2023 AMC 12A Problems/Problem 14

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 $\boxed{\textbf{(E)} 7}$.

~plasta

Solution 2

Let $z = re^{i\theta}.$ We now have $\overline{z} = re^{-i\theta},$ and want to solve

\[r^5e^{5i\theta} = re^{-i\theta}.\]

From this, we have $r = 0$ as a solution, which gives $z = 0$. If $r\neq 0$, then we divide by it, yielding

\[r^4e^{5i\theta} = e^{-i\theta}.\]

Dividing both sides by $e^{-i\theta}$ yields $r^4e^{6i\theta} = 1$. Taking the magnitude of both sides tells us that $r^4 = 1$, so $r^2 = \pm 1$. However, if $r^2 = -1$, then $r = \pm i$, but $r$ must be real. Therefore, $r^2 = 1$.

Multiplying both sides by $r^2$,

\[r^6\cdot e^{6i\theta} = z^6 = 1.\]

Each of the $6$th roots of unity is a solution to this, so there are $6 + 1 = \boxed{\textbf{(D)}\ 7}$ solutions.

-Benedict T (countmath 1)

Video Solution by OmegaLearn

https://youtu.be/rbdIrmOyczk

Video Solution

https://youtu.be/m627Mjp3PkM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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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