Question

设求 $z = \dfrac{1 + i}{\sqrt{2}}.$

$\begin{aligned} &\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \\ &\quad {}\cdot \scriptsize \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)? \end{aligned}$

Let $z = \dfrac{1 + i}{\sqrt{2}}.$ What is

$\begin{aligned} &\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \\ &\quad {}\cdot \scriptsize \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)? \end{aligned}$

$18$

$72 - 36\sqrt{2}$

$36$

$72$

$72 + 36\sqrt{2}$

答案：C

知识点：单位根复数模运算

难度评级：2160

解答：

因为 $z = e^{i\pi/4}$ ，所以 $z^{k^2} = e^{i\pi k^2/4}$ ，只取决于 $k^2 \bmod 8$ 。

对 $k = 1$ 到 $12$ ，余数 $k^2 \bmod 8$ 为 $1$ （给出 $z$ ）出现六次，为 $4$ （给出 $-1$ ）出现三次，为 $0$ （给出 $1$ ）出现三次。所以第一个和为 $6z - 3 + 3 = 6z$ 。

第二个和同理为 $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}$ 。两者乘积为 $6z \cdot \dfrac{6}{z} = 36$ 。

所以正确答案是 C。

Since $z = e^{i\pi/4},$ we have $z^{k^2} = e^{i\pi k^2/4},$ depending only on $k^2 \bmod 8.$

For $k = 1$ to $12,$ the residue $k^2 \bmod 8$ is $1$ (giving $z$ ) six times, $4$ (giving $-1$ ) three times, and $0$ (giving $1$ ) three times. So the first sum is $6z - 3 + 3 = 6z.$

The second sum is likewise $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.$ Their product is $6z \cdot \dfrac{6}{z} = 36.$

Thus, the correct answer is C.

2019 AMC 12A 第 21 题

答案：C

难度评级：2160

解答：

其他年份的第 21 题