I'm having troubles, finding how solution would look like for complex equation of the form $\bar{z} = |z|$. Taking $z = x + iy$, we get the following: $$x - iy = \sqrt{x^2 + y^2},$$ then raising it to the 2nd power we get $$x^2 - 2ixy - y^2 = x^2 + y^2,$$ $$2y^2 = -2ixy.$$ Now, taking complex terms as one equation and real as second, we construct the system of equations: $$-2xy = 0$$ $$2y^2 = 0$$ Which again yields equation: $$y^2 + xy = 0.$$ Here we get that $\Im{(z)} = 0$. But what is $\Re{(z)}$ equal to?
Is my solution correct?
From $2y^2 = 0$, you have $y=0$ or $\Im(z) = 0$. But then you have the problem that $x$ can be anything. This is because by squaring the first equation, you "threw away" the information that the solutions require $x \geq 0$. (I.e., $x^2 = |x|^2$ for all $x \in \Bbb{R}$.)
You would be better served working with $z = r \mathrm{e}^{\mathrm{i}\theta}$ with $r \in \Bbb{R}_{\geq 0}$ and $\theta \in \Bbb{R}$. Then $$ r = |z| = \overline{z} = r \mathrm{e}^{-\mathrm{i}\theta} \text{.} $$ If $r=0$, we have the solution $z=0$. If $r \neq 0$, we can divide through, getting $\mathrm{e}^{-\mathrm{i}\theta} = 1$, so $\theta = 2 \pi k$ for $k \in \Bbb{Z}$. That is, $z \in \Bbb{R}_{>0}$.
Gluing the two solution sets together, we have $z \in \Bbb{R}_{\geq 0}$.