Solving roots in binomial complex equations

In this post I will describe how to solve binomial complex equations of the form \[ z^n = a + bi. \]

This is done by first putting the right side on polar form using the following: \[ \begin{aligned} R &= \sqrt{a^2 + b^2} \\ \theta &= \left\{ \begin{aligned} &\tan^{-1}\frac{b}{a} &\text{if}\; a > 0 \\ &\tan^{-1}\frac{b}{a} + \pi &\text{if}\; a < 0 \\ &\frac{\pi}{2} &\text{if}\; a = 0, b > 0 \\ &-\frac{\pi}{2} &\text{if}\; a = 0, b < 0 \\ &\text{random} &\text{if}\; a = 0, b = 0 \end{aligned} \right. . \end{aligned} \]

The roots are now found using the following: \[ z = \left( \sqrt[n]{R} \right)_{\frac{a}{n} + p\frac{2\pi}{n}} \qquad p = 0, 1, \ldots, n-1 \] where p is altered from $0$ to $n-1$, to find all roots.

Worked example

Given: \[ z^4 = 3+4i \]

Finding $R$: \[ \begin{aligned} R &= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \\ R &= 5 \end{aligned} \]

Finding $\theta$ (note: $a = 3 \Leftrightarrow a > 0$): \[ \theta = \tan^{-1}\frac{4}{3} \]

The roots are therefore: \[ \begin{aligned} z_0 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4}} \approx 1.495_{0.75}\\ z_1 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 1\frac{2\pi}{4}} \approx 1.495_{2.32}\\ z_2 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 2\frac{2\pi}{4}} \approx 1.495_{3.89}\\ z_4 &= \left( \sqrt[4]{5} \right)_{\frac{3}{4} + 3\frac{2\pi}{4}} \approx 1.495_{5.46} \end{aligned} \] in polar coordinates. To reverse to regular coordinates, just use the following: \[ \begin{aligned} a &= R\cos{\theta} \\ b &= R\sin{\theta} \end{aligned} \]

Looking at the complex plane the roots are distributed as follows: