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 (De Moivre's Theorem: Degrees)== | ==Solution 1 (De Moivre's Theorem: Degrees)== | ||
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<math>\textbf{(E): }(2i)^5=32i</math>, whose real part is <math>0</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>. | + | Thus, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}</math>. |
~JHawk0224 | ~JHawk0224 | ||
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\textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] | \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 | ~MRENTHUSIASM | ||
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* For <math>\textbf{(D)},</math> we have <math>\operatorname{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</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> | + | Therefore, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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* 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> | * 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> | + | Therefore, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 23:27, 18 December 2024
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.