Difference between revisions of "2021 AMC 12A Problems/Problem 13"
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Rewriting each choice to the polar form <math>z=re^{i\theta},</math> we know that by the <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\text{Re}(z^5)=r^5\cos{(5\theta)}.</cmath> We make a table as follows: | Rewriting each choice to the polar form <math>z=re^{i\theta},</math> we know that by the <b>De Moivre's Theorem</b>, the real part of <math>z^5</math> is <cmath>\text{Re}(z^5)=r^5\cos{(5\theta)}.</cmath> We make a table as follows: | ||
<cmath>\begin{array}{c|ccc|cclclclcc} | <cmath>\begin{array}{c|ccc|cclclclcc} | ||
− | \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \textbf{Re}\boldsymbol{(z^5)} & & & & \\ \hline | + | \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \textbf{\ Re}\boldsymbol{(z^5)} & & & & \\ \hline |
& & & & & & & & & & & & \\ [-1ex] | & & & & & & & & & & & & \\ [-1ex] | ||
\textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] | \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] |
Revision as of 01:47, 7 March 2021
Contents
Problem
Of the following complex numbers , which one has the property that
has the greatest real part?
Solution 1 (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 (Radians)
For every complex number where
and
are real numbers and
its magnitude is
For each choice, we get that the magnitude is
Rewriting each choice to the polar form we know that by the De Moivre's Theorem, the real part of
is
We make a table as follows:
Clearly, 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
Solution 5
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.