Difference between revisions of "2021 AMC 12A Problems/Problem 13"
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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> |
− | ==Solution 1 (Degrees)== | + | ==Solution 1 (De Moivre's Theorem: Degrees)== |
− | First, <math>\textbf{(B)} | + | 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>. |
− | Taking the real part of the | + | 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( | + | <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> | + | <math>\textbf{(D): }32\cos(600)=32\cos(240)<0</math> |
− | <math>\textbf{(E): }(2i)^5</math> | + | <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>. | ||
− | |||
~JHawk0224 | ~JHawk0224 | ||
− | ==Solution 2 (Radians)== | + | ==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|cclclclcc} | + | <cmath>\begin{array}{c|ccc|ccc|cclclclcc} |
− | & & & & & & & & & & & & \\ [-2ex] | + | & & & & & & & & & & & & & & & \\ [-2ex] |
− | \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \ | + | \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{Re}\left(z^5\right)}} & & & & \\ [0.5ex] |
\hline | \hline | ||
− | & & & & & & & & & & & & \\ [-1ex] | + | & & & & & & & & & & & & & & & \\ [-1ex] |
− | \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] | + | \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] |
− | \textbf{(B)} & & \ | + | \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)} & & \ | + | \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)} & & \ | + | \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)} & & \ | + | \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> | \end{array}</cmath> | ||
− | Clearly, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> | + | 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 | ~MRENTHUSIASM | ||
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https://www.youtube.com/watch?v=AjQARBvdZ20 | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
− | == Video Solution by OmegaLearn (Using Polar Form and | + | == 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 | ||
− | |||
− | |||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Latest revision as of 17:36, 4 December 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, is
,
is
,
is
.
Taking the real part of the th power of each we have:
, whose real part is
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
such that
and
is the argument of
such that
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
More generally, let for some real numbers
and
Two solutions follow from here:
Solution 3.1 (Real Parts Only)
To find the real part of we only need the terms with even powers of
We find the real parts of
and
directly:
- For
we have
- For
we have
- For
we have
Therefore, the answer is
~MRENTHUSIASM
Solution 3.2 (Full Expansions)
The full expansion of is
We find the full expansions of
and
then extract their real parts:
- 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.