AMC12 2017 B

AMC12 2017 B · Q12

AMC12 2017 B · Q12. It mainly tests Complex numbers (rare), Trigonometry (basic).

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

$z^{12} = 64$ 中，具有正实部的根的和是多少？

(A) 2 2

(B) 4 4

(D) $2\sqrt{2} + \sqrt{6}$ $2\sqrt{2} + \sqrt{6}$

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The principal root of the equation $z^{12}=64$ is $z=64^{\frac{1}{12}}\cdot\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)=\sqrt{2}\cdot\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right).$ The 12 roots lie in the complex plane on the circle of radius $\sqrt{2}$ centered at the origin. The roots with positive real part make angles of $0,\pm\frac{\pi}{6},\text{ and }\pm\frac{\pi}{3}$ with the positive real axis. When these five numbers are added, the imaginary parts cancel out and the sum is $\sqrt{2}+2\sqrt{2}\cdot\cos\frac{\pi}{6}+2\sqrt{2}\cdot\cos\frac{\pi}{3}=\sqrt{2}\cdot(1+\sqrt{3}+1)=2\sqrt{2}+\sqrt{6}.$

答案（D）：方程 $z^{12}=64$ 的主根为 $z=64^{\frac{1}{12}}\cdot\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)=\sqrt{2}\cdot\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right).$ 这 12 个根位于复平面上、以原点为圆心、半径为 $\sqrt{2}$ 的圆上。实部为正的根与正实轴所成的角为 $0,\pm\frac{\pi}{6}$ 和 $\pm\frac{\pi}{3}$。把这五个数相加时，虚部相互抵消，和为 $\sqrt{2}+2\sqrt{2}\cdot\cos\frac{\pi}{6}+2\sqrt{2}\cdot\cos\frac{\pi}{3}=\sqrt{2}\cdot(1+\sqrt{3}+1)=2\sqrt{2}+\sqrt{6}.$

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