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Difference between revisions of "2023 AMC 12A Problems/Problem 14"
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==Solution 1==
 
==Solution 1==
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When <math>z^5=\overline{z}</math>, there are two conditions: either <math>z=0</math> or <math>z\neq 0</math>. When <math>z\neq 0</math>, since <math>z^5=\overline{z}</math>, <math>|z|=1</math>. <math>z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1</math>. Consider the <math>r(\cos \theta +i\sin \theta)</math> form, when <math>z^6=1</math>, there are 6 different solutions for <math>z</math>. Therefore, the number of complex numbers satisfying <math>z^5=\bar{z}</math> is <math>\boxed{\textbf{(E)} 7}</math>.
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When <math>z^5=\overline{z}</math>, there are two conditions: either <math>z=0</math> or <math>z\neq 0</math>. When <math>z\neq 0</math>, since <math>|z^5|=|\overline{z}|</math>, <math>|z|=1</math>. <math>z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1</math>. Consider the <math>r(\cos \theta +i\sin \theta)</math> form, when <math>z^6=1</math>, there are 6 different solutions for <math>z</math>. Therefore, the number of complex numbers satisfying <math>z^5=\bar{z}</math> is <math>\boxed{\textbf{(E)} 7}</math>.
  
 
~plasta
 
~plasta
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==Solution 2==
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Let <math>z = re^{i\theta}.</math> We now have <math>\overline{z} = re^{-i\theta},</math> and want to solve
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<cmath>r^5e^{5i\theta} = re^{-i\theta}.</cmath>
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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
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<cmath>r^4e^{5i\theta} = e^{-i\theta}.</cmath>
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Dividing both sides by <math>e^{-i\theta}</math> yields <math>r^4e^{6i\theta} = 1</math>.
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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>.
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Multiplying both sides by <math>r^2</math>,
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<cmath>r^6\cdot e^{6i\theta} = z^6 = 1.</cmath>
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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.
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-Benedict T (countmath 1)
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==Solution 3 (Rectangular Form, similar to Solution 1)==
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Let <math>z = a+bi</math>.
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Then, our equation becomes:
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<math>(a+bi)^5=a-bi</math>
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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.
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 +
Now, considering the other case that either <math>a</math> or <math>b</math> does not equal <math>0</math>:
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Multiplying both sides by <math>a+bi</math> (or <math>z</math>), we get:
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<math>(a+bi)^6=a^2+b^2</math> (since <math>i^2=-1</math>).
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Substituting <math>z</math> back into the left hand side, we get:
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<math>z^6=a^2+b^2</math>
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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.
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Adding up the solutions, we get <math>1+6=</math> <math>\boxed{\textbf{(E)} 7}</math>
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-SwordOfJustice
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==Solution 4==
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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>.
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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.
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Therefore, the answer is <math>6+1=</math> <math>\boxed{\textbf{(E)} 7}</math>.
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- xHypotenuse
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==Video Solution (easy to digest) by Power Solve==
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https://youtu.be/YXIH3UbLqK8?si=l8Ay2f0dMqSkjuQH&t=1975
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==

Revision as of 19:37, 10 June 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


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 (easy to digest) 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)

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

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