2023 AMC 12A Problem 14 — Solution | LIVE by Po-Shen Loh

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2023 AMC 12A Problem 14

Attempt Problem 14 of the 2023 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

How many complex numbers satisfy the equation $z^5=\overline{z},$ where $\overline{z}$ is the conjugate of the complex number $z?$

$2$

$3$

$5$

$6$

$7$

Answer: E

Concepts:complex number roots of unity

Difficulty rating: 1730

Solution:

Taking magnitudes gives $|z|^5=|z|,$ so $|z|=0$ or $|z|=1.$ The value $z=0$ works, giving one solution.

If $|z|=1,$ multiply the equation by $z$ to get $z^6=z\overline{z}=|z|^2=1.$ This has $6$ distinct roots, all of modulus $1.$

Altogether there are $1+6=7$ solutions.

Thus, the correct answer is E.

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