Difference between revisions of "2021 AMC 12A Problems/Problem 13"

(Solution 2 (Radians))
m
Line 51: Line 51:
  
 
Solution 5
 
Solution 5
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/ySWSHyY9TwI?t=568
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:21, 19 February 2021

Problem

Of the following complex numbers $z$, which one has the property that $z^5$ has the greatest real part?

$\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i$

Solution 1 (Degrees)

First, $\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)$$, \textbf{(D)} =2\text{cis}(120)$.

Taking the real part of the 5th power of each we have:

$\textbf{(A): }(-2)^5=-32$,

$\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}$

$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$

$\textbf{(D): }32\cos(600)=32\cos(240)$ which is negative

$\textbf{(E): }(2i)^5$ which is zero

Thus, the answer is $\boxed{\textbf{(B)}}$. ~JHawk0224

Solution 2 (Radians)

For every complex number $z=a+bi,$ where $a$ and $b$ are real numbers and $i=\sqrt{-1},$ its magnitude is $|z|=\sqrt{a^2+b^2}.$ For each choice, we get that the magnitude is $2.$

Rewriting each choice to the polar form $z=re^{i\theta},$ we know that by the De Moivre's Theorem, the real part of $z^5$ is \[\text{Re}(z^5)=r^5\cos{(5\theta)}.\] We make a table as follows: \[\begin{array}{c|c|c} \textbf{Choice} & \boldsymbol{\theta} & \textbf{Re}\boldsymbol{(z^5)} \\ \hline & & \\ [-1ex] \textbf{(A)} & \pi & 32\cos{(5\pi)}=32\cos\pi=32(-1) \\ [2ex] \textbf{(B)} & \frac{5\pi}{6} & 32\cos{\frac{25\pi}{6}}=32\cos{\frac{\pi}{6}}=32\left(\frac{\sqrt3}{2}\right) \\ [2ex] \textbf{(C)} & \frac{3\pi}{4} & 32\cos{\frac{15\pi}{4}}=32\cos{\frac{7\pi}{4}}=32\left(\frac{\sqrt2}{2}\right) \\ [2ex] \textbf{(D)} & \frac{2\pi}{3} & 32\cos{\frac{10\pi}{3}}=32\cos{\frac{4\pi}{3}}=32\left(-\frac{1}{2}\right) \\ [2ex] \textbf{(E)} & \frac{\pi}{2} & 32\cos{\frac{5\pi}{2}}=32\cos{\frac{\pi}{2}}=32\left(0\right) \end{array}\] Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$

~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)

https://youtu.be/2qXVQ5vBKWQ

~ pi_is_3.14

Solution 5

Video Solution by TheBeautyofMath

https://youtu.be/ySWSHyY9TwI?t=568

~IceMatrix

See Also

2021 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png