When is $(2cis\frac{2\pi}{3})^n$ real?
Using de Moivre's theorem:
$$(2cis\frac{2\pi}{3})^n = 2^ncos(\frac{2\pi}{3}n) + i2^nsin(\frac{2\pi}{3}n)$$
$$\therefore sin(\frac{2\pi}{3}n) = 0 = sin(0), sin(\pi), sin(2\pi)...$$
$$\therefore \frac{2\pi}{3}n = 0, \pi, 2\pi...$$
$$\therefore n = 0, \frac{3}{2}, 3...$$
$$\therefore n = \frac{3}{2}k, \ k \geqslant 0 $$
However, the answer is stated to be "the complex number is real only when n is a multiple of 3." I have not been able to spot the error above.