Roots of complex numbers
Master roots of complex numbers for Year 12 Specialist Mathematics in Queensland (QCAA, Unit 3). Using De Moivre’s theorem, a non-zero complex number has exactly n distinct nth roots that sit equally spaced around a circle in the complex plane.
You will learn to find the roots in polar form, work with the nth roots of unity on the unit circle, and plot each root as the vertex of a regular polygon — a core skill for factorising over the complex numbers and solving polynomial equations later in the course.
Theory
Roots of complex numbers extend De Moivre’s theorem in Year 12 Specialist Mathematics (Unit 3, QCAA Queensland). A non-zero complex number has exactly \(n\) distinct \(n\)th roots, equally spaced around a circle of radius \(r^{1/n}\). This page shows how to find them in polar form, locate them on an Argand diagram, and use the special case of the \(n\)th roots of unity.
An \(n\)th root of a complex number \(w\) is any \(z\) with \(z^n=w\). Over the complex numbers this equation always has solutions, and a non-zero \(w\) has exactly \(n\) distinct \(n\)th roots.
Write \(w\) in polar form \(w=r\operatorname{cis}\theta\), where \(r=|w|\) and \(\theta=\arg w\). Every root then has the same modulus \(r^{1/n}\) (the real \(n\)th root of \(|w|\)), so all \(n\) roots lie on a single circle of radius \(r^{1/n}\) centred at the origin.
The arguments start at \(\dfrac{\theta}{n}\) and step round by \(\dfrac{2\pi}{n}\) each time. The roots are therefore equally spaced and form the vertices of a regular \(n\)-gon.
The \(n\)th roots of unity are the solutions of \(z^n=1\): they are \(\operatorname{cis}\dfrac{2k\pi}{n}\) for \(k=0,1,\dots,n-1\), all on the unit circle with one root at \(z=1\). For \(n\ge 2\) their sum is \(0\).
If \(w=r\operatorname{cis}\theta\) with \(r>0\), the \(n\) \(n\)th roots of \(w\) are:
The special case \(w=1\) gives the \(n\)th roots of unity:
Finding the \(n\)th roots of \(w\)
- Convert to polar form: write \(w=r\operatorname{cis}\theta\), finding \(r=|w|\) and \(\theta=\arg w\).
- Take the modulus: every root has modulus \(r^{1/n}\), the real \(n\)th root of \(|w|\) — do not divide by \(n\).
- Set the arguments: the first is \(\dfrac{\theta}{n}\); add \(\dfrac{2\pi}{n}\) repeatedly for \(k=0,1,\dots,n-1\).
- Write and locate: state each root \(r^{1/n}\operatorname{cis}\!\left(\dfrac{\theta+2k\pi}{n}\right)\), reduce arguments to the principal range, convert to \(a+bi\) if asked, and mark them on the circle.
Write \(w\) in polar form (it is in the second quadrant):
| \(|w|\) | \(=\) | \(\sqrt{(-2)^2+(2\sqrt{3})^2}\) |
| \(=\) | \(\sqrt{4+12}\) | |
| \(=\) | \(4\) | |
| \(\arg w\) | \(=\) | \(\dfrac{2\pi}{3}\) |
| \(w\) | \(=\) | \(4\operatorname{cis}\dfrac{2\pi}{3}\) |
Take square roots \((n=2)\): halve the argument, and the second root is \(\pi\) further round:
| \(r\) | \(=\) | \(4^{1/2}=2\) |
| \(\theta_0\) | \(=\) | \(\dfrac{1}{2}\times\dfrac{2\pi}{3}=\dfrac{\pi}{3}\) |
| \(z_1\) | \(=\) | \(2\operatorname{cis}\dfrac{\pi}{3}\) |
| \(=\) | \(2\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)\) | |
| \(=\) | \(1+\sqrt{3}\,i\) | |
| \(z_2\) | \(=\) | \(2\operatorname{cis}\left(\dfrac{\pi}{3}+\pi\right)\) |
| \(=\) | \(-1-\sqrt{3}\,i\) |
\(\pm(1+\sqrt{3}\,i)\).
Write \(27i\) in polar form:
| \(27i\) | \(=\) | \(27\operatorname{cis}\dfrac{\pi}{2}\) |
Apply \(z_k=r^{1/n}\operatorname{cis}\dfrac{\theta+2k\pi}{n}\) with \(n=3\); the modulus is \(27^{1/3}=3\):
| \(z_1\) | \(=\) | \(3\operatorname{cis}\dfrac{\pi}{6}\) |
| \(z_2\) | \(=\) | \(3\operatorname{cis}\left(\dfrac{\pi}{6}+\dfrac{2\pi}{3}\right)\) |
| \(=\) | \(3\operatorname{cis}\dfrac{5\pi}{6}\) | |
| \(z_3\) | \(=\) | \(3\operatorname{cis}\left(\dfrac{\pi}{6}-\dfrac{2\pi}{3}\right)\) |
| \(=\) | \(3\operatorname{cis}\left(-\dfrac{\pi}{2}\right)\) |
\(3\operatorname{cis}\dfrac{\pi}{6},\ 3\operatorname{cis}\dfrac{5\pi}{6},\ 3\operatorname{cis}\left(-\dfrac{\pi}{2}\right)\).
Write \(1=\operatorname{cis}0\); the roots are \(\operatorname{cis}\dfrac{2k\pi}{5}\), \(k=0,1,2,3,4\):
| \(z_1\) | \(=\) | \(\operatorname{cis}0=1\) |
| \(z_2\) | \(=\) | \(\operatorname{cis}\dfrac{2\pi}{5}\) |
| \(z_3\) | \(=\) | \(\operatorname{cis}\dfrac{4\pi}{5}\) |
| \(z_4\) | \(=\) | \(\operatorname{cis}\left(-\dfrac{4\pi}{5}\right)\) |
| \(z_5\) | \(=\) | \(\operatorname{cis}\left(-\dfrac{2\pi}{5}\right)\) |
The roots are the vertices of a regular pentagon centred at the origin, so they cancel in a ring:
| \(\text{sum}\) | \(=\) | \(0\) |
Five roots \(\operatorname{cis}\dfrac{2k\pi}{5}\); their sum is \(0\).
Write \(-81\) in polar form:
| \(-81\) | \(=\) | \(81\operatorname{cis}\pi\) |
Fourth roots \((n=4)\): modulus \(81^{1/4}=3\), arguments \(\dfrac{\pi+2k\pi}{4}\):
| \(z_1\) | \(=\) | \(3\operatorname{cis}\dfrac{\pi}{4}=\dfrac{3\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}i\) |
| \(z_2\) | \(=\) | \(3\operatorname{cis}\dfrac{3\pi}{4}=-\dfrac{3\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}i\) |
| \(z_3\) | \(=\) | \(3\operatorname{cis}\left(-\dfrac{3\pi}{4}\right)=-\dfrac{3\sqrt{2}}{2}-\dfrac{3\sqrt{2}}{2}i\) |
| \(z_4\) | \(=\) | \(3\operatorname{cis}\left(-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{2}-\dfrac{3\sqrt{2}}{2}i\) |
\(\pm\dfrac{3\sqrt{2}}{2}\pm\dfrac{3\sqrt{2}}{2}i\) (all four sign combinations).
Common pitfalls
Frequently asked questions
How many nth roots does a complex number have?
A non-zero complex number has exactly \(n\) distinct \(n\)th roots. They are equally spaced by \(\dfrac{2\pi}{n}\) on a circle of radius \(r^{1/n}\), forming a regular \(n\)-gon.
What is the formula for the nth roots of a complex number?
If \(w=r\operatorname{cis}\theta\), the roots are \(z_k=r^{1/n}\operatorname{cis}\dfrac{\theta+2k\pi}{n}\) for \(k=0,1,\dots,n-1\).
What are the nth roots of unity?
They are the solutions of \(z^n=1\), namely \(\operatorname{cis}\dfrac{2k\pi}{n}\) for \(k=0,\dots,n-1\). They lie on the unit circle with one root at \(z=1\).
Why do the nth roots of unity add up to zero?
They are the vertices of a regular \(n\)-gon centred at the origin, so their position vectors cancel; for \(n\ge 2\) the sum is \(0\).
How do you find the modulus of each root?
Take the real \(n\)th root of the modulus of \(w\): each root has modulus \(|w|^{1/n}\). Do not divide the modulus by \(n\).
How do you locate the roots on an Argand diagram?
Draw the circle of radius \(r^{1/n}\), mark the first root at argument \(\dfrac{\theta}{n}\), then step round by \(\dfrac{2\pi}{n}\) to place the rest evenly.