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2021 AMC 12A Problems/Problem 13: Difference between revisions

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==Solution==
==Solution==
<cmath>\textbf{(B)} = 2\text{cis}(150)</cmath>
<math>\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)</math><math>, \textbf{(D)} =2\text{cis}(120)</math>
<cmath>\textbf{(C)} =2\text{cis}(135)</cmath>
Taking the real part of the 5th power of each we have:
<cmath>\textbf{(D)} =2\text{cis}(120)</cmath>
<math>\textbf{(A):} (-2)^5=-32</math>,
Taking the real part of the 5th power of each we have
<math>\textbf{(B):} 32\cos(650)=32\cos(30)=16\sqrt{3}</math>,
<cmath>\textbf{(A):} (-2)^5=-32</cmath>
<math>\textbf{(C):} 32\cos(675)=32\cos(-45)=16\sqrt{2}</math>,
<cmath>\textbf{(B):} 32\cos(650)=32\cos(30)=16\sqrt{3}</cmath>
<math>\textbf{(D):} 32\cos(600)=32\cos(240)</math> which is negative, and
<cmath>\textbf{(C):} 32\cos(675)=32\cos(-45)=16\sqrt{2}</cmath>
<math>\textbf{(E):} (2i)^5</math><math> which is imaginary.
<cmath>\textbf{(D):} 32\cos(600)=32\cos(240)</cmath> which is negative.
Thus, the answer is </math>\boxed{\textbf{(B)}}$.
<cmath>\textbf{(E):} (2i)^5</cmath> which is imaginary.
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
~JHawk0224
~JHawk0224


Revision as of 15:44, 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

$\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 (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

These problems are copyrighted © by the Mathematical Association of America.

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