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

Latest revision as of 17:17, 10 October 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 $6 + 1 = \boxed{\textbf{(E)}\ 7}$ 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

Solution 4

Using the fact that $z\bar{z}=|z|^2$ , we rewrite our equation as $z^6=|z|^2$ . Now, let $r$ represent $|z|$ . We know that $r^6 = r^2$ ; hence, we have $r^6-r^2=r^2(r^4-1)=0$ .

From here, we have two cases: $r=0$ , or $r=1$ . In the case that $r=0$ , we have $z^6=0$ hence $z=0$ . This gives one solution. Alternatively, if $r=1$ , then we have $z^6=1$ , giving $6$ solutions for each root of unity.

Therefore, the answer is $6+1=$ $\boxed{\textbf{(E)} 7}$ .

- xHypotenuse

Video Solution by Power Solve

https://youtu.be/YXIH3UbLqK8?si=l8Ay2f0dMqSkjuQH&t=1975

Video Solution by OmegaLearn

https://youtu.be/rbdIrmOyczk

Video Solution

https://youtu.be/m627Mjp3PkM

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

@@ Line 27: / Line 27: @@
 <cmath>r^6\cdot e^{6i\theta} = z^6 = 1.</cmath>
-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.
+Each of the <math>6</math>th roots of unity is a solution to this, so there are <math>6 + 1 = \boxed{\textbf{(E)}\ 7}</math> solutions.
 -Benedict T (countmath 1)
@@ Line 53: / Line 53: @@
 -SwordOfJustice
+==Solution 4==
+Using the fact that <math>z\bar{z}=|z|^2</math>, we rewrite our equation as <math>z^6=|z|^2</math>.  Now, let <math>r</math> represent <math>|z|</math>. We know that <math>r^6 = r^2</math>; hence, we have <math>r^6-r^2=r^2(r^4-1)=0</math>.
+From here, we have two cases: <math>r=0</math>, or <math>r=1</math>. In the case that <math>r=0</math>, we have <math>z^6=0</math> hence <math>z=0</math>. This gives one solution. Alternatively, if <math>r=1</math>, then we have <math>z^6=1</math>, giving <math>6</math> solutions for each root of unity.
+Therefore, the answer is <math>6+1=</math> <math>\boxed{\textbf{(E)} 7}</math>.
+- xHypotenuse
+==Video Solution by Power Solve==
+https://youtu.be/YXIH3UbLqK8?si=l8Ay2f0dMqSkjuQH&t=1975
 ==Video Solution by OmegaLearn==

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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