Fundamental theorem of algebra (complex analysis)

Statement

Every nonconstant polynomial $p\in\mathbb C[z]$ has a zero in $\mathbb C$ . Consequently, a degree- $n$ polynomial factors as

p(z)=c\prod_{j=1}^{n}(z-\alpha_j)

for complex numbers $\alpha_j$ , counted with multiplicity.

Complex-analytic proof

If $p$ had no zero, then $1/p$ would be entire. Since $|p(z)|\to\infty$ as $|z|\to\infty$ , the reciprocal is bounded outside a large disc; continuity bounds it on the disc. The Liouville theorem would make $1/p$ , and hence $p$ , constant, a contradiction.

Scope

The theorem says that $\mathbb C$ is algebraically closed. The proof recorded here is analytic; algebraic and topological proofs establish the same statement by different methods. The theorem does not say that polynomial roots can always be expressed by radicals.

References

John B. Conway, Functions of One Complex Variable I, 2nd ed., Springer, 1978. Publisher record. Relevant: Chapter III, §4.