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

m (Solution 2 (Radians))
m (Solution 2 (Radians))
Line 25: Line 25:
 
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>
  
Rewriting each answer 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>\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.</cmath> We make a table as follows:
+
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

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 answer choice, we get that the magnitude is $2.$

We rewrite each answer choice to the polar form $z=re^{i\theta}.$ By De Moivre's Theorem, the real part of $z^5$ is \[\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\] We construct a table as follows: \[\begin{array}{c|ccc|cclclclcc} & & & & & & & & & & & &  \\ [-2ex] \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \hspace{0.75mm} \boldsymbol{\mathrm{Re}\left(z^5\right)} & & & & \\ [0.5ex] \hline & & & & & & & & & & & &  \\ [-1ex] \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & \frac{5\pi}{6} & & & &32\cos{\frac{25\pi}{6}}&=&32\cos{\frac{\pi}{6}}&=&32\left(\frac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & \frac{3\pi}{4} & & & &32\cos{\frac{15\pi}{4}}&=&32\cos{\frac{7\pi}{4}}&=&32\left(\frac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & \frac{2\pi}{3} & & & &32\cos{\frac{10\pi}{3}}&=&32\cos{\frac{4\pi}{3}}&=&32\left(-\frac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & \frac{\pi}{2} & & & &32\cos{\frac{5\pi}{2}}&=&32\cos{\frac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\] 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

Invalid username
Login to AoPS