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

(Solution 3)
(Solution 3)
Line 25: Line 25:
  
 
==Solution 3==
 
==Solution 3==
From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) 12}</math> such <math>z</math>.
+
From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) \space 12}</math> such <math>z</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 00:47, 13 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 $\boxed{(D)}$.

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 $\boxed{(D) \space 12}$ such $z$.

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS