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

Revision as of 12:18, 18 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: $\begin{array}{ccc} \text{Choice} & \theta & \text{Re}(z^5) \\ \hline \textbf{(A)} & \pi & 32\cos{(5\pi)}=32\cos\pi=32(-1) \\ [2ex] \textbf{(B)} & \frac{5\pi}{6} & 32\cos{\frac{25\pi}{6}}=32\cos{\frac{\pi}{6}}=32\left(\frac{\sqrt3}{2}\right) \\ [2ex] \textbf{(C)} & \frac{3\pi}{4} & 32\cos{\frac{15\pi}{4}}=32\cos{\frac{7\pi}{4}}=32\left(\frac{\sqrt2}{2}\right) \\ [2ex] \textbf{(D)} & \frac{2\pi}{3} & 32\cos{\frac{10\pi}{3}}=32\cos{\frac{4\pi}{3}}=32\left(-\frac{1}{2}\right) \\ [2ex] \textbf{(E)} & \frac{\pi}{2} & 32\cos{\frac{5\pi}{2}}=32\cos{\frac{\pi}{2}}=32\left(0\right) \end{array}$ Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$