Complex Analysis

Fundamental Theorem of Algebra

Fundamental Theorem of Algebra

Every nonconstant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$

The core structure of the proof

$$p\text{ has no zeros}⟹\frac{1}{p}\text{ is entire}⟹\frac{1}{p}\text{ is bounded}⟹\frac{1}{p}\text{ is constant}⟹p\text{ is constant}$$

Let $p(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_0,a_n\ne 0,n\ge 1.$ Assume, for contradiction, that $p$ has no zeros in $\mathbb{C}$. Then the function

$$f(z)=\frac{1}{p(z)}$$

is holomorphic on the entire complex plane; that is, it is an entire function.

Next, we show that $f$ is bounded. For sufficiently large $|z|$, the highest-degree term dominates the polynomial. Specifically,

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

The expression in parentheses tends to $a_n\ne 0$ as $|z|\to \infty$. Therefore, there exists $R>0$ such that whenever $|z|\ge R$,

$$\left|a_n+\frac{a_{n-1}}{z}+\cdots +\frac{a_0}{z^n}\right|\ge \frac{|a_n|}{2}.$$

Hence,

$$|p(z)|\ge \frac{|a_n|}{2}|z{|}^n,$$

and therefore

$$\left|\frac{1}{p(z)}\right|\le \frac{2}{|a_n||z{|}^n}.$$

In particular, on the region $|z|\ge R$, the function $1/p(z)$ is bounded, and in fact tends to $0$ as $|z|\to \infty$.

On the closed disk $\{z\in \mathbb{C}:|z|\le R\}$ the function $1/p(z)$ is continuous. Since the closed disk is compact, $1/p(z)$ is bounded there as well.

Therefore, $1/p(z)$ is bounded on all of $\mathbb{C}$. By Liouville’s Theorem, $\frac{1}{p(z)}$ must be constant. Consequently, $p(z)$ is also constant, contradicting the assumption that $\mathrm{deg}p=n\ge 1.$ Thus, $p$ must have at least one zero in $\mathbb{C}$.

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which leads to a contradiction.

The key analytic result used here is

$$\boxed{\text{Every bounded entire function is constant.}}$$

This is precisely Liouville’s Theorem.