<s>[INST] Complex numbers - solving for smallest positive value of $n$ Given that $z_1=2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)$ and $z_2=2\operatorname{cis}\left(\frac{2\pi}{3}\right)$ find the smallest positive value of $n$ such that $\left(\frac{z_1}{z_2}\right)^n \in \Bbb{R}^+$.
My attempt:
$$\frac{z_1}{z_2}=\frac{2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)}{2\operatorname{cis}
\left(\frac{2\pi}{3}\right)}=\sqrt{3}\operatorname{cis}\left(\frac{5\pi}{6}\right)$$
Hence,
$$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^n\operatorname{cis}\frac{5n\pi}{6}$$
Since $(\frac{z_1}{z_2})^n \in \Bbb{R}^+$, so
$$\sqrt{3}^ni\sin\frac{5n\pi}{6}=0$$
$$5n\pi=0,6\pi$$
$$n=0,1.2$$
At the same time, I also know that
$$\sqrt{3}^n\cos\frac{5n\pi}{6}>0$$
Solving for the above will lead to $n>\frac{3}{5}$
However, the textbook answer is $n=12$.
Somehow, I could not reconcile the my two solutions to $n=12$.
May I know where did I did wrongly? Thank you in advance.
 [/INST] For $n\in\Bbb Z$ we have
$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^{\,n}\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $(\,\cos \frac{5n\pi}{6}>0 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $(\,\cos \frac{5n\pi}{6}=1 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $\frac {5n}{6}\,$ is twice an integer
iff $\,12|n$.
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