## Fundamental theorem of algebra

### Theorem

Let \begin{equation} P=c_0z^n+c_1z^{n-1}+\ldots+c_{n-1}z+c_n \label{eq-1} \end{equation} where $n\ge 1$, $c_j\in \mathbb{C}$ and $c_0\ne 0$.

Then there exists $a\in \mathbb{C}$ such that $P(a)=0$.

#### Proof

##### Step 1

Let
$$R=\displaystyle{\max_{1\le k\le n} \bigl|\frac{c_k}{c_0}\bigr|^{1/k}}.$$ Assume $|z|\ge 2R$; then $|P(z)|\ge |c_n|=|P(0)|$. Indeed, using $$\bigl(\frac{|z|}{2}\bigr)^k|c_0|\ge R^k|c_0|\ge |c_k|$$ for $k=1,2,\ldots,n$ we conclude that \begin{multline*} |\sum_{k=1}^n c_k z^{n-k}|\le \sum_{k=1}^n |c_k|\cdot |z|^{n-k} \le \\ \sum_{k=1}^n |z|^n \bigl(\frac{1}{2}\bigr)^k|c_0|= |c_0|\bigl(1-\bigl(\frac{1}{2}\bigr)^n\bigr)|z|^n. \end{multline*} Therefore \begin{multline*} |P(z)|=|c_0z^n +\sum_{k=1}^n c_k z^{n-k}|\ge \\ |c_0|\cdot|z|^n-|\sum_{k=1}^n c_k z^{n-k}|\ge \bigl(\frac{1}{2}\bigr)^n|c_0|\ge |c_n|=|P(0)| \end{multline*} as required.

##### Step 2.

Since $P(x+yi)=A(x,y)+iB(x,y)$ where $A(x,y)$ and $B(x,y)$ are polynomials in two variables with real coefficients, it implies $|P(x+yi)|=\sqrt{|A(x,y)|^2+|B(x,y)|^2}$. Therefore map $\mathbb{C}\ni z\to |P(z)|\in \mathbb{R}^+$ is continuous and hence (Calculus II) attains a minimal value on any closed bounded set (such sets are called compacts), in particular on $D=\{z\in \mathbb{C}: |z|\le 2R\}$.

Due to Step 1 such value is also a minimal value on all complex plane $\mathbb{C}$. Thus, there is $z_0\in \mathbb{C}$ with $|z_0|\le 2R$ such that $|P(z_0)|=\min_{z\in \mathbb{C}}|P(z)|$.

#### Step 3

We may assume that $z_0=0$ by considering a polynomial $Q(z)=P(z+z_0)$ instead of $P(z)$.

#### Lemma

Given a local minima at $0$ of $z\to |P(z)|$ we claim that $P(0)=0$ (which suffices).

##### Proof

Assume that $P(0)=c_n\ne 0$. Let us rewrite $P(z)$ in inverse order and skip terms with coefficients equal $0$: $P(z)=c_n+c_{n-k} z^k+\ldots$ where $c_n\ne 0$, $c_{n-k}\ne 0$, and $k\ge 1$; recall that $c_0\ne 0$. Then $$P(z)=c_n+c_{n-k}z^k +c_{n-k}z^k Q(z)$$ with $$Q(z)=\Bigl(\frac{c_{n-k+1}}{c_{n-k}} + \frac{c_{n-k+2}}{c_{n-k}}z^2+\ldots \frac{c_0}{c_{n-k}}z^{n-k}\Bigr).$$ By Moivre formula there exists $z_*\in \mathbb{C}$ such that $c_n+c_{n-k}z_*^k=0$. Also there is $\delta_0\in (0,1)$ such that if $|z|< \delta_0|z_*|$ then $|Q(z)|\le \frac{1}{4}$.

Let $z_\delta =\delta \cdot z_*$ and $0<\delta\le \delta_0$. Then $$|P(z_\delta)|=|c_n +\delta^k c_{n-k}z_*^k+ \delta^k c_{n-k}z_*^k Q(\delta z_*)|\le |c_n-\delta^kc_k|+\frac{1}{4}\delta^k|c_n|$$ where we use $c_{n-k} z_*^k = - c_n$.

Hence $$|P(z_\delta)|\le |c_n|(1-\delta^k)+\frac{1}{4}\delta^k|c_n|=|c_n|(1-\frac{3}{4}\delta^k)<|c_n|=|P(0)|.$$ This contradicts to the assumption that $z_0=0$ is a minima. Done: Lemma has been proven and Theorem follows from it.