Let $z=x+iy \, ,\quad x,y\in \mathbb R.$ Then $|z|^2=x^2+y^2,$ so $|z|^4=(x^2+y^2)^2.$

We are given $z^4-|z|^4=4iz^2.$ We must find all such $z$, then minimize $|z_1-z_2|^2$ for $z_1,z_2\in S$ with $\operatorname{Re}(z_1)>0$ and $\operatorname{Re}(z_2)<0$.

1. Use polar form

Let $z=re^{i\theta},\quad r\ge 0.$ Then $z^2=r^2e^{i2\theta},\qquad z^4=r^4e^{i4\theta},\qquad |z|^4=r^4.$ So the equation becomes $r^4(e^{i4\theta}-1)=4ir^2e^{i2\theta}.$

Case 1: $r=0$

Then $z=0$, which satisfies the equation. But it cannot be used in the final condition since its real part is neither positive nor negative.

Case 2: $r>0$

Divide by $r^2e^{i2\theta}$: $r^2\frac{e^{i4\theta}-1}{e^{i2\theta}}=4i.$ Now $\frac{e^{i4\theta}-1}{e^{i2\theta}}=e^{i2\theta}-e^{-i2\theta}=2i\sin 2\theta.$ Hence $r^2(2i\sin 2\theta)=4i.$ Cancelling $2i$, $r^2\sin 2\theta=2.$ Thus every nonzero solution satisfies r^2\sin 2\theta=2.\tag{1}

Since $\sin 2\theta\le 1$, we get $r^2\ge 2.$

2. Convert condition into Cartesian form

Using $\sin 2\theta=\frac{2xy}{r^2},$ equation (1) gives $r^2\cdot \frac{2xy}{r^2}=2$ which simplifies to $2xy=2\implies xy=1.$

So the set of all nonzero solutions is $S\setminus\{0\}=\{x+iy:xy=1\}.$ Including $0$, the full solution set is $S=\{0\}\cup\{x+iy:xy=1\}.$

For the required condition:

• $\operatorname{Re}(z_1)>0 \Rightarrow x_1>0$,
• $\operatorname{Re}(z_2)<0 \Rightarrow x_2<0$.

Since $xy=1$, $x$ and $y$ have the same sign. Therefore:

• if $x_1>0$, then $y_1=1/x_1>0$,
• if $x_2<0$, then $y_2=1/x_2<0$.

So write $z_1=x_1+i\frac1{x_1},\quad x_1>0,$ $z_2=x_2+i\frac1{x_2},\quad x_2<0.$

3. Compute $|z_1-z_2|^2$

We have $|z_1-z_2|^2=(x_1-x_2)^2+\left(\frac1{x_1}-\frac1{x_2}\right)^2.$ Let $x_2=-t,\quad t>0.$ Also let $x_1=s,\quad s>0.$ Then $z_1=s+i\frac1s,\qquad z_2=-t-i\frac1t.$ Hence $|z_1-z_2|^2=(s+t)^2+\left(\frac1s+\frac1t\right)^2.$ So we must minimize $f(s,t)=(s+t)^2+\left(\frac1s+\frac1t\right)^2,\qquad s,t>0.$

4. Minimize the expression

Set $u=s+t,\qquad v=\frac1s+\frac1t.$ By AM-GM, $s+t\ge 2\sqrt{st},$ and also $\frac1s+\frac1t\ge \frac{2}{\sqrt{st}}.$ Thus $f(s,t)=u^2+v^2\ge 4st+\frac{4}{st}.$ Now let $p=st>0.$ Then $f(s,t)\ge 4\left(p+\frac1p\right).$ By AM-GM, $p+\frac1p\ge 2,$ so $f(s,t)\ge 8.$ Equality holds when simultaneously $s=t$ and $p=1\Rightarrow st=1.$ Therefore $s=t=1.$

This gives $z_1=1+i,\qquad z_2=-1-i,$ and indeed both satisfy $xy=1$. Then $|z_1-z_2|^2=|(1+i)-(-1-i)|^2=|2+2i|^2=4+4=8.$

5. Final answer

The minimum possible value is $\boxed{8}.$

This matches the stored correct answer.