Difference between revisions of "2021 AMC 12A Problems/Problem 13"
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For every complex number <math>z=a+bi,</math> where <math>a</math> and <math>b</math> are real numbers and <math>i=\sqrt{-1},</math> its magnitude is <math>|z|=\sqrt{a^2+b^2}.</math> For each answer choice, we get that the magnitude is <math>2.</math> | For every complex number <math>z=a+bi,</math> where <math>a</math> and <math>b</math> are real numbers and <math>i=\sqrt{-1},</math> its magnitude is <math>|z|=\sqrt{a^2+b^2}.</math> For each answer choice, we get that the magnitude is <math>2.</math> | ||
− | + | We rewrite each answer choice to the polar form <math>z=re^{i\theta}.</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|cclclclcc} | <cmath>\begin{array}{c|ccc|cclclclcc} | ||
& & & & & & & & & & & & \\ [-2ex] | & & & & & & & & & & & & \\ [-2ex] |
Revision as of 19:23, 14 May 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 answer choice, we get that the magnitude is
We rewrite each answer choice to the polar form By De Moivre's Theorem, the real part of is We construct 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.