2017 AMC 12B Problem 12

Question

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

The twelve roots lie on a circle of radius

\sqrt{2},

spaced

30^\circ

apart

The roots with positive real part are at angles

0, \pm 30^\circ, \pm 60^\circ;

imaginary parts cancel in pairs

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.

Problem 12 in other years