Real Complex Roots of Unity for Even Index

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Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is even.

Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.


The only $x \in U_n$ such that $x \in \R$ are:

$x = 1$ or $x \in -1$


That is, $1$ and $-1$ are the only complex $n$th roots of unity which are real number.


Proof

From Positive Real Complex Root of Unity, we have that $1$ is the only element of $U_n$ which is a positive real number.

We note that $\paren {-1}^n = 1$ as $n$ is even.

Thus $-1$ is also an element of $U_n$.

Now let $z \in U_n$ such that $\cmod z \ne 1$.


Let $z > 0$.

From Positive Power Function on Non-negative Reals is Strictly Increasing it follows that:

$z < 1 \implies z^n < 1$

and:

$z > 1 \implies z^n > 1$


Let $z < 0$.

From Positive Power Function on Negative Reals is Strictly Decreasing it follows that:

$z < -1 \implies z^n > 1$

and:

$z > -1 \implies z^n < 1$

That is, in all cases where $\cmod z \ne 1$ we have that $z^n \ne 1$.

$\blacksquare$


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