Complex Polynomial Equations

Theorem: Complex Conjugate Root Theorem

If $z\in \mathbb{C}$ is a root of the complex polynomial equation

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

and $a_k$ are all real numbers, then the complex conjugate $\bar{z}$ is also a root of the equation.

PROOF

TODO

Theorem: Cauchy's Bound

If $r\in \mathbb{C}$ is a root of the complex polynomial equation

$$\sum _{k=0}^n{a}_k{z}^k=0,$$

then

$$|r|\le 1+\max \left\{\left|\frac{a_{n-1}}{a_n}\right|,\left|\frac{a_{n-2}}{a_n}\right|,\dots ,\left|\frac{a_0}{a_n}\right|\right\}$$

PROOF

TODO

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Theorem: Second-Degree Polynomial Equations

The solutions to the complex polynomial equation

$$a{z}^2+bz+c=0$$

are given by

$$z=\frac{-b\pm \left(\sqrt{\frac{\sqrt{{\delta }_x^2+{\delta }_y^2}+{\delta }_x}{2}}+\mathrm{i}\mathrm{sgn}({\delta }_y)\sqrt{\frac{\sqrt{{\delta }_x^2+{\delta }_y^2}-{\delta }_x}{2}}\right)}{2a},$$

where ${\delta }_x=\Re (b^2-4ac)$ and ${\delta }_y=\Im (b^2-4ac)$.

PROOF

TODO

Tip: Complex Roots of Polynomial Equations

If $a,b,c$ are real numbers and $b^2-4ac<0$, then the above formula reduces to

$$x=-\frac{b}{2a}\pm \frac{\mathrm{i}}{2a}\sqrt{-b^2+4ac}$$