Proofs of the Fundamental Theorem of Algebra

published by Luca Wellmeier on Jan 26, 2025

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Contents:

1. Notation and statement
2. Proofs
   1. Liouville via a global minimum of $|p|$

Notation and statement

Denote by $\C[z]$ the ring of univariate polynomials with complex coefficients. Let $p \in \C[z]$ be non-constant of degree $n \geq 1$ with coeficients

$$p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$$

and $a_n \neq 0$. Then $p$ has a complex root.

Proofs

Liouville via a global minimum of $|p|$

Write

$$p(z)=a_n z^n\left(1+\frac{a_{n-1}}{a_n}\frac{1}{z}+\cdots+\frac{a_0}{a_n}\frac{1}{z^n}\right).$$

The bracket tends to $1$ as $|z|\to\infty$, so there exists $R>0$ and $c>0$ such that for $|z|\ge R$,

$$|p(z)|\ge c|z|^n \to \infty.$$

On the closed disk $B_R=\set{z}{|z| \le R}$, the continuous function $z\mapsto |p(z)|$ attains a minimum $m$ by compactness. For $|z|>R$, $|p(z)|>|p(0)|\ge m$ by taking $R$ large enough, hence this minimum on $B_R$ is also the global minimum on $\C$. Choose $z_0$ with $|p(z_0)|=m$.

If $m>0$, then $p(z)\neq 0$ for all $z$, hence

$$f(z)=\frac{1}{p(z)}$$

is entire. Also $|p(z)|\ge m$ for all $z$, so $|f(z)|\le 1/m$: $f$ is bounded. By Liouville’s theorem, a bounded entire function is constant, so $f$ is constant, hence $p$ is constant - contradiction. Therefore $m=0$, i.e. $|p(z_0)|=0$, so $p(z_0)=0$. $\square$