Tag Archives: cube roots of unity

June 4, 2026 · 11:33 am

Complex Numbers – Roots of Unity

$\begin{equation*}z^n=1, z\in\mathbb{C}\end{equation}$

In polar form $1=cis (2\pi)$

Hence,

$\begin{equation*}z^n=cis(2\pi k), k\in\mathbb{Z}\end{equation}$

$\begin{equation*}z^n=cis(\frac{2\pik}{n})\end{equation}$

For example,

$\begin{equation*}z^4=1\end{equation}$

$\begin{equation*}z^4=cis(2\pi k)\end{equation}$

$\begin{equation*}z_k=cis(\frac{2\pi k}{4})=cis(\frac{\pi k}{2})\end{equation}$

$\begin{equation*}z_0=1\end{equation}$

$\begin{equation*}z_1=cis(\frac{\pi}{2})=i\end{equation}$

$\begin{equation*}z_2=cis(\pi)=-1\end{equation}$

$\begin{equation*}z_3=cis(\frac{3\pi}{2})=-i\end{equation}$

The roots of unity are spread evenly $(\frac{2\pi}{n}$ or $\frac{360}{n}$ apart) around a circle of radius $1$ .

Properties of the roots of Unity

All the roots of unity can be generated by powers of a single root $\omega=cis(\frac{2\pi}{n})$ . The roots form the sequence $1, \omega, \omega^2, \omega^3, ..., \omega^{n-1}$ Note: $\omega \ne 1$

For example,

$\begin{equation*}z^4=1\end{equation}$

$\begin{equation*}\omega_1=cis(\frac{\pi}{2})\end{equation}$

$\begin{equation*}\omega_1^2=cis(\pi)=-1=\omega_2\end{equation}$

$\begin{equation*}\omega_1^3=cis(\frac{3\pi}{2})=-i=\omega_3\end{equation}$

$\begin{equation*}\omega_1^4=cis(2\pi)=1=\omega_0\end{equation}$

A root that can generate the remaining roots is a primitive root.

$\omega^k$ is a primitive root if $n$ and $k$ are coprime.

The sum of the $n^{th}$ roots of unity is always zero.

$\begin{equation*}\Sigma_{k=0}^{n-1}1+\omega+\omega^2+...+\omega^{n-1}=0\end{equation}$

The product of the $n^{th}$ roots of unity is

$\begin{equation*}\Pi_{k=0}^{n-1}\omega^k=(-1)^{n-1}\end{equation}$

It is $1$ when $n$ is odd and $-1$ when $n$ is even.

Example WATP 2024 Question 7a

(a) Evaluate $(4\omega^2+3)(4\omega+3)$ where $\omega$ is a complex root of unity, $\omega\ne 1$ .

$\begin{equation*}(4\omega^2+3)(4\omega+3)=16\omega^3+12\omega^2+12\omega+9\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16(1)+9(\omega^2+\omega+1)+3\omega^2+3\omega\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+9(0)+3(\omega^2+\omega)\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+3(-1)\end{equation}$

$\begin{equation*}=13\end{equation}$

Remember the sum of the roots is zero, $\omega^2+\omega+1=0$ hence $\omega^2+\omega=-1$

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Filed under Algebra, Complex Numbers, Roots of Unity, Year 12 Specialist Mathematics

Tagged as complex numbers, complex roots of unity, cube roots of unity