For a complex number $z,$ let $\text{Re}\left(z\right)$ denote the real part of $z.$ Let $S$ be the set of all complex numbers $z$ satisfying $z^4-{\left|z\right|}^4=4i{z}^2,$ where $i=\sqrt{-1}.$ Then the minimum possible value of ${|z_1-z_2|}^2,$ where $z_1,z_2\in S$ with $\text{Re}\left(z_1\right)>0$ and $\text{Re}\left(z_2\right)<0,$ is

$z^4-{\left|z\right|}^4=4i{z}^2$
$\Rightarrow z^4-z^2{\overline{z}}^2=4i{z}^2(\because z\overline{z}={\left|z\right|}^2)$
$\Rightarrow z^2(z^2-{\left(\overline{z}\right)}^2-4i)=0$
$\Rightarrow z^2=0$ or $z^2-{\left(\overline{z}\right)}^2=4i$
Let $z=x+iy$
$z^2-{\left(\overline{z}\right)}^2=4i$
$\Rightarrow {(x+iy)}^2-{(x-iy)}^2=4i$
$\Rightarrow x^2-y^2+2ixy-(x^2-y^2-2ixy)=4i$
$\Rightarrow 4ixy=4i$
$\Rightarrow xy=1$

Now, ${|z_1-z_2|}^2={(x_1-x_2)}^2+{(y_1-y_2)}^2$
$=x_1^2+x_2^2+y_1^2+y_2^2-2x_1{x}_2-2y_1{y}_2$
$=x_1^2+x_2^2+y_1^2+y_2^2+2x_1(-x_2)+2y_1(-y_2)$
Applying $\text{A.M.}\ge \text{G.M.},$
$\frac{(x_1^2+x_2^2+y_1^2+y_2^2+x_1(-x_2)+x_1(-x_2)+y_1(-y_2)+y_1(-y_2))}{8}$
$\ge {\left(x_1^2{x}_2^2{y}_1^2{y}_2^2{x}_1^2{x}_2^2{y}_1^2{y}_2^2\right)}^{1/8}$
$\ge {\left[\left(x_1{y}_1\right)\left(x_2{y}_2\right)\right]}^{1/2}=1$

$\therefore {|z_1-z_2|}^2\ge 8$