ed} $$

$\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text { (Given) } \\\\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\\\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned}$

Solving (1) and (2), we get

$$ \begin{aligned} & (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\\\ & \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\\\ & x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\\\ & x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\\\ & x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0 \end{aligned} $$

$\begin{aligned} & \Rightarrow x=1 \text { and } x=\frac{1}{2-\sqrt{2}} ..........(3) \\\\ & \text { When } x=1, y=1 \Rightarrow z_2=1+i \\\\ & \text { When } x=\frac{1}{2-\sqrt{2}}, y=\sqrt{2}-\frac{1}{\sqrt{2}} \\\\ & \Rightarrow z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}\end{aligned}$

Now,

$$ \begin{aligned} & \left|\sqrt{2} z_1-z_2\right|^2 \\\\ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\\\ & =(\sqrt{2})^2 \\\\ & =2 \end{aligned} $$19:T513,Let $z=x+i y$

$$ \begin{aligned} & |z-1|=1 \Rightarrow|x+i y-1|=1 \\\\ & (x-1)^2+y^2=1 .......(1) \end{aligned} $$

$\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text { (दिया है) } \\\\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\\\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned}$

(1) और (2) को हल करने पर, हमें प्राप्त होता है

$$ \begin{aligned} & (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\\\ & \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\\\ & x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\\\ & x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\\\ & x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0 \end{aligned} $$

$\begin{aligned} & \Rightarrow x=1 \text { और } x=\frac{1}{2-\sqrt{2}} ..........(3) \\\\ & \text { जब } x=1, y=1 \Rightarrow z_2=1+i \\\\ & \text { जब } x=\frac

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