<s>[INST] 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
 [/INST] First of all, clearly $z = 0$ works, so let's just assume $z \neq 0$.
Let $z = r e^{i \theta}$. We have:
\begin{align}
\bar{z} - z^2 &= i(\bar{z} + z^2) \\
(1 - i)\bar{z} &= (1 + i)z^2 \\
\frac{1 - i}{1 + i} &= \frac{z^2}{\bar{z}} \\
e^{-i \pi/2} &= r e^{3i\theta}
\end{align}
This shows that $r = 1$ and $\theta = -\frac{\pi}{6} + \frac{2 \pi}{3} n$ for some $n \in \mathbb{Z}$.
In summary, the solutions are $0, e^{-i\pi/6}, e^{i 3\pi / 6}$ and $e^{i 7\pi / 6}$.
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