### 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?

Letβs begin by writing π in its
exponential form. Itβs 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 thatβs of course the same as π
squared.

Letβs 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 theyβre evenly
spaced about the origin. In fact, they form the vertices of
an equilateral triangle inscribed within a unit circle. But letβs 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.