Solve complex equations with absolute value $|z|^2+z|z|+\bar{z} = 0$ I literally have no idea on how to solve equations like this:
$$|z|^2+z|z|+\bar{z} = 0$$
Even if I try to rewrite it as 
$$(x^2+y^2)+(x+iy)\sqrt{x^2+y^2}+(x-iy) = 0$$
how should I proceed?
 A: From the given $|z|^2+z|z|+\bar{z} = 0$, 
$z |z|+\bar{z} $ is real, i.e.
$$z |z|+\bar{z} = \bar{z} |z|+z$$
Factorize to get
$$(|z|-1)(z-\bar{z}) =0$$
So, two cases to examine:
Case 1) $z-\bar{z} =0$ yields the solution $z=0$.
Case 2) $|z|-1=0$ leads to
$$\frac{z^3-1}{z-1} = z^2+z+1= \frac1{\bar z}[(|z|-1) z|z| +(|z|^2+z|z|+\bar z)]=0$$
or $z^3=1$ and $z\ne1$, 
which yields the solutions $z=e^{i2\pi/3}$ and $z=e^{i4\pi/3}$.
A: Looking at imaginary parts:
$$y\sqrt{x^2+y^2}-y=0,$$
so $y=0$ or $\sqrt{x^2+y^2}-1=0$. 
In the case $y=0$, you then just have the equation $2x^2+x=0$, so $x=0$ or $-\frac{1}{2}$, so $z=0$ or $z=-\frac{1}{2}$. But we find if $z=-\frac{1}{2}$, the real part is not zero, so we reject this solution.
In the case $\sqrt{x^2+y^2}=1$, we have $|z|=1$, so the equation becomes $1+z+\overline{z}=0$, which implies $x=-\frac{1}{2}$, so $y=\pm\sqrt{1-x^2}=\pm\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.
Therefore, the solutions are $z=0,-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.
A: Hint: Use polar coordinates: $z=re^{i\theta}$:
$$r^2 +re^{i\theta}\cdot r + re^{-i\theta} = 0 \\ r^2 +r^2 e^{i\theta}+ re^{-i\theta} = 0$$
A: Equate the real & the imaginary parts 
$$y\sqrt{x^2+y^2}-y=0\iff y(\sqrt{x^2+y^2}-1)=0\ \ \ \  (1)$$
$$x^2+y^2+x\sqrt{x^2+y^2}+x=0\  \  \ \  (2)$$
Case $\#1:$
By $(1)$ if $y=0$  by $(2),$ 
$$x^2+|x|+x=0, x=?$$
Check when $x\ge0$ and when $x<0?$
Case $\#2:$
By $(1)$ if $\sqrt{x^2+y^2}=1$  by $(2),1+x+x=0$
A: Write $z=x+iy$ and take the imaginary part:
$$|z|y-y=0$$
so either (i) $y=0$ or (ii) $|z|=1$. In (ii) you get $1+z+\overline z=0$.
Then $z=\cos t+i\sin t$, so that $1+2\cos t=0$. Then $\cos t=-1/2$
and $\sin t=\pm\sqrt 3/2$. In (i), $z$ is real, and I'll leave that to you.
