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: $$-z + \frac {3} {\overline z}=2$$
 A: Hint: Multiplying both sides by $\bar z$ and using $z\bar z = |z|^2$, you will see that $2\bar z$ is a real number. So $\bar z = z$ and it becomes just a quadratic equation.
A: Multiply by $\bar{z}$ to get 
$$\vert z \vert^2 + 2z^* - 3 = 0$$
Let $z = x + jy$, so
$$x^2 + y^2 + 2(x-jy) - 3 =0$$
which gives real part equal zero
$$x^2 + y^2 +2x - 3 = 0$$
and imaginary part is zero
$$-2y = 0$$
Hence $y = 0$
and 
$$x^2 + 2x - 3 = 0$$
Solve the above in $\mathbb{R}$ to get $x = -3$ and $x = 1$. So $z = -3$ and $z = 1$
A: We have that $z\not=0$ and by multiplying both sides by $\overline{z}$ we obtain
$$-z\cdot\overline {z}+3=2\overline{z},$$
Now let $z=x+iy$ with $x,y\in\mathbb{R}$ and we get
$$-(x+iy)(x-iy)+3=2(x-iy).$$
Can you take it from here?
A: $$-z + \frac {3} {\overline z}=2$$
$$-z \overline z+3=2\overline z$$
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 $z=1$.
A: Multiply both sides by $\overline z$ and since $z\overline z=|z|^2$, you'll get
$$
2\overline z +|z|^2=3
$$
put then $z=a+ib$, from which last equation turns into
$$
2(a-ib)+a^2+b^2-3=0
$$
so you get $b=0$ and $a^2+2a-3=0$ once solved gives $a=1,-3$.
Thus the solutions to your equation are $z=1$ and $z=-3$.
