Difference between revisions of "2023 AMC 12A Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
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>. | 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 | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC12 box|year=2023|ab=A|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Revision as of 19:23, 9 November 2023
Problem
How many complex numbers satisfy the equation , where is the conjugate of the complex number ?
Solution 1
When , there are two conditions: either or . When , since , . . Consider the form, when , there are 6 different solutions for . Therefore, the number of complex numbers satisfying is .
~plasta
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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