Video Transcript
Let 𝜔 be the primitive cubic
roots of unity. 1) Find 𝜔 to the power of
negative one. How is this related to the
other cubic roots of unity? 2) Find 𝜔 to the power of
negative two. How is this related to the
other cubic roots of unity?
Let’s begin by writing 𝜔 in
its exponential form. It’s 𝑒 to the 𝑖 two 𝜋 by
three. This means that 𝜔 to the power
of negative one is 𝑒 to the 𝑖 two 𝜋 by three to the power of negative
one. And if we apply the laws of
exponents, we see that 𝜔 to the power of negative one is equal to 𝑒 to the
negative 𝑖 two 𝜋 by three. And that’s of course the same
as 𝜔 squared.
Let’s repeat this process for
part two. This time, 𝜔 to the power of
negative two is 𝑒 to the 𝑖 two 𝜋 by three, all to the power of negative
two. And again, applying the laws of
exponents, we see that 𝜔 to the power of negative two is equal to 𝑒 to the
negative 𝑖 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 𝜔 to the power
of negative two is equal to 𝑒 to the 𝑖 two 𝜋 by three, which is equal to
𝜔. And since 𝜔 to the power of
negative one is the reciprocal of 𝜔, 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 they’re evenly
spaced about the origin. In fact, they form the vertices
of an equilateral triangle inscribed within a unit circle. But let’s have a look at this
carefully.
The visual representation of 𝜔
and 𝜔 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 𝜔 squared
must be equal to the conjugate of 𝜔.
