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माना $S=\{z \in C:|z-1|=1$ तथा $(\sqrt{2}-1)(\mathrm{z}+\overline{\mathrm{z}})-\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})=2 \sqrt{2}\}$ है
माना $z_1, z_2 \in S$ के लिए $\left|z_1\right|=\max _{z \in S}|z|$ तथा $\left|z_2\right|=\min _{z \in S}|z|$ है, तो $\left|\sqrt{2} z_1-z_2\right|^2$ बराबर है :
$1$
$4$
$3$
$2$
Solution
Let $Z=x+i y$
Then $(x-1)^2+y^2=1 \rightarrow(1)$
$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$
$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$
Solving $(1) and (2)$ we get
Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$ $(3)$
On solving $(3)$ with $(2)$ we get
For $x=1 \Rightarrow y=1 \Rightarrow Z_2=1+i$ and for
$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$
Now
$\left|\sqrt{2} z_1-z_2\right|^2$
$=\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2$
$=(\sqrt{2})^2$
$=2$
