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|=undersetz∈ s max|z| and |z_2|=undersetz∈ s min|z|. Then |√2…

Given,

$$|z-1|=1$$∣z−1∣=1

Now, let $z=x+iy$z=x+iy

$$\Rightarrow {(x-1)}^{2}+{y}^{2}=1...(i)$$⇒(x−1)2+y2=1...(i)

And

$$(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2\sqrt{2}$$(2​−1)(z+zˉ)−i(z−zˉ)=22​

$$\Rightarrow (\sqrt{2}-1)(2x)-i(2iy)=2\sqrt{2}$$⇒(2​−1)(2x)−i(2iy)=22​

$$\Rightarrow (\sqrt{2}-1)x+y=\sqrt{2}...(ii)$$⇒(2​−1)x+y=2​...(ii)

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

$$\Rightarrow {(x-1)}^{2}+{[\sqrt{2}-(\sqrt{2}-1)x]}^{2}=1$$⇒(x−1)2+[2​−(2​−1)x]2=1

$$\Rightarrow {x}^{2}+1-2x+2+{(\sqrt{2}-1)}^{2}{x}^{2}-2\sqrt{2}(\sqrt{2}-1)x=1$$⇒x2+1−2x+2+(2​−1)2x2−22​(2​−1)x=1

$$\Rightarrow {x}^{2}+1-2x+2+(2+1-2\sqrt{2}){x}^{2}-4x+2\sqrt{2}x=1$$⇒x2+1−2x+2+(2+1−22​)x2−4x+22​x=1

$$\Rightarrow {x}^{2}-2x+2+3{x}^{2}-2{x}^{2}\sqrt{2}-4x+2\sqrt{2}x=0$$⇒x2−2x+2+3x2−2x22​−4x+22​x=0

$$\Rightarrow 4{x}^{2}-2{x}^{2}\sqrt{2}-6x+2\sqrt{2}x+2=0$$⇒4x2−2x22​−6x+22​x+2=0

$$\Rightarrow (4-2\sqrt{2}){x}^{2}+(-6+2\sqrt{2})x+2=0$$⇒(4−22​)x2+(−6+22​)x+2=0

$$\Rightarrow (4-2\sqrt{2}){x}^{2}+(-4+2\sqrt{2})x-2x+2=0$$⇒(4−22​)x2+(−4+22​)x−2x+2=0

$$\Rightarrow (4-2\sqrt{2})x(x-1)-2(x-1)=0$$⇒(4−22​)x(x−1)−2(x−1)=0

$$\Rightarrow (x-1)((4-2\sqrt{2})x-2)=0$$⇒(x−1)((4−22​)x−2)=0

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

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

For $x=1\Rightarrow y=1\Rightarrow {z}_{2}=1+i$x=1⇒y=1⇒z2​=1+i and for $x=\frac{1}{2-\sqrt{2}}\Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}}\Rightarrow {z}_{1}=(1+\frac{1}{\sqrt{2}})+\frac{i}{\sqrt{2}}$x=2−2​1​⇒y=2​−2​1​⇒z1​=(1+2​1​)+2​i​

$$\Rightarrow {|\sqrt{2}{z}_{1}-{z}_{2}|}^{2}={|(\frac{1}{\sqrt{2}}+1)\sqrt{2}+i-(1+i)|}^{2}$$⇒∣2​z1​−z2​∣2=∣(2​1​+1)2​+i−(1+i)∣2

$$\Rightarrow {|\sqrt{2}{z}_{1}-{z}_{2}|}^{2}={|1+\sqrt{2}+i-(1+i)|}^{2}$$⇒∣2​z1​−z2​∣2=∣1+2​+i−(1+i)∣2

$$\Rightarrow {|\sqrt{2}{z}_{1}-{z}_{2}|}^{2}={(\sqrt{2})}^{2}$$⇒∣2​z1​−z2​∣2=(2​)2

$$\Rightarrow {|\sqrt{2}{z}_{1}-{z}_{2}|}^{2}=2$$⇒∣2​z1​−z2​∣2=2