Let S={z ∈ C:|z-1|=1 and (√ 2 -1)(z+z̄)-i(z-z̄)=2 √ 2 }. Let z_1, z_2 ∈ S be such that |z_1|= max_Z ∈ S |z| and |z_2|= min_Z ∈ S |z|. Then |√ 2…

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