<s>[INST] Solve: $ z^5 = \bar{z} * (- \frac{1}{2} + \frac{\sqrt3}{2} i)$. P.S. if $z = x+y*i$, $\bar{z} = x - y*i$. Solve: $ z^5 = \bar{z} \left(- \frac{1}{2} + \frac{\sqrt3}{2}i\right)$.
P.S. if $z = x+y i$, $\bar{z} = x - y i$.
Seems trivial but I can not solve it. I tried to write $z, \bar{z}$ using $x$ and $y$ and I got nothing. Then I tried by multiplying everything by $z$ and still I got nowhere. Any hint helps.
 [/INST] You can use polar coordinates. So let $r$ and $\theta$ be such that $z=re^{i\theta}$. Then $\bar{z}=re^{-i\theta}$. Furthermore, 
$$-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{\frac{2\pi}{3}i}.$$
Thus, we get
$$r^5e^{5i\theta}=re^{-i\theta}\cdot e^{\frac{2\pi}{3}i}\Leftrightarrow r^4e^{6i\theta}=e^{\frac{2\pi}{3}},$$
so that $r^4=1\Leftrightarrow r=1$ and $6i\theta=\frac{2\pi}{3}i\Leftrightarrow \theta=\frac{\pi}{6}$. Hence, $z=e^{\frac{\pi}{6}i}$.
If you want, you can convert this back to a $x+yi$ form by solving $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}\left(\frac{y}{x}\right)$.
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