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Difference between revisions of "2017 AMC 12A Problems/Problem 17"

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<math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math>
 
<math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math>
  
−
By [[Euler's identity]], <math>1 = e^{0 \times i} = cos (0+2k\pi) + i sin(0+2k\pi)</math>, where <math>k</math> is an integer.
+
By [[Euler's identity]], <math>1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)</math>, where <math>k</math> is an integer.
  
−
Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {cos (\frac{k\pi}{12}) + i sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math> that produce <math>24</math> unique results.
+
Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math> that produce <math>24</math> unique results.
  
−
Using De Moivre's Theorem again, we have <math>z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}</math>
+
Using De Moivre's Theorem again, we have <math>z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}</math>
  
−
For <math>z^6</math> to be real, <math>sin(\frac{k\pi}{2})</math> has to equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is an integer multiple of <math>\pi</math>, requiring that <math>k</math> is even. There are exactly <math>\boxed{12}</math> even values of <math>k</math> on the interval <math>0 \leq k<24</math>, so the answer is <math>\boxed{(D)}</math>.
+
For <math>z^6</math> to be real, <math>\sin(\frac{k\pi}{2})</math> has to equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is an integer multiple of <math>\pi</math>, requiring that <math>k</math> is even. There are exactly <math>\boxed{12}</math> even values of <math>k</math> on the interval <math>0 \leq k<24</math>, so the answer is <math>\boxed{(D)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:19, 12 February 2017

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 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, $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$, 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 $\boxed{(D)}$.

See Also

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

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