Let $w_1 = 3\cdot cis(\frac{4\pi}{3})$ and $w_2=1/3 \cdot cis(\alpha)$. Determine $\alpha$ so that $w_1\cdot w_2$ is a real number I tried:
$$3 \operatorname*{cis}(\frac{4\pi}{3})\cdot \frac{1}{3} \operatorname*{cis}(\alpha) = \operatorname*{cis}(\frac{4\pi}{3}+\alpha) = \cos(\frac{4\pi}{3}+\alpha) + \sin(\frac{4\pi}{3}+\alpha)i$$
Since this has to be a real number I did:
$$\sin(\frac{4\pi}{3}+\alpha) = 0 \Leftrightarrow \frac{4\pi}{3}+\alpha = 2k\pi \lor \frac{4\pi}{3}+\alpha = \pi+2k\pi \Leftrightarrow \\
\alpha = 2k\pi-\frac{4\pi}{3} \lor \alpha = \pi+2k\pi-\frac{4\pi}{3}$$
But my book states the solution is $\alpha = \frac{2\pi}{3}+k\pi$. What went wrong?
By the way I know theres a shorter way of representing the solution and it goes something like $\alpha = \text{something }+(-1)^k\arcsin(\text{number})$, anyone knows it?

Nothing went wrong, except you made things more complicated for yourself.
Hint: if I have $x = 2k\pi \vee x = 2k\pi + \pi$, that's the same as saying $x = k\pi$.
