The Fundamental Theorem of Algebra

We invented i to give x^2 + 1 = 0 a root. A natural worry: will we have to keep inventing new numbers every time a harder equation comes along? The astonishing answer is no. Once you have the complex numbers, you have all the roots you will ever need. That promise is the fundamental theorem of algebra:

\text{every non-constant polynomial has a complex root.}

Apply it repeatedly and a degree-n polynomial splits into n linear factors over \mathbb{C}. The complex numbers are the end of the line.

What the theorem says

The single existence statement above has powerful consequences once you feed it back through the factor theorem: a root r means a factor (x - r), which you divide out and repeat.

Let p(z) be a polynomial of degree n \ge 1 with complex coefficients. Then:

p has exactly n complex roots, counted with multiplicity;
it factors completely into linear factors over \mathbb{C}: p(z) = a(z - r_1)(z - r_2)\cdots(z - r_n);
\mathbb{C} is algebraically closed — no larger number system is ever needed to solve a polynomial;
if p has real coefficients, its non-real roots occur in conjugate pairs: z a root \implies \bar{z} a root.

The conjugate-pair theorem

The last bullet is one we can prove right now, using only the conjugate rules. Suppose p has real coefficients and z is a root, so p(z) = 0. Write

p(z) = c_n z^n + c_{n-1} z^{n-1} + \cdots + c_1 z + c_0 = 0,

where every coefficient c_k is real.

Step 1 — conjugate both sides. The conjugate of 0 is 0:

\overline{p(z)} = \overline{0} = 0.

Step 2 — push the conjugate inside the sum. Conjugate of a sum is the sum of conjugates:

\overline{c_n z^n} + \overline{c_{n-1} z^{n-1}} + \cdots + \overline{c_1 z} + \overline{c_0} = 0.

Step 3 — push it through each product. Conjugate of a product is the product of conjugates, and \overline{z^k} = \bar{z}^{\,k}:

\overline{c_k z^k} = \bar{c}_k\,\bar{z}^{\,k}.

Step 4 — use that the coefficients are real. A real number is its own conjugate, so \bar{c}_k = c_k, leaving

c_n \bar{z}^{\,n} + c_{n-1}\bar{z}^{\,n-1} + \cdots + c_1 \bar{z} + c_0 = 0.

Step 5 — recognise the result. The left side is exactly p(\bar{z}). So

p(\bar{z}) = 0.

If z is a root of a real polynomial, so is its conjugate \bar{z} — non-real roots come in pairs. A neat corollary: an odd-degree real polynomial has an odd number of roots, and since the non-real ones pair up, at least one root must be real.

The factoring and conjugate-pair consequences are easy given that a root exists — but proving existence in the first place is genuinely deep. Every known proof reaches outside pure algebra. The cleanest uses complex analysis: if a polynomial p had no root, then 1/p would be a bounded function that is complex-differentiable everywhere, and Liouville's theorem forces such a function to be constant — a contradiction, since p is non-constant. So a root must exist. Gauss gave the first rigorous proof in 1799, and several more followed using topology and analysis — but never algebra alone.

n roots, conjugate pairs and all

The polynomial z^n - 1 has real coefficients, so its roots — the roots of unity — must respect the conjugate-pair rule. Move the degree slider and watch: there are always exactly n roots, and every non-real root has its mirror image across the real axis (the dashed links). The real roots (1, and -1 when n is even) sit alone on the horizontal axis.