LetS $=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $\left|z_1\right|= \max _{z \in S}|z|$ and $\left|z_2\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals:
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