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

Revision as of 15:45, 11 February 2021

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

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

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, and

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

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

2021 AMC 12A (Problems • Answer Key • Resources)
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

@@ Line 10: / Line 10: @@
 <math>\textbf{(A):} (-2)^5=-32</math>,
 <math>\textbf{(B):} 32\cos(650)=32\cos(30)=16\sqrt{3}</math>,
 <math>\textbf{(C):} 32\cos(675)=32\cos(-45)=16\sqrt{2}</math>,
 <math>\textbf{(D):} 32\cos(600)=32\cos(240)</math> which is negative, and
-<math>\textbf{(E):} (2i)^5</math><math> which is imaginary.
-Thus, the answer is </math>\boxed{\textbf{(B)}}$.
+<math>\textbf{(E):} (2i)^5</math> which is imaginary.
+Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 ~JHawk0224