# 2017 AMC 12A Problems/Problem 17

## Problem

There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number? $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

## Solution 1

Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$. So $z^6=e^{\frac{ni\pi}{2}}$

This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even $)$. Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{(D)=\ 12}$.

## Solution 2 $z = \sqrt{1} = 1^{\frac{1}{24}}$

By Euler's identity, $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$, where $k$ is an integer.

Using De Moivre's Theorem, we have $z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}$, where $0 \leq k<24$ that produce $24$ unique results.

Using De Moivre's Theorem again, we have $z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}$

For $z^6$ to be real, $\sin(\frac{k\pi}{2})$ has to equal $0$ to negate the imaginary component. This occurs whenever $\frac{k\pi}{2}$ is an integer multiple of $\pi$, requiring that $k$ is even. There are exactly $\boxed{12}$ even values of $k$ on the interval $0 \leq k<24$, so the answer is $\boxed{(D)}$.

## Solution 3

From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be $24$. Notice that $1 = z^{24} = (z^6)^4$, so for any solution $z$, $z^6$ will be one of the 4th roots of unity ( $1$, $i$, $-1$, or $-i$). Then $6$ solutions $z$ will satisfy $z^6 = 1$, $6$ will satisfy $z^6 = -1$ (and this is further justified by knowledge of the 6th roots of unity), so there must be $\boxed{(D) \: 12}$ such $z$.

## Solution 4 (Quick)

Let $a\in\mathbb{R}$ and $a = z^6.$ We have $$a^4 = 1 \implies a = 1,-1.$$ $z^6 = \pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\boxed{(D)\ 12}.$ $$ -svyn

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