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

Revision as of 21:35, 22 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 $\boxed{\textbf{(E)} 7}$ .

~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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 <cmath>r^4e^{5i\theta} = e^{-i\theta}.</cmath>
-Taking the magnitude of both sides tells us that <math>r^4 = 1</math>, so
+Dividing both sides by <math>e^{-i\theta}</math> yields <math>r^4e^{6i\theta} = 1</math>.
+Taking the magnitude of both sides tells us that <math>r^4 = 1</math>, so <math>r^2 = \pm 1</math>. However, if <math>r^2 = -1</math>, then <math>r = \pm i</math>, but <math>r</math> must be real. Therefore, <math>r^2 = 1</math>.
-<cmath>e^{5i\theta} = e^{-i\theta}.</cmath>
+Multiplying both sides by <math>r^2</math>,
-Dividing by <math> e^{-i\theta},</math> we have <math>e^{6i\theta} = 1</math> so <math>z^6 = 1</math>. Each of the <math>6</math>th roots of unity satisfy this condition.
+<cmath>r^6\cdot e^{6i\theta} = z^6 = 1.</cmath>
-In total, there are <math>1 + 6 = \boxed{\textbf{(E)} 7}</math> numbers that work.
+Each of the <math>6</math>th roots of unity is a solution to this, so there are <math>6 + 1 = \boxed{\textbf{(D)}\ 7}</math> solutions.
 -Benedict T (countmath 1)

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