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 $\boxed{(D)}$ .

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2017 AMC 12A Problems/Problem 17

Contents

Problem

Solution

Solution 2

See Also