Question Video: The Primitive Cubic Root of Unity Mathematics

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.