Let $z\in \mathbb{C}$. How do I determine all complex solutions of $|z|z=-i(\bar{z})$?
My approach: We can see that $$|x+yi|(x+yi)=y-ix$$ Then I came up with the real part $$-y^2+x^2+x^4-y^4$$ and the imaginary part $$2ixy+x^2(2iyx)+y^2(2xiy)$$
I don't know how to go on and I think that is wrong anyway.I would appreciate every attempt to help!
Beside $z=0$, polar form gives $r^2e^{i\theta}=re^{-i(\theta+\pi/2)}$ so $r^2=r\implies r=1$, and $2\theta+\pi/2=2n\pi$ for $n\in\Bbb Z$, i.e. $\theta=-\pi/4+n\pi$. There are two nonzero solutions, $\pm e^{-\pi i/4}$.
Hint:
If $|z|\, z=-i\bar z$, the moduli of both sides are the same, so $|z|^2=|z|$, which implies that $|z|=0$ (so $z=0$) or $|z|=1$.
For the latter case, the best is to use the exponential form: $z=\mathrm e^{i\theta}$. The equation can be rewritten, since the modulus is $1$, as $$\mathrm e^{i\theta}=-i\mathrm e^{-i\theta}=\mathrm e^{-i\bigl(\theta+\tfrac\pi 2\bigr)}.$$ Can you take it from there?
If $|z|=1$, then $\overline{z} = 1/z$ and the equation becomes $$ z = -i/z, \\ z^2 = -i $$ with two solutions $$ z = \frac{-1+i}{\sqrt{2}},\qquad z=\frac{1-i}{\sqrt{2}} $$
\bar{z}. – Bernard Dec 18 '20 at 15:48