TheoremProved

Fundamental theorem of algebra

Statement

Every non-constant single-variable polynomial with complex coefficients has at least one complex root; equivalently, a polynomial of degree $n \ge 1$ has exactly $n$ complex roots counted with multiplicity.

Why is it true?

Real polynomials like $x^2+1$ can fail to touch the x-axis, so they seem to have no roots. Once you allow the coefficients and the root to live in the complex plane instead of just the real line, every polynomial — no matter how wild — is guaranteed to cross zero somewhere. Adding the imaginary direction closes a gap that the real numbers leave open.

Proof sketch

One standard route (d'Alembert–Argand / Gauss) uses Liouville's theorem: if a polynomial $p(z)$ had no root, $1/p(z)$ would be a bounded entire function, hence constant by Liouville, contradicting that $p$ is non-constant. A more elementary route shows $|p(z)|$ attains a global minimum on $\mathbb{C}$ (since $|p(z)|\to\infty$ as $|z|\to\infty$ ) and that this minimum must be zero, otherwise a suitable perturbation would make $|p|$ smaller.

Proved by

proved 1799Carl Friedrich Gauss

Topics that use this theorem

Complex numbers

Related theorems

Euler's formula Abel–Ruffini theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Benjamin Fine, Gerhard Rosenberger (1997). The Fundamental Theorem of Algebra