Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation  $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as follow
$z = r{e^{i\theta }}$& $\overline z  = r{e^{ - i\theta }}$
$r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = i\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right) \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = {e^{i\frac{\pi }{2}}}\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right)$
$ \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = \left( {r{e^{ - i\left( {\theta  - \frac{\pi }{2}} \right)}} + {r^2}{e^{i\left( {2\theta  + \frac{\pi }{2}} \right)}}} \right)$
$ \Rightarrow r\left( {\cos \theta  - i\sin \theta } \right) - {r^2}\left( {\cos 2\theta  + i\sin 2\theta } \right) = \left( {r\left( {\sin \theta  + i\cos \theta } \right) + {r^2}\left( { - \sin 2\theta  + i\cos 2\theta } \right)} \right)$
$ \Rightarrow r\cos \theta  - {r^2}\cos 2\theta  - r\sin \theta  + {r^2}\sin 2\theta  - i\left( {r\sin \theta  - {r^2}\sin 2\theta  - r\cos \theta  - {r^2}\cos 2\theta } \right) = 0$
Not able to proceed further

We have
$$\bar{z} - z^2 = i(\bar{z} + z^2)$$
and multiplying by $i$
$$-\bar{z} - z^2 = i(\bar{z} -z^2)$$
then summing
$$z^2=-i\bar z\iff r^2e^{i2\theta}=re^{i\left(-\theta+\frac 3 2\pi\right)}$$
from which we can conclude that $r=0$ or $r=1$ with
$$2\theta=-\theta+\frac 3 2\pi+2k\pi \iff \theta =\frac \pi 2 +\frac 2 3 k\pi$$
