nth Roots of Unity
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nth Roots of Unity

Recall from the nth Roots of Complex Numbers page that any nonzero complex number, say $z \in \mathbb{C}$ ($n \neq 0$) where $z = r (\cos \theta + i \sin \theta)$ has $n$ many $n^{\mathrm{th}}$ roots given by the formula:

\begin{align} \quad r^{1/n} \left ( \cos \left ( \frac{\theta + 2k \pi}{n} \right ) + i \sin \left ( \frac{\theta + 2k \pi}{n} \right ) \right ) \end{align}

Where each $k \in \{ 0, 1, ..., n - 1\}$ yields a distinct $n^{\mathrm{th}}$ root.

We will now look at a special case of this result. Suppose that $z = 1$. Then $r = \mid z \mid = 1$, so $r^{1/n} = 1$. So for each $n \in \mathbb{N}$, the $n$ many $n^{\mathrm{th}}$ roots of $1$ all lie on the unit circle. Furthermore, by De Moivre's Formula for the Polar Representation of Powers of Complex Numbers we must have that these $n$ many $n^{\mathrm{th}}$ roots are equally spaced on the unit circle. These roots are called the $n^{\mathrm{th}}$ roots of unity which we define below.

Definition: The $n^{\mathrm{th}}$ Roots of Unity are the $n$ many complex $n^{\mathrm{th}}$ roots of $1$.

The following diagram illustrates the $n^{\mathrm{th}}$ roots of unity for $n = 1,2, 3, 4$:


Of course we could generalize this important case to any complex number $z = r (\cos \theta + i \sin \theta)$ with $\mid r \mid = 1$. That is, if $\mid r \mid = 1$ then the $n$ many $n^{\mathrm{th}}$ of $z$ are equally spaced on the unit circle.

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