Fundamental theorem of algebra by Liouville's theorem

Suppose, for contradiction, that $p$ has no complex root.

The denominator $p$ is then nonzero throughout $\mathbb{C}$, so [L1] makes $g:=1\mathrm{/}p$ entire.

By [L2], choose $R\ge 1$ such that $\mathrm{\mid }p(z)\mathrm{\mid }\ge 1$ whenever $\mathrm{\mid }z\mathrm{\mid }\ge R$; then step 2.1 and [L4] give $\mathrm{\mid }g(z)\mathrm{\mid }=1\mathrm{/}\mathrm{\mid }p(z)\mathrm{\mid }\le 1$ on that exterior region.

By step 2.1 and [L3], $g$ is continuous; the inequality $\mathrm{\mid }\mathrm{\mid }u\mathrm{\mid }-\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$ derived from [L4] makes the real-valued function $\mathrm{\mid }g\mathrm{\mid }$ continuous.

The closed disc $K=\{z:\mathrm{\mid }z\mathrm{\mid }\le R\}$ contains $0$, is bounded, and is closed because $\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }-\mathrm{\mid }w\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }z-w\mathrm{\mid }$; by [L5] it is a nonempty closed bounded subset of ${\mathbb{R}}^2$, so [L6] makes it compact.

Applying [L7] to the continuous function $\mathrm{\mid }g\mathrm{\mid }$ from step 3.2 on the compact set from step 4.1 gives a finite maximum $M\ge 0$ with $\mathrm{\mid }g(z)\mathrm{\mid }\le M$ for $z\in K$.

If $\mathrm{\mid }z\mathrm{\mid }\le R$, step 5.1 gives $\mathrm{\mid }g(z)\mathrm{\mid }\le M$, while if $\mathrm{\mid }z\mathrm{\mid }\ge R$, step 3.1 gives $\mathrm{\mid }g(z)\mathrm{\mid }\le 1$; hence $\mathrm{\mid }g(z)\mathrm{\mid }\le \max \{M,1\}$ on the whole plane, with the boundary $\mathrm{\mid }z\mathrm{\mid }=R$ covered by both estimates.

The function $g$ is entire by step 2.1 and bounded by step 6.1, so [L8] makes it constant.

The constant value of $g=1\mathrm{/}p$ is nonzero by [L4], so $p=1\mathrm{/}g$ is constant, contradicting the given nonconstancy; the assumption of step 1.1 is false, and $p$ has a complex root.