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

Let $z=x+i y$ <br/><br/>$$ \begin{aligned} & |z-1|=1 \Rightarrow|x+i y-1|=1 \\\\ & (x-1)^2+y^2=1 .......(1) \end{aligned} $$ <br/><br/>$\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text { (Given) } \\\\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\\\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned}$ <br/><br/>Solving (1) and (2), we get <br/><br/>$$ \begin{aligned} & (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\\\ & \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\\\ & x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\\\ & x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\\\ & x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0 \end{aligned} $$ <br/><br/>$\begin{aligned} & \Rightarrow x=1 \text { and } x=\frac{1}{2-\sqrt{2}} ..........(3) \\\\ & \text { When } x=1, y=1 \Rightarrow z_2=1+i \\\\ & \text { When } x=\frac{1}{2-\sqrt{2}}, y=\sqrt{2}-\frac{1}{\sqrt{2}} \\\\ & \Rightarrow z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}\end{aligned}$ <br/><br/>Now, <br/><br/>$$ \begin{aligned} & \left|\sqrt{2} z_1-z_2\right|^2 \\\\ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\\\ & =(\sqrt{2})^2 \\\\ & =2 \end{aligned} $$