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

Revision as of 22:15, 3 April 2024

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 $0$ .

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

-SwordOfJustice

Video Solution by OmegaLearn

https://youtu.be/rbdIrmOyczk

Video Solution

https://youtu.be/m627Mjp3PkM

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

@@ Line 51: / Line 51: @@
 Adding up the solutions, we get <math>1+6=</math> <math>\boxed{\textbf{(E)} 7}</math>
+-SwordOfJustice
 ==Video Solution by OmegaLearn==

2023 AMC 12A (Problems • Answer Key • Resources)
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