Equation in the complex plane $8z=i|z|^3\bar{z}$? I do not know what to do with this equation. I tried to make both sides in trigonometric form, but after I don't know how to move forward to solve it:
$$8z=i|z|^3\bar{z}$$
In trigonometric form
$$8\varphi\left[\cos(\theta) + i \sin(\theta)\right]
=\varphi^4\left[\cos\big(\tfrac{\pi}{2}-\theta\big) + i \sin\big(\tfrac{\pi}{2}-\theta\big)\right]$$
Now I don't know how to continue.
 A: Let $r=|z|$ and $\theta=\arg(z)$ so that $z=re^{i\theta}$. Then $\overline{z}=re^{-i\theta}$ and so the equation becomes
$$8re^{i\theta}=ir^4e^{-i\theta}.$$
Note that $i=e^{i\tfrac{\pi}{2}}$, so the right-hand side equals $r^4e^{i(\tfrac{\pi}{2}-\theta)}$. Rearranging terms the above becomes
$$re^{i\theta}\cdot\left(8-r^3e^{i(\tfrac{\pi}{2}-2\theta)}\right)=0,$$
which shows that either $r=0$ or
$$8-r^3e^{i(\tfrac{\pi}{2}-2\theta)}=0,$$
in which case $r=2$ and $\tfrac{\pi}{2}-2\theta=2k\pi$ for some $k\in\Bbb{Z}$. Then $\theta=(\tfrac14-k)\pi$, so $\theta=\tfrac14\pi$ or $\theta=\tfrac54\pi$ as $\theta\in[0,2\pi)$. This shows that there are precisely three solutions, which are
$$z=\pm\sqrt{2}(1+i)\qquad\text{ or }\qquad z=0.$$
A: We see that $z=0$ is a trivial solution. Let $z=re^{i\theta+i2\pi k }$. Then
$$8re^{i\theta+i2\pi k}=ir^4e^{-i\theta-i2\pi k},$$
giving $$8e^{i2\theta+i4\pi k}=r^3e^{i\pi/2+i2\pi\ell}.$$
Hence
$r=2$ and $2\theta+4\pi k=\pi/2+2\pi\ell\implies\theta=\pi/4+\pi\ell-2\pi k\equiv \pi/4+\pi\ell$.
So we have $z=2e^{i(\pi/4+\pi\ell)}$, where $\ell\in\mathbb{Z}$. This gives the solutions
$$\left\{0,2 e^{\frac{i \pi }{4}},2 e^{-\frac{i3}{4}\pi}\right\}.$$
