Complex Root of Unity Generator

Math Algebra • Complex Numbers

Written by STEM Calculators Team Published December 20, 2025 Updated February 24, 2026

8. Complex Root Of Unity Generator

Generate the \(n\)-th roots of unity (solutions of \(z^n=1\)), list them in polar/rectangular form, highlight primitive roots, and visualize them on the unit circle.

Inputs

Order \(n\)

Integer \(n \ge 1\). For large \(n\), labels are thinned to avoid clutter.

Angle unit

Polar notation

In the table/output, the polar form is still rendered as proper LaTeX.

Show only primitive roots Highlight primitive roots Connect roots in order (regular \(n\)-gon) Show labels \(k\)

Cyclic group walk (generator)

Show cycle edges using a primitive step

If \(k\) is coprime to \(n\), then \(\omega^k\) is primitive and its powers visit all roots: \(1,\omega^k,\omega^{2k},\dots\).

Ready

Unit circle plot

Drag to pan • wheel/pinch to zoom • hover a point to see its value • Auto fit keeps the unit circle visible

x: 0, y: 0, zoom: 1

All roots lie on \(|z|=1\). The root for index \(k\) is \(z_k=e^{i 2\pi k/n}\) for \(k=0..n-1\).

Results

Click Generate.

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Theory: 8. Complex Root Of Unity Generator

The \(n\)-th roots of unity are the complex solutions of \[ z^n=1. \] Geometrically, they are points equally spaced on the unit circle \(|z|=1\).

1) Formula for the roots

Write \(1=e^{i2\pi m}\) for any integer \(m\). Then \[ z^n = e^{i2\pi m} \quad\Rightarrow\quad z = e^{i\frac{2\pi m}{n}}. \] Taking \(m=k\) with \(k=0,1,\dots,n-1\) gives all distinct roots:

\[ z_k = e^{i\frac{2\pi k}{n}} = \cos\!\left(\frac{2\pi k}{n}\right) + i\sin\!\left(\frac{2\pi k}{n}\right), \qquad k=0,1,\dots,n-1. \]

2) Primitive roots and the cyclic group idea

A root \(z_k\) is called primitive if it has order \(n\), meaning its powers generate all roots. This happens exactly when \(k\) is coprime to \(n\):

\[ z_k \text{ is primitive } \iff \gcd(k,n)=1. \]

There are \(\varphi(n)\) primitive \(n\)-th roots of unity, where \(\varphi\) is Euler’s totient function. If you pick a primitive root (for example \(\omega=e^{i2\pi/n}\) when \(\gcd(1,n)=1\)), then \[ 1,\ \omega,\ \omega^2,\ \dots,\ \omega^{n-1} \] cycles through every root exactly once.

3) Sum of all roots

For \(n>1\), the roots form a geometric progression with ratio \(\omega=e^{i2\pi/n}\neq 1\). Therefore:

\[ \sum_{k=0}^{n-1}\omega^k = \frac{\omega^n-1}{\omega-1}=\frac{1-1}{\omega-1}=0. \]

This explains why the roots are “balanced” around the origin (the centroid is at \(0\)).

Solved example: 4th roots of unity

Let \(n=4\). Then \[ z_k=e^{i\frac{2\pi k}{4}}=e^{i\frac{\pi k}{2}}. \] Evaluating for \(k=0,1,2,3\) gives:

\[ z_0=1,\quad z_1=i,\quad z_2=-1,\quad z_3=-i. \]

Cyclotomic link: the primitive roots are exactly the zeros of \(\Phi_n(x)\), the \(n\)-th cyclotomic polynomial.

Frequently Asked Questions

What are the n-th roots of unity?

They are the complex numbers that satisfy z^n = 1. They lie equally spaced on the unit circle and can be written as z_k = e^(i 2pi k / n) for k = 0, 1, ..., n-1.

How do I know which roots of unity are primitive?

A root z_k is primitive exactly when gcd(k, n) = 1. Primitive roots have order n, meaning their powers generate all n roots.

Why do the roots of unity add up to zero?

For n > 1, the roots form a geometric series with ratio omega = e^(i 2pi / n) not equal to 1. The sum 1 + omega + ... + omega^(n-1) equals (omega^n - 1) / (omega - 1) = 0.

Should I use radians or degrees for the arguments of the roots?

Either unit works; the roots are determined by angles 2pi k / n in radians or 360 k / n in degrees. The unit choice only changes how the angle is displayed in the output.

What does connecting the roots in order show?

Connecting consecutive roots draws a regular n-gon inscribed in the unit circle. It helps visualize the equal spacing and symmetry of the roots around the origin.

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Frequently Asked Questions

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