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Difference between revisions of "2021 AMC 12A Problems/Problem 13"

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<math>\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i</math>
 
<math>\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i</math>
  
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==Solution==
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==Solution 1 (Degrees)==
 
First, <math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>.
 
First, <math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>.
  
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Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 
~JHawk0224
 
~JHawk0224
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==Solution 2 (Radians)==
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For every complex number <math>z=a+bi,</math> where <math>a</math> and <math>b</math> are real numbers and <math>i=\sqrt{-1},</math> its magnitude is <math>|z|=\sqrt{a^2+b^2}.</math> For each choice, we get that the magnitude is <math>2.</math>
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Rewriting each choice to the polar form <math>z=re^{i\theta},</math> we know that by the <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\text{Re}(z^5)=r^5\cos{(5\theta)}.</cmath> We make a table as follows:
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<cmath>\begin{array}{ccc}
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\text{Choice} & \theta & \text{Re}(z^5) \\ \hline
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\textbf{(A)} & \pi & 32\cos{(5\pi)}=32\cos\pi=32(-1) \\
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\textbf{(B)} & 5\pi/6 & 32\cos{\frac{25\pi}{6}}=32\cos{\frac{\pi}{6}}=32\left(\frac{\sqrt3}{2}\right) \\
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\textbf{(C)} & 3\pi/4 & 32\cos{\frac{15\pi}{4}}=32\cos{\frac{7\pi}{4}}=32\left(\frac{\sqrt2}{2}\right) \\
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\textbf{(D)} & 2\pi/3 & 32\cos{\frac{10\pi}{3}}=32\cos{\frac{4\pi}{3}}=32\left(-\frac{1}{2}\right) \\
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\textbf{(E)} & \pi/2 & 32\cos{\frac{5\pi}{2}}=32\cos{\frac{\pi}{2}}=32\left(0\right)
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\end{array}</cmath>
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Clearly, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math>
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~MRENTHUSIASM
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 17:18, 13 February 2021

Problem

Of the following complex numbers $z$, which one has the property that $z^5$ has the greatest real part?

$\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i$

Solution 1 (Degrees)

First, $\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)$$, \textbf{(D)} =2\text{cis}(120)$.

Taking the real part of the 5th power of each we have:

$\textbf{(A): }(-2)^5=-32$,

$\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}$

$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$

$\textbf{(D): }32\cos(600)=32\cos(240)$ which is negative

$\textbf{(E): }(2i)^5$ which is zero

Thus, the answer is $\boxed{\textbf{(B)}}$. ~JHawk0224

Solution 2 (Radians)

For every complex number $z=a+bi,$ where $a$ and $b$ are real numbers and $i=\sqrt{-1},$ its magnitude is $|z|=\sqrt{a^2+b^2}.$ For each choice, we get that the magnitude is $2.$

Rewriting each choice to the polar form $z=re^{i\theta},$ we know that by the De Moivre's Theorem, the real part of $z^5$ is \[\text{Re}(z^5)=r^5\cos{(5\theta)}.\] We make a table as follows: \[\begin{array}{ccc} \text{Choice} & \theta & \text{Re}(z^5) \\ \hline \textbf{(A)} & \pi & 32\cos{(5\pi)}=32\cos\pi=32(-1) \\ \textbf{(B)} & 5\pi/6 & 32\cos{\frac{25\pi}{6}}=32\cos{\frac{\pi}{6}}=32\left(\frac{\sqrt3}{2}\right) \\ \textbf{(C)} & 3\pi/4 & 32\cos{\frac{15\pi}{4}}=32\cos{\frac{7\pi}{4}}=32\left(\frac{\sqrt2}{2}\right) \\ \textbf{(D)} & 2\pi/3 & 32\cos{\frac{10\pi}{3}}=32\cos{\frac{4\pi}{3}}=32\left(-\frac{1}{2}\right) \\ \textbf{(E)} & \pi/2 & 32\cos{\frac{5\pi}{2}}=32\cos{\frac{\pi}{2}}=32\left(0\right) \end{array}\] Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$

~MRENTHUSIASM

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=FD9BE7hpRvg

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLearn (Using Polar Form and de Moivre's Theorem)

https://youtu.be/2qXVQ5vBKWQ

~ pi_is_3.14

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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