Question

$z^{12} = 64$ 的所有根中，实部为正的根之和是多少？

What is the sum of the roots of $z^{12} = 64$ that have a positive real part?

$2$

$4$

$2 + 2\sqrt{3}$

$2\sqrt{2} + \sqrt{6}$

$(1 + \sqrt{3}) + (1 + \sqrt{3})i$

答案：D

知识点：单位根复数对称性

难度评级：1630

小提示：

十二个根位于半径为 $\sqrt{2}$ 的圆上，相邻根相隔 $30^\circ$ 。

The twelve roots lie on a circle of radius $\sqrt{2},$ spaced $30^\circ$ apart

大提示：

实部为正的根对应角度 $0, \pm 30^\circ, \pm 60^\circ$ ；虚部会成对抵消

The roots with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ;$ imaginary parts cancel in pairs

解答：

$z^{12} = 64$ 的根位于半径 $64^{\frac{1}{12}} = \sqrt{2}$ 的圆上，角度为 $30^\circ$ 的整数倍。实部为正的根对应角度 $0, \pm 30^\circ, \pm 60^\circ$ 。它们的虚部相互抵消，所以和为 $\begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6} \end{aligned}\text{。}$

所以正确答案是 D。

The roots of $z^{12} = 64$ lie on the circle of radius $64^{\frac{1}{12}} = \sqrt{2},$ at angles that are multiples of $30^\circ.$ Those with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ.$ Their imaginary parts cancel, so the sum is $\begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6}. \end{aligned}$

Thus, the correct answer is D.

2017 AMC 12B 第 12 题

答案：D

难度评级：1630

小提示：

大提示：

解答：

其他年份的第 12 题