Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Solution 2)
Line 14: Line 14:
 
Let <math>z = re^{i\theta}.</math> We now have <math>\overline{z} = re^{-i\theta},</math> and want to solve  
 
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>
+
<cmath>r^5e^{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  
 
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>
+
<cmath>r^4e^{5i\theta} = e^{-i\theta}.</cmath>
  
−
Multiplying by <math>e^{i\theta}</math> cancels the righthand side's exponents:
+
Taking the magnitude of both sides tells us that <math>r^4 = 1</math>, so
  
−
<cmath>e^{6i\theta} = 1.</cmath>
+
<cmath>e^{5i\theta} = e^{-i\theta}.</cmath>
−
 
 
−
Hence, <math>z^6 = 1</math> and each of the <math>6</math>th roots of unity satisfies this case.
 
  
 +
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.
 
In total, there are <math>1 + 6 = \boxed{\textbf{(E)} 7}</math> numbers that work.  
 
In total, there are <math>1 + 6 = \boxed{\textbf{(E)} 7}</math> numbers that work.  
  

Revision as of 20:07, 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

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

Taking the magnitude of both sides tells us that $r^4 = 1$, so

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

Dividing by $e^{-i\theta},$ we have $e^{6i\theta} = 1$ so $z^6 = 1$. Each of the $6$th roots of unity satisfy this condition. 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)

See also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_14&oldid=205267"