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How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$

$$z=dfrac{|z|^2}{|z|-1}$$ which is real

If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable

If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$

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 $z=-1/2$.

WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real

$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$

If $rne0,$ equating the real & the imaginary parts

$r-rcos t+cos t=0=(1-r)sin t$

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 $#2A:cos t=-1,r+r-1=0iff r=?$

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
  $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
  

Let $r = e^{itheta}$.

We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.

Factorising, we get $r = 0$, giving $z = 0$ as one solution or:

$r - re^{itheta} + e^{itheta} = 0$

$e^{itheta} = frac{r}{r-1}$

From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.

$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.

Therefore the two solutions are $z = 0$ and $z = -frac 12$.