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

Revision as of 11:07, 11 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

$re^{5i\theta} = re^{-i\theta}.$

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

$e^{5i\theta} = e^{-i\theta}.$

Multiplying by $e^{i\theta}$ cancels the righthand side's exponents:

$e^{6i\theta} = 1.$

Hence, $z^6 = 1$ and each of the $6$ th roots of unity satisfies this case.

In total, there are $1 + 6 = \boxed{\textbf{(E)} 7}$ numbers that work.

-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 9: / Line 9: @@
 ~plasta
+==Solution 2==
+Let <math>z = re^{i\theta}.</math> We now have <math>\overline{z} = re^{-i\theta},</math> and want to solve
+<cmath>re^{5i\theta} = re^{-i\theta}.</cmath>
+From this, we have <math>r = 0</math> as a solution, which gives <math>z = 0</math>. If <math>r\neq 0</math>, then we divide by it, yielding
+<cmath>e^{5i\theta} = e^{-i\theta}.</cmath>
+Multiplying by <math>e^{i\theta}</math> cancels the righthand side's exponents:
+<cmath>e^{6i\theta} = 1.</cmath>
+Hence, <math>z^6 = 1</math> and each of the <math>6</math>th roots of unity satisfies this case.
+In total, there are <math>1 + 6 = \boxed{\textbf{(E)} 7}</math> numbers that work.
+-Benedict T (countmath 1)
 ==Video Solution by OmegaLearn==

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