Complex Polynomial Equations#

Theorem:

Every complex polynomial equation

\[\sum_{k = 0}^n a_k z^k = 0,\]

where \(\sum_{k = 0}^n a_k z^k\) is not constant, has at least one solution \(z \in \mathbb{C}\).

Proof

TODO

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

\[ |r| \le 1 + \max \left\{\left|\frac{a_{n-1}}{a_n}\right|, \left|\frac{a_{n-2}}{a_n}\right|, \dotsc, \left|\frac{a_{0}}{a_n}\right|\right\} \]

Proof

TODO

Theorem: Second-Degree Polynomial Equations

The solutions to the complex polynomial equation

\[ az^2 + bz + c = 0 \]

are given by

\[ z = \frac{-b \pm \left(\sqrt{\frac{\sqrt{\delta_x^2 + \delta_y^2} + \delta_x}{2}}+ \mathrm{i} \mathop{\operatorname{sgn}}(\delta_y) \sqrt{\frac{\sqrt{\delta_x^2 + \delta_y^2}-\delta_x}{2}}\right)}{2a}, \]

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 \(b^2 - 4ac \lt 0\), then the above formula reduces to

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