Difference between revisions of "2023 AMC 12A Problems/Problem 14"

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==Video Solution by OmegaLearn==
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https://youtu.be/rbdIrmOyczk
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==See also==
 
==See also==
 
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{{AMC12 box|year=2023|ab=A|num-b=13|num-a=15}}
 
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Revision as of 23:50, 9 November 2023

Problem

How many complex numbers satisfy the equation $z^5=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$?

$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$

Solution 1

When $z^5=\overline{z}$, there are two conditions: either $z=0$ or $z\neq 0$. When $z\neq 0$, since $z^5=\overline{z}$, $|z|=1$. $z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1$. Consider the $r(\cos \theta +i\sin \theta)$ form, when $z^6=1$, there are 6 different solutions for $z$. Therefore, the number of complex numbers satisfying $z^5=\bar{z}$ is $\boxed{\textbf{(E)} 7}$.

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Video Solution by OmegaLearn

https://youtu.be/rbdIrmOyczk


See also

2023 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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