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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

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    Hint: just factor out $z$ and write it as $z(z+ \bar z - 2)=0\,$. – 2017-02-14

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$z^2+z\overline z-2z=0\implies (z-1)^2=1-|z|^2\in \mathbb R\implies$ $$(i) \quad z-1=A\in \mathbb R$$ $$OR$$ $$(ii) \quad z-1=iA \text { with } A\in \mathbb R.$$ 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 $\{0\}\cup \{1+iA: A\in \mathbb R\}.$

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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.

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    Beautiful answer! – 2017-02-14
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    thanks @SimplyBeautifulArt – 2017-02-14
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    @Dr.MV It looks fine now ;-) – 2017-02-14
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    @Dr. MV, that is ok. We don't need to worry about that. – 2017-02-14
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    @Arnaldo Thank you my friend! Much appreciative. -Mark – 2017-02-14
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    @Dr. MV, you are very welcome my friend! – 2017-02-14
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    One has to be careful here. This approach gives the entire real line as solutions to the final equation. but $0$ and $1$ are the only real solutions to the original equation. Since difference of (1) and (2) is $0$ any time their LHS are the same, not just when they each equal $0$, it introduces false solutions. So it necessary to check all solutions in the original equation. – 2017-02-15
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    @PaulSinclair: In fact, the values that you find in the final equation one has to test in the original, like any other implication (see that it isn't an equivalence), that is why a called a "Hint". Dr. MV had the same doubt but after a while he realized that there wasn't any problem with the "Hint". – 2017-02-15
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    I just don't understand why the downvote! – 2017-02-15
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    @Arnaldo - I didn't downvote. Nor did I have a problem with the solution per se. But not every student of mathematics has yet developed the ability to spot when false solutions are introduced in a calculation, (which as we both know is a common occurrence). So I thought it useful to add a warning to said students that there is something more here that they need to be aware of. – 2017-02-15
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    @PaulSinclair: I appreciate your comment. I agreed with it. That is why I edited my answer. I put the downvote comment as a separe comment because I don't know who did that. I just would like to know why. – 2017-02-15
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    @PaulSinclair: I'm sorry if you though that I was talking to you about the downvote. – 2017-02-15
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    I agree with you about the downvoting - blind downvoting is something I consider childish. If you don't tell someone what you disagree with, what is the point? It just becomes a meaningless insult. Though it is possible here that someone interpreted my comment as a complaint, not the addendum I intended it to be, and downvoted because they felt introducing false solutions was a mistake. For myself, while I find the other solution cleaner, it is very specific to this problem. Your technique is more broadly applicable. – 2017-02-15
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One obvious solution is $z=0$,. The other ( for $z=x+iy$) is the solution of: $$ z+\bar z-2=0 \iff 2x-2=0 \iff x=1 $$

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Note that $z^2+z\bar z-2z=0\implies (2x^2-2x)+i2(x-1)y=0$.

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    ,MV $2i(x-1)y$ ?? – 2017-02-14
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    @DonaldSplutterwit Thank you for catching the erratum. I've edited accordingly. – 2017-02-14
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\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*}