solutions of $\bar z = |z-2\Im(z)|^2$.

Question

I need to find all the solutions of $\bar z = |z-2\Im(z)|^2$.

I know that $z=x+iy$ and $\bar z=x-iy$ and then $2\Im(z)=2y$. But can someone show the algebra for what I do next?

Answer 1

Since the right side of $\bar z = |z-2\Im(z)|^2$ is a real number, so we should have $y=0$ for $z=x+iy$ on the left side. Now, the equation will be $x=|x|^2$ with the only answers of $x=0,1$.

Answer 2

You get by your ideas that $$\overline z = \vert z - \Im(z)\vert^2 \iff x - iy = \vert x + iy - 2y\vert^2$$ and hence $$x - iy = (x - 2y)^2 + y^2.$$ By comparing these terms you get that $y = 0$ and it follows by substitution that $x = x^2$. Hence you have $x \in \{0, 1\}$ and $y = 0$. So as your solution you have $z \in \{0 ,1\}$. Please check my answer if you are satisfied with it :)