AMC12 2017 A

AMC12 2017 A · Q17

AMC12 2017 A · Q17. It mainly tests Complex numbers (rare), Powers & residues.

There are 24 different complex numbers $z$ such that $z^{24} = 1$. For how many of these is $z^{6}$ a real number?

存在 24 个不同的复数 $z$ 使得 $z^{24} = 1$。其中有多少个满足 $z^{6}$ 是实数？

(A) 0 0

(B) 4 4

(D) 12 12

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The complex numbers $z$ such that $z^{24}=1$ are the roots of $z^{24}-1=(z^6-1)(z^6+1)((z^6)^2+1)$. The factors can have at most $6$, $6$, and $12$ roots, respectively. Because $z^{24}-1$ has $24$ distinct roots, the factors do actually have $6$, $6$, and $12$ distinct roots, respectively. The six roots of the first factor satisfy $z^6=1$, and the six roots of the second factor satisfy $z^6=-1$. The twelve roots of the third factor satisfy $(z^6)^2=-1$, so $z^6$ is never real in this case. There are $6+6=12$ roots such that $z^6$ is real.

答案（D）：满足 $z^{24}=1$ 的复数 $z$ 是 $z^{24}-1=(z^6-1)(z^6+1)((z^6)^2+1)$ 的根。各因式分别最多有 $6$、$6$、$12$ 个根。由于 $z^{24}-1$ 有 $24$ 个互不相同的根，因此这些因式实际上分别有 $6$、$6$、$12$ 个互不相同的根。第一个因式的六个根满足 $z^6=1$，第二个因式的六个根满足 $z^6=-1$。第三个因式的十二个根满足 $(z^6)^2=-1$，所以在这种情况下 $z^6$ 从不为实数。使得 $z^6$ 为实数的根共有 $6+6=12$ 个。

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