2017 AMC 12B Problem 12

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.