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AMC12 2023 A

AMC12 2023 A · Q14

AMC12 2023 A · Q14. It mainly tests Exponents & radicals, Complex numbers (rare).

How many complex numbers satisfy the equation $z^5=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$?
有几个复数满足方程 $z^5=\overline{z}$,其中 $\overline{z}$ 是复数 $z$ 的共轭?
(A) 2 2
(B) 3 3
(C) 5 5
(D) 6 6
(E) 7 7
Answer
Correct choice: (E)
正确答案:(E)
Solution
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 $\boxed{\textbf{(E)} 7}$.
当 $z^5=\overline{z}$ 时,有两种情况:$z=0$ 或 $z\neq 0$。当 $z\neq 0$ 时,由 $|z^5|=|\overline{z}|$,得 $|z|=1$。则 $z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1$。用 $r(\cos \theta +i\sin \theta)$ 表示,当 $z^6=1$ 时,有 $6$ 个不同解。因此,满足 $z^5=\bar{z}$ 的复数个数为 $\boxed{\textbf{(E)} 7}$。
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