Complex Roots of Unity in Exponential Form

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z$ be an integer such that $n > 0$.

Let $z \in \C$ be a complex number such that $z^n = 1$.

Then:

$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$

where $U_n$ is the set of $n$th roots of unity.

That is:

$z \in \set {1, e^{2 i \pi / n}, e^{4 i \pi / n}, \ldots, e^{2 \paren {n - 1} i \pi / n} }$


Thus for every integer $n$, the number of $n$th roots of unity is $n$.


Setting $\omega := e^{2 i \pi / n}$, $U_n$ can then be written as:

$U_n = \set {1, \omega, \omega^2, \ldots, \omega^{n - 1} }$


Proof

Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$.

Then:

$z^n \in \set {e^{2 i k \pi}: k \in \N_n}$

Hence $z^n = 1$.


Now suppose $z^n = 1$. We determine the possible values of $z$ using Roots of Complex Number.

Let $z = r e^{i \theta}$.

Then $\cmod {z^n} = 1 \implies \cmod z = 1$.

Similarly, we have $n \theta = 0 \bmod 2 \pi$.

So $\theta = \dfrac {2 k \pi} n$ for $k \in \Z$.

Hence the result.

$\blacksquare$


Also see


Sources