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

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(Adding a new solution to this problem using the visual representation of roots of unity.)
 
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<math>z^6=e^{\frac{ni\pi}{2}}</math>
 
<math>z^6=e^{\frac{ni\pi}{2}}</math>
  
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This is real iff <math>\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n</math> is even<math>)</math>. Thus, the answer is the number of even <math>0\leq n<24</math> which is <math>\boxed{(D)=\ 12}</math>.
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This is real if <math>\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n</math> is even<math>)</math>. Thus, the answer is the number of even <math>0\leq n<24</math> which is <math>\boxed{(D)=\ 12}</math>.
  
 
==Solution 2==
 
==Solution 2==
 
<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 \cdot i} = cos (0+2k\pi) + i sin(0+2k\pi)</math>, where <math>k</math> is an integer.
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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>.
  
 +
==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>.
 +
 +
==Solution 4 (Quick)==
 +
Let <math>a\in\mathbb{R}</math> and <math>a = z^6.</math> We have
 +
<cmath>a^4 = 1 \implies a = 1,-1.</cmath>
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<math>z^6 = \pm 1</math> has 6 solutions for <math>1</math> and <math>-1</math> respectively, so <math>6+6=\boxed{(D)\ 12}.</math> <cmath> </cmath>
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-svyn
 +
 +
==Solution 5 (Visual Roots of Unity)==
 +
Because <math>z^{24} = 1</math>, we can plot these points on the Argand plane as a regular 24-gon, as shown:
 +
[[Image:2017AMC12aP17.png]]
 +
These are a graphical representation of all 24 values of z, as stated in the problem. Now, we want <math>z^6</math> to be real. The only 2 cases where this happens are if <math>z^6 = 1</math> or <math>z^6 = -1</math>. Squaring both sides for the latter equation, we get <math>z^{12}=1</math>, which, if one were to square root it, would give us a system of both <math>z^6 = 1</math> and <math>z^6 = -1</math>, just as we desire. We can plot the points for <math>z^{12}=1</math> on an Argand plane again, giving us:
 +
[[Image:2017AMC12aP17-p2.png]]
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We note that all of these points are also on the first Argand plane, and counting the points nets us <math>\boxed{(D)\ 12}</math> total values for <math>z</math>. <cmath> </cmath>
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-yingkai_0_ <cmath> </cmath>
 +
Credit to Michael Andrejkovics for providing the GeoGebra widget used to make these diagrams!
 
==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}}

Latest revision as of 00:42, 19 September 2024

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

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

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 $\boxed{(D) \: 12}$ 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

Solution 5 (Visual Roots of Unity)

Because $z^{24} = 1$, we can plot these points on the Argand plane as a regular 24-gon, as shown: 2017AMC12aP17.png These are a graphical representation of all 24 values of z, as stated in the problem. Now, we want $z^6$ to be real. The only 2 cases where this happens are if $z^6 = 1$ or $z^6 = -1$. Squaring both sides for the latter equation, we get $z^{12}=1$, which, if one were to square root it, would give us a system of both $z^6 = 1$ and $z^6 = -1$, just as we desire. We can plot the points for $z^{12}=1$ on an Argand plane again, giving us: 2017AMC12aP17-p2.png We note that all of these points are also on the first Argand plane, and counting the points nets us $\boxed{(D)\ 12}$ total values for $z$. \[\] -yingkai_0_ \[\] Credit to Michael Andrejkovics for providing the GeoGebra widget used to make these diagrams!

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

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

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