nth Roots of Unity

To find the nth roots of a complex number \(z\), we look for all complex numbers \(w\) such that:

\(w^n = z\).

If \(z\) is written in polar form as:

\(z = r(\cos \theta + i \sin \theta)\),

then its nth roots are obtained by taking the nth root of the modulus and dividing the argument by n.

Roots of unity are special nth roots of the number 1.

We want all complex numbers \(w\) satisfying:

\(w^n = 1\).

Since \(1 = 1(\cos 2\pi k + i \sin 2\pi k)\), for any integer \(k\), the nth roots are:

General formula:

\(w_k = \cos \left(\frac{2\pi k}{n}\right) + i \sin \left(\frac{2\pi k}{n}\right), \,\ k = 0, 1, 2, ..., n-1\).

These give exactly n distinct roots.

All nth roots of unity lie on the unit circle because their modulus is 1.

They are equally spaced and divide the circle into n equal arcs.

So each root corresponds to a point that is rotated by:

\(\frac{2\pi}{n}\) radians from the previous one.

This creates a beautiful regular n-sided polygon when points are joined in order.

For cube roots of unity, take \(n = 3\).

The general formula gives:

• \(w_0 = \cos 0 + i \sin 0 = 1\)
• \(w_1 = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\)
• \(w_2 = \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\)

Let:

\(\omega = w_1\),   \(\omega^2 = w_2\).

Important properties:

• \(1 + \omega + \omega^2 = 0\)
• \(\omega^3 = 1\)
• \(\omega \neq 1\),   \(\omega^2 \neq 1\)

For any \(n\), the roots:

\(w_k = \cos \left(\frac{2\pi k}{n}\right) + i \sin \left(\frac{2\pi k}{n}\right)\)

form a regular n-gon on the Argand plane.

Examples:

• 4th roots of unity form a square.
• 5th roots form a regular pentagon.
• 6th roots form a regular hexagon.

The nth roots of unity have beautiful algebraic structures.