13 - Fundamental Theorem of Algebra

Fundamental Theorem of Algebra

Suppose that

$$p(z)=\sum _{k=0}^n{a}_k{z}^k$$

is a non-constant polynomial where $a_k\in \mathbb{C}$ and $a_n\ne 0$ then

$$\exists z_0\in \mathbb{C}\text{ such that }p(z_0)=0$$

Proof

By rescaling $p$ then assume that $a_n=1$

If $p(z)\ne 0$ for all $z\in \mathbb{C}$ then

$$f(z)=\frac{1}{p(z)}\text{ is an entire function }$$

since $p$ is entire

Claim that $f$ is bounded
Since $f$ is continuous then it is bounded on any disc $\overline{B}(0,R)$ so need to show

$$|f(z)|\to 0\text{ as }z\to \infty$$

In other words

$$|p(z)|\to \infty \text{ as }z\to \infty$$

However

$$|p(z)|=\left|z^n+\sum _{k=0}^{n-1}{z}_k{z}^k\right|=|z^n|\left(\left|1+\sum _{k=0}^{n-1}\frac{a_k}{z^{n-k}}\right|\right)\ge |z^n|\left(1-\sum _{k=0}^{n-1}\frac{|a_k|}{|z{|}^{n-k}}\right)$$

As $\frac{1}{|z{|}^m}\to 0$ as $|z|\to \infty$ then for sufficiently large $|z|$ so for $|z|>R$

$$1-\sum _{k=0}^{n-1}\frac{|a_k|}{|z{|}^{n-k}}\ge \frac{1}{2}$$

So for $|z|\ge R$ then $|p(z)|\ge \frac{1}{2}|z{|}^n$ so as $|z{|}^n\to \infty$ as $|z|\to \infty$ then $|p(z)|\to \infty$

So $f(z)$ is bounded and hence constant so $p(z)$ is constant which is contradiction