I'm trying to solve $z^2 = 2\bar{z}$. Using the polar form, I obtain $r^2e^{2i\theta} =2re^{-i\theta}$. Setting $r^2=2r$, I obtain $r=0$ or $r=2$. Furthermore: $2\theta = -\theta + 2\pi m$ gives $\theta = \frac{2\pi m}{3}$ for which I get the real solution $z=2$ as well as two complex solutions $z=2e^{2\pi /3}$ and $z=2e^{4\pi /3}$ when I let $m$ equal to $0$, $1$ and $2$.
I want to challenge myself slightly more by ''imagining'' that we have $z^2=1+\bar{2z}$ instead. In the other case I compare $r^2$ to $2r$, aswell as the angles directly, but in this case I don't see how to solve the equation appropriately
I would rewrite in rectangular form. You get $$ x^2 + 2ixy + (iy)^2 = 1 + 2x - 2iy $$ Equate the real and imaginary components and solve the system of 2 equations in 2 unknowns.