Algebra/Fundamental Theorem of Algebra

For every non-constant polynomial with complex coefficients, there exists at least one complex root.

Furthermore, its degree is also the amount of its roots (with multiplicity).

Proof
Let there be a non-constant polynomial
 * $$p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\quad(a_n\ne0)$$

Then we have $$\lim_{z\to\infty}|p(z)|=\infty$$. Since the function $$|p(z)|$$ is continuous, there exists a $$z_0\in\C$$ such that $$|p(z_0)|=\min_{z\,\in\,\C}|p(z)|$$.

Let us write $$p(z)=p(z_0)+(z-z_0)^mq(z)$$, for $$m\in\N$$ and a polynomial $$q(z)$$ such that $$q(z_0)\ne0$$.

Let $$\overline{p(z_0)}$$ be the complex conjugate of $$p(z_0)$$. Then for all $$z\in\C$$ we get:
 * $$\begin{align}&|p(z_0)|^2\le|p(z)|^2=|p(z_0)|^2+|z-z_0|^{2m}|q(z)|^2+2\,\text{Re}\!\left[\,\overline{p(z_0)}(z-z_0)^mq(z)\,\right]\\[5pt]&|z-z_0|^{2m}|q(z)|^2+2\,\text{Re}\!\left[\,\overline{p(z_0)}(z-z_0)^mq(z)\,\right]\ge0\end{align}$$

Let $$z=z_0+re^{\theta i}$$ for $$r>0$$:
 * $$\begin{align}&r^{2m}\bigl|q(z_0+re^{\theta i})\bigr|^2+2\,\text{Re}\!\left[\,\overline{p(z_0)}\,r^me^{m\theta i}q(z_0+re^{\theta i})\,\right]\ge0\\[5pt]&r^m\bigl|q(z_0+re^{\theta i})\bigr|^2+2\,\text{Re}\!\left[\,\overline{p(z_0)}q(z_0+re^{\theta i})e^{m\theta i}\,\right]\ge0\end{align}$$

Taking the limit as $$r\to0^+$$ yields:
 * $$\text{Re}\!\left[\,\overline{p(z_0)}q(z_0)e^{m\theta i}\,\right]\ge0$$

Let $$c=\overline{p(z_0)}q(z_0)$$ and $$\omega=\text{cis}\!\left(\frac{\pi}{2m}\right)$$.

By plugging $$e^{\theta i}=1,\omega,\omega^2,\omega^3$$ into the inequality and by de Moivre's formula, we get:
 * $$\begin{align}&\text{Re}(\pm c)=\pm\,\text{Re}(c)\ge0\\&\text{Re}(\pm ci)=\mp\,\text{Im}(c)\ge0\end{align}$$

hence $$\text{Re}(c)=\text{Im}(c)=0$$ and so $$\overline{p(z_0)}q(z_0)=0$$.

Therefore, from $$q(z_0)\ne0$$ we get $$\overline{p(z_0)}=p(z_0)=0$$.

$$\blacksquare$$

Reference and Authors

 * McDougal Algebra 2
 * Holt Algebra 2
 * Lial, Hornspy, Schenider Precalculus
 * Alvin Ling (starter)