Let $S$ be the set of all complex numbers $z$ satisfying $z^{4}-|z|^{4}=4 i z^{2}$, where $i=\sqrt{-1}$. Then the minimum possible value of $|z_{1}-z_{2}|^{2}$, where $z_{1}, z_{2} \in S$ with $\operatorname{Re}(z_{1})>0$ and $\operatorname{Re}(z_{2})$ $<0$, is

$\text{Given: } z^{4} - |z|^{4} = 4iz^{2}, \quad i=\sqrt{-1}$ $\text{To find: } \min |z_{1}-z_{2}|^{2}, \ z_{1},z_{2} \in S,$ $\operatorname{Re}(z_{1})>0, \ \operatorname{Re}(z_{2})<0$ $|z|^{2} = z\bar z \Rightarrow |z|^{4} = z^{2}\bar z^{2}$ $z^{4}-|z|^{4} = z^{2}\big(z^{2}-\bar z^{2}\big)$ $z^{2}-\bar z^{2} = (z-\bar z)(z+\bar z)$ $z^{4}-|z|^{4} = z^{2}(z-\bar z)(z+\bar z)$ $z-\bar z = 2iy, \quad z+\bar z = 2x$ $z^{4}-|z|^{4} = 4ixy\,z^{2}$ $4ixy\,z^{2} = 4i z^{2}$ $4iz^{2}(xy-1)=0$ $z=0 \quad \text{or} \quad xy=1$ $S = \{\,x+iy : xy=1\,\} \cup \{0\}$ $\operatorname{Re}(z_{1})>0 \Rightarrow z_{1} \text{ on branch } xy=1,\ x>0$ $\operatorname{Re}(z_{2})<0 \Rightarrow z_{2} \text{ on branch } xy=1,\ x<0$ $z_{1} = a + \dfrac{i}{a}, \quad a>0$ $z_{2} = b + \dfrac{i}{b}, \quad b<0$ $|z_{1}-z_{2}|^{2} = (a-b)^{2} + \left(\dfrac{1}{a}-\dfrac{1}{b}\right)^{2}$ $b=-c \Rightarrow d = (a+c)^{2} + \left(\dfrac{1}{a}+\dfrac{1}{c}\right)^{2}$ $d = \left(a^{2}+\dfrac{1}{a^{2}}\right) + \left(c^{2}+\dfrac{1}{c^{2}}\right) + 2\left(ac+\dfrac{1}{ac}\right)$ $a^{2}+\dfrac{1}{a^{2}} \geq 2, \quad c^{2}+\dfrac{1}{c^{2}} \geq 2, \quad ac+\dfrac{1}{ac} \geq 2$ $d \geq 2+2+2(2) = 8$ $z_{1}=1+i,\ z_{2}=-1-i \Rightarrow |z_{1}-z_{2}|^{2} = 8$ $\boxed{8}$