Product of nth Roots of Unity

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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:

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

Then the product of all of the elements of $U_n$ is $(-1)^{n-1}$


Proof

\(\ds \prod_{k \mathop = 0}^{n - 1} e^{2 i k \pi / n}\) \(=\) \(\ds e^{2 i \paren {0} \pi / n} e^{2 i \paren {1} \pi / n} e^{2 i \paren {2} \pi / n} \dotsm e^{2 i \paren {n - 1} \pi / n}\)
\(\ds \) \(=\) \(\ds e^{\paren {2 i \pi / n} \paren {0 + 1 + \dotsm + \paren {n - 1} } }\) Product of Powers
\(\ds \) \(=\) \(\ds e^{\paren {2 i \pi / n} \paren {\frac {n \paren {n - 1} } 2 } }\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds e^{\paren {i \pi } \paren {n - 1 } }\)
\(\ds \) \(=\) \(\ds \paren {-1}^{\paren {n - 1 } }\) $e^{i \pi} = -1$


$\blacksquare$