2021 AMC 12A Problems/Problem 13: Difference between revisions
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==Solution 2 (De Moivre's Theorem: Radians)== | ==Solution 2 (De Moivre's Theorem: Radians)== | ||
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: | We rewrite each answer choice to the polar form <math>z=re^{i\theta},</math> 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} | <cmath>\begin{array}{c|ccc|ccc|cclclclcc} | ||
& & & & & & & & & & & & & & & \\ [-2ex] | & & & & & & & & & & & & & & & \\ [-2ex] | ||
Revision as of 23:23, 21 May 2021
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 ![\[\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\]](http://latex.artofproblemsolving.com/9/7/d/97d52497bc6031d24d59e4d6f90f3eaa1fd39890.png)
We construct a table as follows:
![\[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \hspace{0.75mm} \boldsymbol{\mathrm{Re}\left(z^5\right)} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\]](http://latex.artofproblemsolving.com/e/f/a/efa03c52f0ffba6c95b283abe0ad1f70be889cae.png)
Clearly, the answer is 
~MRENTHUSIASM
Solution 3 (Binomial Theorem)
We evaluate the fifth power of each answer choice:
- For
we have 
from which 
we have 
from which 
- For
we have 
from which 
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 
we have 
we have 
we have 
Clearly, the answer is
~MRENTHUSIASM
Solution 3.2 (Full Expansions)
- For we have
from which 
from which 
- For we have
from which 
from which 
- For we have
from which 
from which 
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
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
