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

Revision as of 23:17, 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 $0$ , 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

Video Solution by OmegaLearn

https://youtu.be/rbdIrmOyczk

Video Solution

https://youtu.be/m627Mjp3PkM

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

@@ Line 30: / Line 30: @@
 -Benedict T (countmath 1)
+==Solution 3 (Rectangular Form, similar to Solution 1)==
+Let <math>z = a+bi</math>.
+Then, our equation becomes:
+<math>(a+bi)^5=a-bi</math>
+Note that since every single term in the expansion contains either an <math>a</math> or <math>b</math>, simply setting <math>a=0</math> and <math>b=0</math> yields a solution.
+Now, considering the other case that either <math>a</math> or <math>b</math> does not equal <math>0</math>:
+Multiplying both sides by <math>a+bi</math> (or <math>z</math>), we get:
+<math>(a+bi)^6=a^2+b^2</math> (since <math>i^2=-1</math>).
+Substituting <math>z</math> back into the left hand side, we get:
+<math>z^6=a^2+b^2</math>
+Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either <math>a</math> or <math>b</math> is not <math>0</math>, and these are simply the sixth roots of a positive real number.
+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)
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

Page

Toolbox

Search