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

## 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

~ pi_is_3.14

Solution 5

~IceMatrix

## See Also

 2021 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 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

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