# 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

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

## Solution 2

$z = \sqrt[24]{1} = 1^{\frac{1}{24}}$

By Euler's identity, $1 = e^{0 \cdot i} = cos (2k\pi) + i sin(2k\pi)$, 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 $0 \leq k<24$ and $k$ is an integer.

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

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