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

(→‎Solution 3 (Binomial Theorem): Solution 3 completed.)
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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 1 (Degrees)==
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==Solution 1 (De Moivre's Theorem: 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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~JHawk0224
 
~JHawk0224
  
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==Solution 2 (Radians)==
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==Solution 2 (De Moivre's Theorem: Radians)==
 
We rewrite each answer choice to the polar form <math>z=re^{i\theta}.</math> By <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.</cmath> We construct a table as follows:
 
We rewrite each answer choice to the polar form <math>z=re^{i\theta}.</math> By <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.</cmath> We construct a table as follows:
 
<cmath>\begin{array}{c|ccc|ccc|cclclclcc}
 
<cmath>\begin{array}{c|ccc|ccc|cclclclcc}

Revision as of 16:08, 21 May 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 (De Moivre's Theorem: 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 (De Moivre's Theorem: Radians)

We rewrite each answer choice to the polar form $z=re^{i\theta}.$ By De Moivre's Theorem, the real part of $z^5$ is \[\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\] We construct a table as follows: \[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \hspace{0.75mm} \boldsymbol{\mathrm{Re}\left(z^5\right)} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\] Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$

~MRENTHUSIASM

Solution 3 (Binomial Theorem)

We evaluate the fifth power of each answer choice:

  • For $\textbf{(A)},$ we have $(-2)^5=-32,$ from which $\mathrm{Re}\left((-2)^5\right)=-32.$
  • For $\textbf{(E)},$ we have $(2i)^5=32i,$ from which $\mathrm{Re}\left((2i)^5\right)=0.$

We will apply the Binomial Theorem to each of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)}.$

Two solutions follow from here:

Solution 3.1 (Real Parts Only)

To find the real parts, we only need the terms with even powers of $i:$

  • For $\textbf{(B)},$ we have \begin{align*} \mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)&=\sum_{k=0}^{2}\binom{5}{2k}\left(-\sqrt3\right)^{5-2k}i^{2k} \\ &=\binom50\left(-\sqrt3\right)^{5}i^{0} + \binom52\left(-\sqrt3\right)^{3}i^{2} + \binom54\left(-\sqrt3\right)^{1}i^{4} \\ &=1\left(-9\sqrt3\right)(1)+10\left(-3\sqrt3\right)(-1)+5\left(-\sqrt3\right)(1) \\ &=16\sqrt3.\end{align*}
  • For $\textbf{(C)},$ we have \begin{align*} \mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)&=\sum_{k=0}^{2}\binom{5}{2k}\left(-\sqrt2\right)^{5-2k}\left(\sqrt2i\right)^{2k} \\ &=\binom50\left(-\sqrt2\right)^5\left(\sqrt2i\right)^0+\binom52\left(-\sqrt2\right)^3\left(\sqrt2i\right)^2+\binom54\left(-\sqrt2\right)^1\left(\sqrt2i\right)^4 \\ &=1\left(-4\sqrt2\right)(1)+10\left(-2\sqrt2\right)(-2)+5\left(-\sqrt2\right)(4) \\ &=16\sqrt2.\end{align*}
  • For $\textbf{(D)},$ we have \begin{align*} \mathrm{Re}\left(\left(-1+\sqrt3 i\right)^5\right)&=\sum_{k=0}^{2}\binom{5}{2k}(-1)^{5-2k}\left(\sqrt3i\right)^{2k} \\ &=\binom50(-1)^5\left(\sqrt3i\right)^0+\binom52(-1)^3\left(\sqrt3i\right)^2 + \binom54(-1)^1\left(\sqrt3i\right)^4 \\ &=1(-1)(1)+10(-1)(-3)+5(-1)(9) \\ &=-16.\end{align*}

Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$

~MRENTHUSIASM

Solution 3.2 (Full Expansions)

  • For $\textbf{(B)},$ we have \begin{align*} \left(-\sqrt3+i\right)^5&=\sum_{k=0}^{5}\binom5k\left(-\sqrt3\right)^{5-k}i^k \\ &=\binom50\left(-\sqrt3\right)^{5}i^0+\binom51\left(-\sqrt3\right)^{4}i^1+\binom52\left(-\sqrt3\right)^{3}i^2+\binom53\left(-\sqrt3\right)^{2}i^3+\binom54\left(-\sqrt3\right)^{1}i^4+\binom55\left(-\sqrt3\right)^{0}i^5 \\ &=1\left(-9\sqrt3\right)(1)+5(9)i+10\left(-3\sqrt3\right)(-1)+10(3)(-i)+5\left(-\sqrt3\right)(1)+1(1)i \\ &=16\sqrt3+16i, \end{align*} from which $\mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.$
  • For $\textbf{(C)},$ we have \begin{align*} \left(-\sqrt2+\sqrt2 i\right)^5&=\sum_{k=0}^{5}\binom5k\left(-\sqrt2\right)^{5-k}\left(\sqrt2i\right)^k \\ &=\binom50\left(-\sqrt2\right)^5\left(\sqrt2i\right)^0+\binom51\left(-\sqrt2\right)^4\left(\sqrt2i\right)^1+\binom52\left(-\sqrt2\right)^3\left(\sqrt2i\right)^2+\binom53\left(-\sqrt2\right)^2\left(\sqrt2i\right)^3+\binom54\left(-\sqrt2\right)^1\left(\sqrt2i\right)^4+\binom55\left(-\sqrt2\right)^0\left(\sqrt2i\right)^5 \\ &=1\left(-4\sqrt2\right)(1)+5(4)\left(\sqrt2i\right)+10\left(-2\sqrt2\right)(-2)+10(2)\left(-2\sqrt2i\right)+5\left(-\sqrt2\right)(4)+1(1)\left(4\sqrt2i\right) \\ &=16\sqrt2-16\sqrt2i, \end{align*} from which $\mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.$
  • For $\textbf{(D)},$ we have \begin{align*} \left(-1+\sqrt3i\right)^5&=\sum_{k=0}^{5}\binom5k(-1)^{5-k}\left(\sqrt3i\right)^k \\ &=\binom50(-1)^5\left(\sqrt3i\right)^0+\binom51(-1)^4\left(\sqrt3i\right)^1+\binom52(-1)^3\left(\sqrt3i\right)^2+\binom53(-1)^2\left(\sqrt3i\right)^3+\binom54(-1)^1\left(\sqrt3i\right)^4+\binom55(-1)^0\left(\sqrt3i\right)^5 \\ &=1(-1)(1)+5(1)\left(\sqrt3i\right)+10(-1)(-3)+10(1)\left(-3\sqrt3i\right)+5(-1)(9)+1(1)\left(9\sqrt3i\right) \\ &=-16-16\sqrt3i, \end{align*} from which $\mathrm{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.$

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

Video Solution by TheBeautyofMath

https://youtu.be/ySWSHyY9TwI?t=568

~IceMatrix

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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