Resolution exercises on complex numbers How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$
 A: Observe that $z=0$ is a solution of the equation 
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z \ne 0$ be a further solution of this equation. We get , since $|z|^2=z \overline{z}:$
$\overline{z}-|z|+1=0$. This gives $\overline{z}=|z|-1 \in \mathbb R$.Hence
$z=|z|-1 \in \mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
A: WLOG $z=r(\cos t+i\sin t)$ where $r>0,t$ are real
$$r(r-r(\cos t+i\sin t)+(\cos t+i\sin t))=0$$
If $r\ne0,$ equating the real & the imaginary parts
$r-r\cos t+\cos t=0=(1-r)\sin t$
Case $\#1:$
If $r=1,1=0$ which is untenable
If $\sin t=0,$
Case $\#2A:\cos t=1,r=0$ which is untenable
Case $\#2A:\cos t=-1,r+r-1=0\iff r=?$
A: You may also proceed as follows:


*

*Rewrite the equation to 
$$|z|^2 - z|z| + z = 0 \Leftrightarrow \boxed{|z|^2 = z(|z|-1)}$$

*Therefore, $\color{blue}{z}$ must be $\color{blue}{\mbox{real}}$ and so we have $\color{blue}{|z|^2 = z^2}$.

*Noting the solution $\boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) \stackrel{z \in \mathbb{R}, z \neq 0}{\Leftrightarrow}z = |z|-1 \Rightarrow \boxed{z = -\frac{1}{2}}$$
A: $$z=\dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+z\iff z=0$$ which is untenable
If $z\le0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$ 
A: Let $r = e^{i\theta}$.
We get $r^2 - re^{i\theta}\cdot r + re^{i\theta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{i\theta} + e^{i\theta} = 0$
$e^{i\theta} = \frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{i\theta} \in \mathbb{R}$. Hence $e^{i\theta} = \pm 1$. 
$\frac{r}{r-1} = 1$ gives no solution, but $\frac{r}{r-1} = -1 \implies r = \frac 12$. Since $e^{i\theta} = -1$ that gives $z = -\frac 12$.
Therefore the two solutions are $z = 0$ and $z = -\frac 12$.
