Fundamental theorem of algebra

The fundamental theorem of algebra states: every non-constant polynomial $p(z)$ with complex coefficients has at least one root in $\mathbb{C}$.

By repeated factoring (peel off $z-z_1$ where $z_1$ is a root, then factor again), this strengthens to: a polynomial of degree $n$ has exactly $n$ complex roots counted with multiplicity. This is the statement that $\mathbb{C}$ is algebraically closed.

What’s “fundamental” about it

In $\mathbb{R}$, the polynomial $x^2+1$ has no roots — you have to step outside the reals. In $\mathbb{C}$, this never happens: every polynomial factors completely into linear factors

$p(z)=a_n(z-z_1)(z-z_2)\cdots (z-z_n).$

The theorem is why complex numbers are the natural place for polynomial algebra. Going to $\mathbb{C}$ is enough; you never need a “more imaginary” extension.

Proof via Liouville’s theorem

The cleanest proof uses complex analysis. Suppose $p(z)$ has no roots in $\mathbb{C}$. Then $1/p(z)$ is entire (analytic on all of $\mathbb{C}$). As $|z|\to \infty$, $|p(z)|\to \infty$, so $1/p\to 0$, and in particular $|1/p|$ is bounded on $|z|>R$ for some $R$. On $|z|\le R$, $1/p$ is continuous on a compact set, hence bounded. So $1/p$ is bounded everywhere.

By Liouville’s theorem, any bounded entire function is constant. So $1/p$ is constant, hence $p$ is constant — contradiction with “non-constant.” So $p$ must have a root.

This proof is striking because it uses an analytic property (bounded entire $\Rightarrow$ constant) to settle a purely algebraic question.

Locating roots

The theorem guarantees existence but not a formula. For degrees 1–4, closed-form solutions exist (linear, quadratic, cubic Cardano, quartic Ferrari). For degree $\ge 5$, no general radical formula exists (Abel–Ruffini). Numerical root-finding takes over for the high-degree case.

Where it shows up in EE

The denominator of a transfer function is a polynomial in $s$. Its roots — the poles of the system — are guaranteed to exist as $n$ complex numbers in $\mathbb{C}$. Same for the numerator and its zeros. Pole-zero diagrams in the $s$-plane rest implicitly on the fundamental theorem: the polynomial has this many poles and zeros, and they live somewhere in $\mathbb{C}$.

A real-coefficient polynomial has complex roots in conjugate pairs (because if $p(z_0)=0$ then $\overline{p(z_0)}=p({\bar{z}}_0)=0$). So a real polynomial of odd degree has at least one real root; an even-degree real polynomial may have all complex roots, but they pair up as conjugates. See Complex conjugate eigenvalues case for the analogous fact in linear ODEs.

Connection to roots of unity

The polynomial $z^n-1$ has the $n$-th roots of unity as its roots — explicit and beautiful. The fundamental theorem says no polynomial does better than this.