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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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    $\begingroup$ Hint: just factor out $z$ and write it as $z(z+ \bar z - 2)=0\,$. $\endgroup$ – dxiv Feb 14 '17 at 21:40
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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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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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  • $\begingroup$ Beautiful answer! $\endgroup$ – Simply Beautiful Art Feb 14 '17 at 20:48
  • $\begingroup$ thanks @SimplyBeautifulArt $\endgroup$ – Arnaldo Feb 14 '17 at 20:48
  • $\begingroup$ @Dr.MV It looks fine now ;-) $\endgroup$ – Simply Beautiful Art Feb 14 '17 at 21:02
  • $\begingroup$ @Dr. MV, that is ok. We don't need to worry about that. $\endgroup$ – Arnaldo Feb 14 '17 at 22:02
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    $\begingroup$ 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. $\endgroup$ – Paul Sinclair Feb 15 '17 at 4:27
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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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  • $\begingroup$ ,MV $2i(x-1)y$ ?? $\endgroup$ – Donald Splutterwit Feb 14 '17 at 20:46
  • $\begingroup$ @DonaldSplutterwit Thank you for catching the erratum. I've edited accordingly. $\endgroup$ – Mark Viola Feb 14 '17 at 20:48
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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*}

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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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