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$. Then there is a complex number $z_0$ with $p(z_0) = 0$.

Remarks

Proof with Liouville's theorem: if $p$ has no zero, then $1/p$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$, $1/p$ is bounded, so it is constant, which is false for $n \ge 1$. By induction on $n$, it follows that $p(z) = a_n (z - z_1) \cdots (z - z_n)$ for some $z_1, \dots, z_n \in \mathbb{C}$. So $p$ has exactly $n$ roots, counted with multiplicity. The theorem holds for polynomials with real coefficients too, but the roots can be non-real: $z^2 + 1$ has the roots $i$ and $-i$.

Explanations

L1 human

Each polynomial equation of degree at least $1$ has a solution in the complex numbers. So we never need to add new numbers to $\mathbb{C}$ to solve polynomial equations, unlike the step from $\mathbb{R}$ to $\mathbb{C}$ for $x^2 = -1$.