# 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)

## Solution 3 (Rectangular Form, similar to Solution 1)

Let $z = a+bi$.

Then, our equation becomes: $(a+bi)^5=a-bi$

Note that since every single term in the expansion contains either an $a$ or $b$, simply setting $a=0$ and $b=0$ yields a solution.

Now, considering the other case that either $a$ or $b$ does not equal $0$:

Multiplying both sides by $a+bi$ (or $z$), we get: $(a+bi)^6=a^2+b^2$ (since $i^2=-1$).

Substituting $z$ back into the left hand side, we get: $z^6=a^2+b^2$

Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either $a$ or $b$ is not $0$, and these are simply the sixth roots of a positive real number.

Adding up the solutions, we get $1+6=$ $\boxed{\textbf{(E)} 7}$

-SwordOfJustice

## Solution 4

Using the fact that $z\bar{z}=|z|^2$, we rewrite our equation as $z^6=|z|^2$. Now, let $r$ represent $|z|$. We know that $r^6 = r^2$; hence, we have $r^6-r^2=r^2(r^4-1)=0$.

From here, we have two cases: $r=0$, or $r=1$. In the case that $r=0$, we have $z^6=0$ hence $z=0$. This gives one solution. Alternatively, if $r=1$, then we have $z^6=1$, giving $6$ solutions for each root of unity.

Therefore, the answer is $6+1=$ $\boxed{\textbf{(E)} 7}$.

- xHypotenuse

## Video Solution

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