How to find results for $-z + \frac {3} {\overline {z}}=2$ Can you just please help me solve this problem, because i don't know how to solve it: $$-z + \frac {3} {\overline z}=2$$

$$-z + \frac {3} {\overline z}=2$$
$$-z \overline z+3=2\overline z$$
As $z \overline{z}=\left|{z^2}\right|$,
$$
\begin{align}
-\left|{z^2}\right|+3&=2 \overline{z}\\
&= 2a-2bi \\
\end{align}
$$
Equating imaginary parts,
$$2\Im{(z)}=0\implies 2b=0\implies b=0$$
Equating real parts,
$$
\begin{align}
2 \Re{(z)} &= -(a^2+b^2)+3 \\
2a &= -a^2+3 \\
a^2+2a-3 &= 0 \\
(a+3)(a-1) &= 0 \\
a &= -3,\,1 \\
\end{align}
$$
So $z=-3$ or $z=1$.
