# Complex Numbers in Solving Cubic Equations

Complex numbers arise naturally from the solving of cubic equations.

Consider, for example, the cubic equation $x^3+ px + q = 0$ (incidentally, all cubic polynomial equations can be rewritten in this form). Solving for $x$ in terms of $p$ and $q$ is possible. The solution is given by:

\begin{aligned} x &= \sqrt{A} + \sqrt{B} \\ A &= -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \\ B &= -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \end{aligned}

(Source: Lectures on Elementary Mathematics by Lagrange. Highly recommended.)

This is the cubic analogue of the famous and familiar "quadratic formula" for calculating roots of quadratic polynomials.

The above equation, however, raises two concerns:

1. All cubic polynomials have a real root (provided the coefficients are real, at least). However, it is clear that if $A$ and $B$ will have imaginary components if $\frac{q^2}{4} + \frac{p^3}{27}$.
2. Cubic polynomials may have up to three roots. But this multiplicity isn't obviously.

We first recall Euler's formula:

$e^{i \theta} = \cos (\theta) + \sin(\theta) i.$

So we can write $A$ and $B$ as

\begin{aligned} A &= r e^{i \theta} \\ B &= r e^{-i \theta}. \end{aligned}

for some unique $r$ and unique $\theta$ (unique modulo $2\pi$).

In the complex field, roots are considered to be multi-valued. Unlike with real roots, we cannot pick the "positive" complex root of $x^n = C$ as the value of $\sqrt[n]{C}$.

$\sqrt{A}=\sqrt{r} \cdot e^{\frac{\theta + 2k\pi}{3}},$

where $k=0,1,2$ give unique roots.

Created 2020-04-28. Last updated 2020-06-22. View source.