How to Find Roots of Unity

Last Updated: December 2, 2021

Complex numbers can be written in the polar form $z=re^{{i\theta }},$ where $r$ is the magnitude of the complex number and $\theta$ is the argument, or phase. It becomes very easy to derive an extension of De Moivre's formula in polar coordinates $z^{{n}}=r^{{n}}e^{{in\theta }}$ using Euler's formula, as exponentials are much easier to work with than trigonometric functions.

We can also extend this to finding roots of the complex number $z.$ Let $\zeta =z^{{1/m}}$ be an mth root of $z.$ Then we can see that $\zeta ^{{m}}=z$ and $\zeta =r^{{1/m}}e^{{i\theta /m}}.$

In this article, we will work with the special case where $\zeta ^{{m}}=1.$ In other words, we are finding numbers that equal 1 when raised to the mth power. These are called roots of unity. Keep reading to learn how to find the roots of unity.

Formula

The formula for finding the mth roots of unity is given below.
- $1^{{1/m}}=e^{{i2\pi k/m}}=\cos {\frac {2\pi k}{m}}+i\sin {\frac {2\pi k}{m}},\ k=0,1,\cdots ,m-1$

Part 1

Part 1 of 2:

Third Roots of Unity

1
Find the third roots of unity. Finding roots of unity means that we find all numbers in the complex plane such that, when raised to the third power, yield 1. When we consider the equation $x^{{3}}-1=0,$ we know that one of the zeroes is 1. But from the fundamental theorem of algebra, we know that every polynomial of degree $n$ has $n$ complex roots. As this is a cubic equation, there are three roots, and two of them are in the complex plane. We can no longer restrict ourselves to dealing with just the real numbers in finding these two remaining roots.
- $z^{{3}}=1$
2
Relate $z$ to its roots.
- We know that a complex number can be written as $z=re^{{i\theta }}.$ But recall from polar coordinates that numbers written in polar form are not uniquely defined. Adding any multiple of $2\pi$ will give the same number as well. Below, the symbols $k\in {\mathbb {Z}}$ mean that $k$ is any integer.
  - $z=re^{{i(\theta +2\pi k)}},\ \ k\in {\mathbb {Z}}$
- Raise $z$ to the one-third power. Since we want to avoid making our function multivalued, we must restrict the domain of the argument to $\theta :[0,2\pi ).$ Therefore, $k=0,1,2.$ In general, the mth roots are found by substituting $k=0,1,\cdots ,m-1.$
  - $z^{{1/3}}=r^{{1/3}}e^{{i\left({\frac {\theta +2\pi k}{3}}\right)}}$
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3
Substitute appropriate values for $r$ and $\theta$ . Since we are finding roots of unity, $r=1$ and $\theta =0.$ In other words, all of the roots lie on the unit circle.
- $1^{{1/3}}=e^{{i2\pi k/3}}=\cos {\frac {2\pi k}{3}}+i\sin {\frac {2\pi k}{3}},\ k=0,1,2$
4
Evaluate. When the roots are plotted on the complex plane, they form an equilateral triangle, where one of the vertices is on the point $z=1.$ Additionally, the complex roots come in conjugate pairs.
- $1^{{1/3}}=1,-{\frac {1}{2}}+{\frac {{\sqrt {3}}}{2}}i,-{\frac {1}{2}}-{\frac {{\sqrt {3}}}{2}}i$
5
Visualize the roots of unity. The plot above is a complex plot of the function $z^{{3}}-1.$ The brightness starts from black and gets brighter as the magnitude increases. The hue starts from red and goes across the color wheel, corresponding to the angle going from $0$ to $2\pi .$ (More precisely, for every $\pi /3,$ the color goes from red, yellow, green, cyan, blue, magenta, to red again.)
- As a starting point in interpreting, we see that on the real axis, the function maps the origin to -1. This is represented on the plot by cyan, as $e^{{i\pi }}=-1,$ and the increasing brightness to the left means that the function is getting smaller and smaller. Meanwhile, the real axis is red for $x>1,$ and gets brighter as well. We can clearly see the zeroes as three black dots which form an equilateral triangle.
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Part 2

Part 2 of 2:

Fifth Roots of Unity

1

Find the fifth roots of unity. As with the third roots, we know that the equation $x^{{5}}-1=0$ has one root, 1, in the reals. Per the fundamental theorem of algebra, there are four other roots, and these roots must be complex.
2
Relate $z$ to its roots.
- $z^{{1/5}}=r^{{1/5}}e^{{i\left({\frac {\theta +2\pi k}{5}}\right)}}$
3
Substitute appropriate values for $r$ and $\theta$ and evaluate. It is fine to leave answers in polar form. As we can see above, the zeros of the function $z^{{5}}-1$ form a regular pentagon, and the complex roots form conjugate pairs, just as with the third roots of unity.
- ${\begin{aligned}1^{{1/5}}&=e^{{i2\pi k/5}},\ k=0,1,2,3,4\\&=1,e^{{i2\pi /5}},e^{{i4\pi /5}},e^{{i6\pi /5}},e^{{i8\pi /5}}\end{aligned}}$
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