2021 AMC 12A Problems/Problem 13
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 ![\[z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),\]](http://latex.artofproblemsolving.com/8/5/9/859d0b5a2f9d1681c8bd37b037983b88bfbc250f.png)
where 
is the magnitude of such that 
and 
is the argument of such that 
By De Moivre's Theorem, the real part of is ![\[\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\]](http://latex.artofproblemsolving.com/0/4/7/047bb436c7c93eac4c9dbe018ff14309c426a280.png)
We construct a table as follows:
![\[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{Re}\left(z^5\right)}} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\]](http://latex.artofproblemsolving.com/8/e/0/8e0a368fc6fe3887324700fa7d57c2224ee22fc3.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 
More generally, let ![\[z=a+bi\]](http://latex.artofproblemsolving.com/9/1/e/91ec3f8718b5dad609b860ea272b81ec7c499347.png)
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 
we have 
- For
we have 
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
from which - For we have
from which
from which - For we have
from which
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 |
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| All AMC 12 Problems and Solutions | |
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