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

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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?

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$$\frac{(x+iy)-\dfrac1{(x+iy)}}{2i}=2$$

$$\iff(x+iy)-\dfrac1{(x+iy)}=2(2i)$$

$$\iff(x+iy)^2-1=4i(x+iy)$$

$$\iff x^2-y^2-1+(2xy)i=(4x)i-4y$$

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, $x^2\ge0$