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.
On
Let $z=re^{it}$ where $t$ is real
Take modulus both sides to find $r=0,1$
If $r=1$
$e^{5it}=e^{-it+2\pi i/3}$
$5it=2n\pi i +2i\pi/3-it$ where $n$ is any integer
On
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 $\theta=\tan^{-1}\left(\frac{y}{x}\right)$.
If$$z^5=\overline z\left(-\frac12+\frac{\sqrt3}2i\right),$$then\begin{align}\lvert z\rvert^5&=\lvert z^5\rvert\\&=\left\lvert\overline z\left(-\frac12+\frac{\sqrt3}2i\right)\right\rvert\\&=\lvert z\rvert.\end{align}Can you take it from here?