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

(Solution)
(Solution)
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==Solution==
 
==Solution==
 
First, <math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>.
 
First, <math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>.
 +
 
Taking the real part of the 5th power of each we have:
 
Taking the real part of the 5th power of each we have:
 +
 
<math>\textbf{(A):} (-2)^5=-32</math>,
 
<math>\textbf{(A):} (-2)^5=-32</math>,
 
<math>\textbf{(B):} 32\cos(650)=32\cos(30)=16\sqrt{3}</math>,
 
<math>\textbf{(B):} 32\cos(650)=32\cos(30)=16\sqrt{3}</math>,

Revision as of 16:45, 11 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

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

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

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