Solve the equation $((x+y i)-\frac{1}{x+y i})/{(2 i)} = 2$ 
Solve the equation $$\frac{\left((x+y i)-\frac{1}{x+y i}\right)}{2 i} = 2$$

So far, I got $(0, 2-\sqrt{3}i)$ and $(0, 2+ \sqrt{3}i)$ as solutions for $x$ and $y$. Do I require $2$ more solutions?

$$\frac{(x+iy)-\dfrac1{(x+iy)}}{2i}=2$$
$$\iff(x+iy)-\dfrac1{(x+iy)}=2(2i)$$
$$\iff(x+iy)^2-1=4i(x+iy)$$
$$\iff x^2-y^2-1+(2xy)i=(4x)i-4y$$
Equating the real parts, $x^2-y^2-1=-4y$
Equating the imaginary parts,  $2xy=4x\iff2x(y-2)=0$ 
If $x=0, -y^2-1=-4y\iff y^2-4y+1=0,\implies y=?$
If  $y-2=0\iff y=2, x^2-2^2-1=-4(2)\implies x^2=?$
But if $x$ is real, $x^2\ge0$
