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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.

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    Your answer is fine; the book's author supposed that using the letter $n$ would automatically suggest to you that $n$ was an integer rather than an arbitrary rational number. In particular, for $n = 3/2$, one value of the expression is $-1$. – John Hughes Jul 10 '16 at 02:23
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    (^That. Also, keep in mind $k$ can be negative.) – anon Jul 10 '16 at 02:23
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    Oooh. Good call, @arctic tern! I missed that the negatives weren't there. – John Hughes Jul 10 '16 at 02:24
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    You should be very careful about exponentiating complex numbers with respect to anything other than integer exponents. You'd like it to be true that $(\text{cis } \theta)^r = \text{cis } r \theta$ for all $r, \theta$, but the problem is that if $r$ is not an integer then changing $\theta$ by $2 \pi$ doesn't change $\text{cis } \theta$ but can change $\text{cis } r \theta$... so the author's convention is very reasonable here. – Qiaochu Yuan Jul 10 '16 at 02:35

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