Equation in complex field $z^2+z\overline{z}-2z =0 $ How can this equation be solved in the complex field as $z=x+iy$
$$ z^2+z\overline{z}-2z =0  $$
I get to one of the solution but not the other one , can someone explain thoroughly 
 A: One obvious solution is $z=0$,. The other ( for $z=x+iy$) is the solution of:
$$
z+\bar z-2=0 \iff 2x-2=0 \iff x=1
$$
A: Hint:
$$z^2+z\overline{z}-2z =0\quad (1)\\
\overline{z}^2+z\overline{z}-2\overline{z} =0\quad (2)$$
Now make $(1)-(2)$ and get
$$(z^2-\overline{z}^2)-2(z-\overline{z})=0\to(z-\overline{z})(z+\overline{z}-2)=0$$
EDIT
The final equation comes from an implication (not an equivalence). It means that once one find the values it is necessary to test in the original equation. 
A: Note that $z^2+z\bar z-2z=0\implies (2x^2-2x)+i2(x-1)y=0$.  
A: \begin{eqnarray*}
2x(x-1)+2iy(x-1)=0
\end{eqnarray*}
Now equate real & imaginary parts. So $x=1$ will satisfy both equations and any value of y will do. alternatively $y=0$ and then $x=0$ will be forced.
So the two solutions are
\begin{eqnarray*}
z=1+iy \\ or  \\ z=0
\end{eqnarray*}
A: $z^2+z\overline z-2z=0\implies (z-1)^2=1-|z|^2\in \mathbb R\implies$  $$(i) \quad z-1=A\in \mathbb R$$ $$OR$$ $$(ii) \quad z-1=iA \text { with } A\in \mathbb R.$$  Observe that for (i) we have  $(z^2+z\overline z-2z=0 \land z\in \mathbb R)\iff 2z^2-2z=0\iff (z=0\lor z=1.)$
For (ii) when $z=1+iA$ with $A\in \mathbb R$ we have $z^2+z\overline z-2z=(1-A^2+2iA)+(1+A^2)-2(1+iA)=0.$
Since the solution $z=1$ belongs to both type (i) and type (ii) we can write the solution set as $\{0\}\cup \{1+iA: A\in \mathbb R\}.$
