QCE Specialist Mathematics - Unit 3 - Further complex numbers

Roots of complex numbers

Learn how Specialist Mathematics finds roots of unity and general complex roots using polar form and equal angular spacing.

Part of the free QCE Specialist Mathematics notes library for Unit 3: Further complex numbers.

Updated 2026-05-18 - 4 min read

Finding roots is the reverse of taking powers. If powers multiply arguments, roots divide arguments. The part that catches students is that angles repeat every $2\pi$, so a complex number has multiple roots.

Start with

$ z^n=w. $

Write $w$ in polar form:

$ w=r\operatorname{cis}(\theta). $

Because $\theta$ and $\theta+2k\pi$ describe the same direction, the $n$ roots are:

$ z_k=r^{1/n}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right), $

where $k=0,1,2,\ldots,n-1$.

Deriving the root formula

The formula is not something to memorise blindly. Let:

$ z=\rho\operatorname{cis}\phi $

and:

$ w=r\operatorname{cis}\theta. $

If $z^n=w$, then De Moivre gives:

$ \rho^n\operatorname{cis}(n\phi)=r\operatorname{cis}(\theta+2k\pi). $

Equal complex numbers in polar form have equal moduli and matching directions up to full rotations. Therefore:

$ \rho^n=r,\quad n\phi=\theta+2k\pi. $

So:

$ \rho=r^{1/n},\quad \phi=\frac{\theta+2k\pi}{n}. $

Use $k=0,1,\ldots,n-1$ because those values give one complete lap of distinct roots. The next value, $k=n$, repeats the first angle plus $2\pi$.

Roots of unity

The roots of unity solve:

$ z^n=1. $

Since $1=\operatorname{cis}(0)$, the roots are:

$ z_k=\operatorname{cis}\left(\frac{2k\pi}{n}\right). $

They all sit on the unit circle, equally spaced. For $z^4=1$, the four roots are:

$ 1,\quad i,\quad -1,\quad -i. $

Another useful property is symmetry. The full set of roots is spaced evenly around the circle, so in many small cases the roots visually balance one another. For example, the cube roots of unity form an equilateral triangle. If your sketch has one angle gap that is different from the others, you have probably skipped or repeated a value of $k$.

For $n>2$, most roots of unity are non-real. That does not make them exotic; it only means their points are not sitting on the real axis. The product and sum of all roots can still behave neatly because the roots are spread symmetrically around the circle. For example, the three cube roots of unity are:

$ 1,\quad \operatorname{cis}\left(\frac{2\pi}{3}\right),\quad \operatorname{cis}\left(\frac{4\pi}{3}\right) $

and their vector sum is $0$. This geometric balancing is a useful quick check on sketches.

Worked example: general complex roots

If $w=16\operatorname{cis}\left(\frac{2\pi}{3}\right)$ and you want the fourth roots, then the modulus of each root is:

$ 16^{1/4}=2. $

The arguments are:

$ \frac{\frac{2\pi}{3}+2k\pi}{4} $

for $k=0,1,2,3$. So the roots are:

$ 2\operatorname{cis}\left(\frac{\pi}{6}\right),\quad 2\operatorname{cis}\left(\frac{2\pi}{3}\right),\quad 2\operatorname{cis}\left(\frac{7\pi}{6}\right),\quad 2\operatorname{cis}\left(\frac{5\pi}{3}\right). $

How to present the final answer

Polar form is usually the cleanest final form because the equal spacing is visible. Cartesian form is also acceptable when exact values are simple. For roots of unity, a quick check is that all roots except the repeated endpoint have been listed once. Using $k=0,1,\ldots,n-1$ gives exactly $n$ roots and prevents the final angle from wrapping back to the first root.

For non-unity roots, the same structure applies but the radius changes. If $z^n=r\operatorname{cis}(\theta)$, every root has modulus $r^{1/n}$ and the same angular spacing $\frac{2\pi}{n}$. A sketch should therefore show all roots on one circle, equally spaced. If your roots have different moduli, something has gone wrong.

When angles look awkward, leave them as exact fractions of $\pi$. Convert to Cartesian form only when the trigonometric values are standard, such as $\frac{\pi}{6}$, $\frac{\pi}{4}$ or $\frac{\pi}{3}$. For angles like $\frac{\pi}{10}$, polar form is usually the more readable exact answer.

Principal arguments and choosing $k$

Using $k=0,1,\ldots,n-1$ always gives one full set of $n$ roots, but those roots may not be written with principal arguments. That is normally fine unless the question specifically asks for principal arguments or asks you to plot points in a preferred interval.

For example, if an angle comes out as $\frac{9\pi}{8}$, the same direction can be written as:

$ \frac{9\pi}{8}-2\pi=-\frac{7\pi}{8}. $

Both represent the same root. The second is in $(-\pi,\pi]$, so it is the principal argument form.

When the roots are plotted, the angular gap should still be $\frac{2\pi}{n}$ even after rewriting some arguments into negative form. If the order looks strange in a list, sketch the points around the circle; the geometry is often clearer than the algebra.

Sum of roots of unity

The roots of unity have a useful algebraic balance. If $\omega$ is an $n$th root of unity and $\omega\ne1$, then:

$ 1+\omega+\omega^2+\cdots+\omega^{n-1}=0. $

This is just a finite geometric series:

$ \frac{\omega^n-1}{\omega-1}=0, $

because $\omega^n=1$. Geometrically, the vectors around the regular polygon cancel. This is a good way to remember why the roots are evenly spread rather than clustered on one side of the Argand plane.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Algebra and Trigonometry 2e: Polar Form of Complex Numbers
• OpenStax Precalculus: Polar Form of Complex Numbers

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