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Difference between revisions of "2021 AMC 12A Problems/Problem 13"
(Undo revision 231164 by Kms888 (talk))
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Of the following complex numbers <math>z</math>, which one has the property that <math>z^5</math> has the greatest real part?
 
Of the following complex numbers <math>z</math>, which one has the property that <math>z^5</math> has the greatest real part?
  
−
<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==
+
==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)}</math> is <math>2\text{cis}(150)</math>, <math>\textbf{(C)}</math> is <math>2\text{cis}(135)</math>, <math>\textbf{(D)}</math> is <math>2\text{cis}(120)</math>.
  
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Taking the real part of the 5th power of each we have:
+
Taking the real part of the <math>5</math>th power of each we have:
  
−
<math>\textbf{(A): }(-2)^5=-32</math>,
+
<math>\textbf{(A): }(-2)^5=-32</math>
  
−
<math>\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}</math>
+
<math>\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}</math>
  
 
<math>\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}</math>
 
<math>\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}</math>
  
−
<math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative
+
<math>\textbf{(D): }32\cos(600)=32\cos(240)<0</math>
  
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<math>\textbf{(E): }(2i)^5</math> which is imaginary
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<math>\textbf{(E): }(2i)^5=32i</math>, whose real part is <math>0</math>
 +
 
 +
Thus, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}</math>.
  
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Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 
 
~JHawk0224
 
~JHawk0224
 +
 +
==Solution 2 (De Moivre's Theorem: Radians)==
 +
We rewrite each answer choice to the polar form <cmath>z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),</cmath> where <math>r</math> is the magnitude of <math>z</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math>
 +
 +
By <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.</cmath> We construct a table as follows:
 +
<cmath>\begin{array}{c|ccc|ccc|cclclclcc}
 +
& & & & & & & & & & & & & & &  \\ [-2ex]
 +
\textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{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}</cmath>
 +
Clearly, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 3 (Binomial Theorem)==
 +
We evaluate the fifth power of each answer choice:
 +
 +
* For <math>\textbf{(A)},</math> we have <math>(-2)^5=-32,</math> from which <math>\operatorname{Re}\left((-2)^5\right)=-32.</math>
 +
 +
* For <math>\textbf{(E)},</math> we have <math>(2i)^5=32i,</math> from which <math>\operatorname{Re}\left((2i)^5\right)=0.</math>
 +
 +
We will apply the Binomial Theorem to each of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)}.</math>
 +
 +
More generally, let <cmath>z=a+bi</cmath> for some real numbers <math>a</math> and <math>b.</math>
 +
 +
Two solutions follow from here:
 +
 +
===Solution 3.1 (Real Parts Only)===
 +
To find the real part of <math>z^5,</math> we only need the terms with even powers of <math>i:</math>
 +
<cmath>\begin{align*}
 +
\operatorname{Re}\left(z^5\right)&=\operatorname{Re}\left((a+bi)^5\right) \\
 +
&=\sum_{k=0}^{2}\binom{5}{2k}a^{5-2k}(bi)^{2k} \\
 +
&=\binom50 a^{5}(bi)^{0} + \binom52 a^{3}(bi)^{2} + \binom54 a^{1}(bi)^{4} \\
 +
&=a^5 - 10a^3b^2 + 5ab^4.
 +
\end{align*}</cmath>
 +
We find the real parts of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)}</math> directly:
 +
 +
* For <math>\textbf{(B)},</math> we have <math>\operatorname{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math>
 +
 +
* For <math>\textbf{(C)},</math> we have <math>\operatorname{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math>
 +
 +
* For <math>\textbf{(D)},</math> we have <math>\operatorname{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</math>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
===Solution 3.2 (Full Expansions)===
 +
The full expansion of <math>z^5</math> is
 +
<cmath>\begin{align*}
 +
z^5&=(a+bi)^5 \\
 +
&=\sum_{k=0}^{5}\binom5k a^{5-k}(bi)^k \\
 +
&=\binom50 a^{5}(bi)^0+\binom51 a^{4}(bi)^1+\binom52 a^{3}(bi)^2+\binom53 a^{2}(bi)^3+\binom54 a^{1}(bi)^4+\binom55 a^{0}(bi)^5 \\
 +
&=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i \\
 +
&=\left(a^5-10a^3b^2+5ab^4\right) + \left(5a^4b-10a^2b^3+b^5\right)i.
 +
\end{align*}</cmath>
 +
We find the full expansions of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)},</math> then extract their real parts:
 +
 +
* For <math>\textbf{(B)},</math> we have <math>\left(-\sqrt3+i\right)^5=16\sqrt3+16i,</math> from which <math>\operatorname{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math>
 +
 +
* For <math>\textbf{(C)},</math> we have <math>\left(-\sqrt2+\sqrt2 i\right)^5=16\sqrt2-16\sqrt2i,</math> from which <math>\operatorname{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math>
 +
 +
* For <math>\textbf{(D)},</math> we have <math>\left(-1+\sqrt3i\right)^5=-16-16\sqrt3i,</math> from which <math>\operatorname{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</math>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/watch?v=FD9BE7hpRvg
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=AjQARBvdZ20
 
https://www.youtube.com/watch?v=AjQARBvdZ20
  
−
==See also==
+
== 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==
 
{{AMC12 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:27, 1 November 2024

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

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

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

$\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}$

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

$\textbf{(D): }32\cos(600)=32\cos(240)<0$

$\textbf{(E): }(2i)^5=32i$, whose real part is $0$

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

~JHawk0224

Solution 2 (De Moivre's Theorem: Radians)

We rewrite each answer choice to the polar form \[z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$

By De Moivre's Theorem, the real part of $z^5$ is \[\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\] We construct a table as follows: \[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{Re}\left(z^5\right)}} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\] 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 $\operatorname{Re}\left((-2)^5\right)=-32.$
  • For $\textbf{(E)},$ we have $(2i)^5=32i,$ from which $\operatorname{Re}\left((2i)^5\right)=0.$

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

More generally, let \[z=a+bi\] for some real numbers $a$ and $b.$

Two solutions follow from here:

Solution 3.1 (Real Parts Only)

To find the real part of $z^5,$ we only need the terms with even powers of $i:$ \begin{align*} \operatorname{Re}\left(z^5\right)&=\operatorname{Re}\left((a+bi)^5\right) \\ &=\sum_{k=0}^{2}\binom{5}{2k}a^{5-2k}(bi)^{2k} \\ &=\binom50 a^{5}(bi)^{0} + \binom52 a^{3}(bi)^{2} + \binom54 a^{1}(bi)^{4} \\ &=a^5 - 10a^3b^2 + 5ab^4. \end{align*} We find the real parts of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)}$ directly:

  • For $\textbf{(B)},$ we have $\operatorname{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.$
  • For $\textbf{(C)},$ we have $\operatorname{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.$
  • For $\textbf{(D)},$ we have $\operatorname{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.$

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

~MRENTHUSIASM

Solution 3.2 (Full Expansions)

The full expansion of $z^5$ is \begin{align*} z^5&=(a+bi)^5 \\ &=\sum_{k=0}^{5}\binom5k a^{5-k}(bi)^k \\ &=\binom50 a^{5}(bi)^0+\binom51 a^{4}(bi)^1+\binom52 a^{3}(bi)^2+\binom53 a^{2}(bi)^3+\binom54 a^{1}(bi)^4+\binom55 a^{0}(bi)^5 \\ &=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i \\ &=\left(a^5-10a^3b^2+5ab^4\right) + \left(5a^4b-10a^2b^3+b^5\right)i. \end{align*} We find the full expansions of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)},$ then extract their real parts:

  • For $\textbf{(B)},$ we have $\left(-\sqrt3+i\right)^5=16\sqrt3+16i,$ from which $\operatorname{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.$
  • For $\textbf{(C)},$ we have $\left(-\sqrt2+\sqrt2 i\right)^5=16\sqrt2-16\sqrt2i,$ from which $\operatorname{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.$
  • For $\textbf{(D)},$ we have $\left(-1+\sqrt3i\right)^5=-16-16\sqrt3i,$ from which $\operatorname{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.$

Therefore, 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
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All AMC 12 Problems and Solutions

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