Roots of unity

The \(n^{\text{th}}\) roots of unity are the complex numbers that satisfy the equation \(z^n = 1\).

From the Fundamental Theorem of Algebra, a polynomial of degree \(n\) has exactly \(n\) roots in \(\mathbb C\), counted with multiplicity.

For example, if \(a\ne0\), then \(ax^3+bx^2+cx+d=0\) has three roots counted with multiplicity.

Hence, for \(x^3 = 1\), \(x = 1\) is not the only answer when we consider complex numbers. There will be \(2\) other complex numbers to work out.

Re-writing \(x\) as \(z\) to account for complex numbers, \(z^3 = 1\) will have \(3\) roots to be found in the method below.

1. Square Roots of Unity
   \[z^2 = 1 \implies z = \pm 1\]
2. Cube Roots of Unity
   \[z^3 = 1 \implies z^3 = \cos 2k\pi + i\sin 2k\pi \implies z = \cos \frac{2k\pi}{3} + i\sin \frac{2k\pi}{3} \text{ for}\hspace{0.5em} k \in \mathbb{Z} \]

When \(k = 0\), \(z_0 = \cos 0 + i\sin 0 = 1\);

When \(k = 1\), \(z_1 = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\);

When \(k = 2\), \(z_2 = \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\).

Note that we would get the same answers even if we had chosen \(k = -1, 0, 1\) or \(k = 1, 2, 3\).