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

(Video Solution by pi_is_3.14 (Using Polar Form and de Moivre's Theorem))
(Solution)
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<math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative
 
<math>\textbf{(D): }32\cos(600)=32\cos(240)</math> which is negative
  
<math>\textbf{(E): }(2i)^5</math> which is imaginary
+
<math>\textbf{(E): }(2i)^5</math> which is zero
  
 
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
 
Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.

Revision as of 03:32, 12 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

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

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

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=FD9BE7hpRvg

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLearn (Using Polar Form and de Moivre's Theorem)

https://youtu.be/2qXVQ5vBKWQ

~ pi_is_3.14

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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