Let S = \ z ∈ C: |z - 1| = 1and( 2 - 1)(z + ¯ z ) - i(z - ¯ z ) = 2...

Given,

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

Now, let $z = x + iy$

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

And 

\left(\sqrt{2} - 1\right)\left(z + \overset{¯}{z}\right) - i\left(z - \overset{¯}{z}\right) = 2\sqrt{2}$</p><p>$⇒ \left(\sqrt{2} - 1\right)\left(2x\right) - i\left(2iy\right) = 2\sqrt{2}$</p><p>$⇒ \left(\sqrt{2} - 1\right)x + y = \sqrt{2} . . . \left(ii\right)$</p><p>Solving \left(i\right)\text{and}\left(ii\right)$,

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

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

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

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

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

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

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

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

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

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

On solving \left(ii\right)\text{and}\left(iii\right)$we get</p><p>For$x = 1 ⇒ y = 1 ⇒ {z}_{2} = 1 + i and for $x = \dfrac{1}{2 - \sqrt{2}} ⇒ y = \sqrt{2} - \dfrac{1}{\sqrt{2}} ⇒ {z}_{1} = \left(1 + \dfrac{1}{\sqrt{2}}\right) + \dfrac{i}{\sqrt{2}}$

$⇒ {\left|\sqrt{2}{z}_{1} - {z}_{2}\right|}^{2} = {\left|\left(\dfrac{1}{\sqrt{2}} + 1\right)\sqrt{2} + i - \left(1 + i\right)\right|}^{2}$

$⇒ {\left|\sqrt{2}{z}_{1} - {z}_{2}\right|}^{2} = {\left|1 + \sqrt{2} + i - \left(1 + i\right)\right|}^{2}$

$⇒ {\left|\sqrt{2}{z}_{1} - {z}_{2}\right|}^{2} = {\left(\sqrt{2}\right)}^{2}$

$⇒ {\left|\sqrt{2}{z}_{1} - {z}_{2}\right|}^{2} = 2$