# Question Video: The Primitive Cubic Root of Unity Mathematics

Let π be the primitive cubic roots of unity. 1. Find π^(β1). How is this related to the other cubic roots of unity? 2. Find π^(β2). How is this related to the other cubic roots of unity?

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### Video Transcript

Let π be the primitive cubic roots of unity. 1) Find π to the power of negative one. How is this related to the other cubic roots of unity? 2) Find π to the power of negative two. How is this related to the other cubic roots of unity?

Letβs begin by writing π in its exponential form. Itβs π to the π two π by three. This means that π to the power of negative one is π to the π two π by three to the power of negative one. And if we apply the laws of exponents, we see that π to the power of negative one is equal to π to the negative π two π by three. And thatβs of course the same as π squared.

Letβs repeat this process for part two. This time, π to the power of negative two is π to the π two π by three, all to the power of negative two. And again, applying the laws of exponents, we see that π to the power of negative two is equal to π to the negative π four π by three. The argument for this complex number is outside of the range for the principal argument. So we add two π. And we see that π to the power of negative two is equal to π to the π two π by three, which is equal to π. And since π to the power of negative one is the reciprocal of π, we can see that the cubic roots of unity also form a cycle under division.

We can even extend the visual representation of the cubic roots of unity by representing them on an Argand diagram. We can see that theyβre evenly spaced about the origin. In fact, they form the vertices of an equilateral triangle inscribed within a unit circle. But letβs have a look at this carefully.

The visual representation of π and π squared on our Argand diagram is as a reflection in the real axis or the horizontal axis. And if we recall, we know that the complex conjugate of a number is represented by a reflection in the horizontal axis. So this means that π squared must be equal to the conjugate of π.