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> | ||
− | ==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)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>. | ||
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<math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative | <math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative | ||
− | <math>\textbf{(E): }(2i)^5</math> which is | + | <math>\textbf{(E): }(2i)^5</math> which is zero |
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>. | 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\mathrm{ \ cis \ }\theta=r(\cos\theta+i\sin\theta),</cmath> where <math>r</math> is the magnitude of <math>z,</math> and <math>\theta</math> is the argument of <math>z.</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} | ||
+ | & & & & & & & & & & & & & & & \\ [-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}</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>\mathrm{Re}\left((-2)^5\right)=-32.</math> | ||
+ | |||
+ | * For <math>\textbf{(E)},</math> we have <math>(2i)^5=32i,</math> from which <math>\mathrm{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> | ||
+ | |||
+ | 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 <math>i:</math> | ||
+ | |||
+ | * For <math>\textbf{(B)},</math> we have <cmath>\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*}</cmath> | ||
+ | * For <math>\textbf{(C)},</math> we have <cmath>\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*}</cmath> | ||
+ | * For <math>\textbf{(D)},</math> we have <cmath>\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*}</cmath> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ===Solution 3.2 (Full Expansions)=== | ||
+ | * For <math>\textbf{(B)},</math> we have <cmath>\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*}</cmath> from which <math>\mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math> | ||
+ | |||
+ | * For <math>\textbf{(C)},</math> we have <cmath>\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*}</cmath> from which <math>\mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math> | ||
+ | |||
+ | * For <math>\textbf{(D)},</math> we have <cmath>\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*}</cmath> from which <math>\mathrm{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== | ==Video Solution by Punxsutawney Phil== | ||
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https://www.youtube.com/watch?v=AjQARBvdZ20 | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
− | == Video Solution by | + | == Video Solution by OmegaLearn (Using Polar Form and De Moivre's Theorem) == |
https://youtu.be/2qXVQ5vBKWQ | https://youtu.be/2qXVQ5vBKWQ | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | ==See | + | ==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}} |
Revision as of 01:08, 22 May 2021
Contents
- 1 Problem
- 2 Solution 1 (De Moivre's Theorem: Degrees)
- 3 Solution 2 (De Moivre's Theorem: Radians)
- 4 Solution 3 (Binomial Theorem)
- 5 Video Solution by Punxsutawney Phil
- 6 Video Solution by Hawk Math
- 7 Video Solution by OmegaLearn (Using Polar Form and De Moivre's Theorem)
- 8 Video Solution by TheBeautyofMath
- 9 See Also
Problem
Of the following complex numbers , which one has the property that has the greatest real part?
Solution 1 (De Moivre's Theorem: Degrees)
First, .
Taking the real part of the 5th power of each we have:
,
which is negative
which is zero
Thus, the answer is . ~JHawk0224
Solution 2 (De Moivre's Theorem: Radians)
We rewrite each answer choice to the polar form where is the magnitude of and is the argument of
By De Moivre's Theorem, the real part of is We construct a table as follows: Clearly, the answer is
~MRENTHUSIASM
Solution 3 (Binomial Theorem)
We evaluate the fifth power of each answer choice:
- For we have from which
- For we have from which
We will apply the Binomial Theorem to each of and
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
- For we have
- For we have
- For we have
Therefore, the answer is
~MRENTHUSIASM
Solution 3.2 (Full Expansions)
- For we have from which
- For we have from which
- For we have from which
Therefore, the answer is
~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)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.