The proof via the minimum of the modulus

Why this theorem is not algebraic

A surprise first of all: no proof of the fundamental theorem of algebra is purely algebraic. All of them use, at some point, an analytic property of $\mathbb{R}$ — the least upper bound, compactness, or the intermediate value theorem.

This is unavoidable: the construction of $\mathbb{R}$ from $\mathbb{Q}$ is itself analytic, and without completeness the analogous statement fails: $\mathbb{Q}(i)$, built on $\mathbb{Q}$, is not algebraically closed ($X^2 - 2$ has no root in it).

Behavior at infinity

Let $P(z) = a_n z^n + \cdots + a_0$ with $a_n \neq 0$ and $n \geq 1$. Factoring out the leading term:

$$P(z) = a_n z^n \left(1 + \frac{a_{n-1}}{a_n z} + \cdots + \frac{a_0}{a_n z^n}\right)$$

The parenthesis tends to $1$ as $\lvert z \rvert \to +\infty$, so

$$\lvert P(z) \rvert \xrightarrow[\lvert z \rvert \to +\infty]{} +\infty$$

There is thus a radius $R$ such that $\lvert P(z) \rvert > \lvert P(0) \rvert$ as soon as $\lvert z \rvert > R$.

The minimum of the modulus is attained

The closed disk $D = \left\{\, z \in \mathbb{C} \;:\; \lvert z \rvert \leq R \,\right\}$ is closed and bounded. The map $z \mapsto \lvert P(z) \rvert$ is continuous on it, so, by the extreme value theorem — itself a consequence of Bolzano–Weierstrass — it attains its minimum on it at a point $z_0$.

This minimum on $D$ is in fact a minimum on the whole of $\mathbb{C}$: outside the disk, $\lvert P \rvert$ exceeds $\lvert P(0) \rvert$, which is already greater than or equal to the minimum on $D$.

d'Alembert's lemma

If $P(z_0) \neq 0$, then every neighborhood of $z_0$ contains a point $z$ such that $\lvert P(z) \rvert < \lvert P(z_0) \rvert$.

Idea of the proof. Near $z_0$, we write $P$ in the form

$$P(z_0 + h) = P(z_0) + c_m\,h^m + \left(\text{terms of degree} > m\right)$$

where $c_m$ is the first nonzero coefficient after the constant term. For small $\lvert h \rvert$, the term $c_m h^m$ dominates everything else. It then suffices to choose the argument of $h$ so that $c_m h^m$ points in the direction opposite to $P(z_0)$, that is

$$\arg(h) = \frac{\arg\!\left(P(z_0)\right) - \arg\left(c_m\right) + \pi}{m}$$

which is always possible in $\mathbb{C}$, since every nonzero complex number has $m$ $m$-th roots there. The modulus of $P$ then strictly decreases.

This is exactly where $\mathbb{C}$ makes the difference: in $\mathbb{R}$, we do not have this freedom over the argument, and the argument collapses.

Concluding the proof

The modulus $\lvert P \rvert$ attains its minimum at $z_0$. If we had $P(z_0) \neq 0$, d'Alembert's lemma would provide a nearby point where the modulus is strictly smaller, contradicting minimality. Therefore

$$P(z_0) = 0$$

and the polynomial does have a root. This is the so-called d'Alembert–Argand proof, the most elementary of the many known proofs.

Classic pitfall

Believing that the minimum of the modulus is attained because "$\mathbb{C}$ is complete". Completeness is not enough: what is at stake is the compactness of the closed bounded disk, exactly as in the extreme value theorem on a closed interval. On an open disk, the minimum might very well never be attained.