Let S = z ∈ C : z − 1 = 1 and 2 − 1 z + z ¯ - i z - z ¯ = 2 2 . Let z 1 , z 2 ∈ S be such that z 1 = max z ∈…

Given,

$\left|z-1\right|=1$

Now, let $z=x+iy$

$\Rightarrow {\left(x-1\right)}^2+y^2=1 ...\left(i\right)$

And 

$\left(\sqrt{2}-1\right)\left(z+\overline{z}\right)-i\left(z-\overline{z}\right)=2\sqrt{2}$

$\Rightarrow \left(\sqrt{2}-1\right)\left(2x\right)-i\left(2iy\right)=2\sqrt{2}$

$\Rightarrow \left(\sqrt{2}-1\right)x+y=\sqrt{2} ...\left(ii\right)$

Solving $\left(i\right) \text{and} \left(ii\right)$,

$\Rightarrow {\left(x-1\right)}^2+{\left[\sqrt{2}-\left(\sqrt{2}-1\right)x\right]}^2=1$

$\Rightarrow x^2+1-2x+2+{\left(\sqrt{2}-1\right)}^2{x}^2-2\sqrt{2}\left(\sqrt{2}-1\right)x=1$

$\Rightarrow x^2+1-2x+2+\left(2+1-2\sqrt{2}\right)x^2-4x+2\sqrt{2}x=1$

$\Rightarrow x^2-2x+2+3x^2-2x^2\sqrt{2}-4x+2\sqrt{2}x=0$

$\Rightarrow 4x^2-2x^2\sqrt{2}-6x+2\sqrt{2}x+2=0$

$\Rightarrow \left(4-2\sqrt{2}\right)x^2+\left(-6+2\sqrt{2}\right)x+2=0$

$\Rightarrow \left(4-2\sqrt{2}\right)x^2+\left(-4+2\sqrt{2}\right)x-2x+2=0$

$\Rightarrow \left(4-2\sqrt{2}\right)x\left(x-1\right)-2\left(x-1\right)=0$

$\Rightarrow \left(x-1\right)\left(\left(4-2\sqrt{2}\right)x-2\right)=0$

Either $x=1$ or $x=\frac{1}{2-\sqrt{2}} ...\left(iii\right)$

On solving $\left(ii\right) \text{and} \left(iii\right)$ we get

For $x=1\Rightarrow y=1\Rightarrow z_2=1+i$ and for $x=\frac{1}{2-\sqrt{2}}\Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}}\Rightarrow z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$

$\Rightarrow {\left|\sqrt{2}{z}_1-z_2\right|}^2={\left|\left(\frac{1}{\sqrt{2}}+1\right)\sqrt{2}+i-\left(1+i\right)\right|}^2$

$\Rightarrow {\left|\sqrt{2}{z}_1-z_2\right|}^2={\left|1+\sqrt{2}+i-\left(1+i\right)\right|}^2$

$\Rightarrow {\left|\sqrt{2}{z}_1-z_2\right|}^2={\left(\sqrt{2}\right)}^2$

$\Rightarrow {\left|\sqrt{2}{z}_1-z_2\right|}^2=2$