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

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

Revision as of 15:40, 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

Clearly $(-2)^5=-32$ and $(2i)^5=32i$. \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 B: $32\cos(650)=32\cos(30)=16\sqrt{3}$. For C: $32\cos(675)=32\cos(-45)=16\sqrt{2}$. For D: $32\cos(600)=32\cos(240)$ which is negative. 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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