Roots of Unity (and Beyond)

In this note, we will explain the mechanics for solving the roots of a complex number. It begins with a simple argument, then many elegant results follow.

We begin with the complex plane, where the vertical axis represents the imaginary numbers and the horizontal axis represents the real numbers (as pictured above). Normally, we define a complex number as z=a+biz = a+bi, where a,ba, b are real numbers. We can also express the number in polar coordinates: r=a2+b2r = \sqrt{{a}^{2}+{b}^{2}} and θ=arctan(ba). \theta = arctan\left( \frac{b}{a} \right).

(Sorry, engineers like to use phi for angles, and j for imaginary numbers)

If we treat aa and bb as the components of a vector with magnitude rr from the origin, we can represent all imaginary numbers in polar form z=r(cosθ+isinθ).z=r\left(cos \theta + isin \theta\right).

By the well-known Taylor expansions of cosxcos x, sinxsin x, and ex{e}^{x}, we can easily show that eiθ=(cosθ+isinθ).{e}^{i\theta} =\left(cos \theta + isin \theta\right).

See proof

By treating this identity from a trigonometric perspective, we discover that ei2πk=1{e}^{i2\pi k} =1; hence, if we solve for zn=1{z}^{n} =1, it follows that rneinθ=e2πik.{r}^{n}{e}^{in\theta} = {e}^{2\pi ik}. This implies that r=1r=1 and θ=2πkn\theta = \frac{2\pi k}{n} where k=0,1,2,...,n1k =0,1,2,...,n-1. Consequently, the roots of unity are cos(2πkn)+isin(2πkn).cos\left(\frac{2\pi k}{n}\right)+isin\left(\frac{2\pi k}{n}\right).

But what do we do to solve the roots of any complex number? Well, we have to make a transformation by finding the nth{n}^{th} root of rr and adding 2πkn\frac{2\pi k}{n} to the angle value for the particular complex number! To visualize this, we look back at the diagram and notice the following:

1)For z=1z = 1, the initial angle θ\theta is zero. However, for z=a+ibz=a+ib we have θ=arctan(ba) \theta = arctan\left( \frac{b}{a} \right).

2)For z=1z = 1, the magnitude is one. However, for z=a+ibz=a+ib we have r=a2+b2r = \sqrt{{a}^{2}+{b}^{2}}.

Hence the roots of a complex number are

r1n[cos(θ+2πkn)+isin(θ+2πkn)]{r}^{\frac{1}{n}}\left[cos\left(\theta +\frac{2\pi k}{n}\right)+isin\left(\theta + \frac{2\pi k}{n}\right)\right]

for k=0,1,2,...,n1k = 0,1,2,...,n-1.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
4 years, 11 months ago

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@Steven Zheng Can you add this to Roots of Unity - Definition or Applications? Thanks!

Calvin Lin Staff - 4 years, 10 months ago

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Thanks. A good presentation.

Niranjan Khanderia - 4 years, 11 months ago

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