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

Question

Let $S=|z\in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\vec{z})-i(z-\vec{z})=2\sqrt{2}|$. Let ${z}_{1},{z}_{2}\in S$ be such that $|{z}_{1}|=\underset{z\in s}{max}|z|$ and $|{z}_{2}|=\underset{z\in S}{min}|z|$. Then ${|\sqrt{2}{z}_{1}-{z}_{2}|}^{2}$ equals :

Accepted Answer

Let $z=x+iy$

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

$\begin{gathered} (\sqrt{2}-1)(z+\vec{z})-i(z-\vec{z})=2\sqrt{2} \ (\sqrt{2}-1)(2x)-i(2iy)=2\sqrt{2} \ (\sqrt{2}-1)x+y=\sqrt{2}.........(2) \end{gathered}$

Solving (1) and (2), we get

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

$\begin{gathered} \Rightarrow x=1x=\frac{1}{2-\sqrt{2}}..........(3) \ x=1,y=1\Rightarrow {z}_{2}=1+i \ x=\frac{1}{2-\sqrt{2}},y=\sqrt{2}-\frac{1}{\sqrt{2}} \ \Rightarrow {z}_{1}=(1+\frac{1}{\sqrt{2}})+\frac{i}{\sqrt{2}} \end{gathered}$

Now,

$\begin{gathered} {|\sqrt{2}{z}_{1}-{z}_{2}|}^{2} \ ={|(\frac{1}{\sqrt{2}}+1)\sqrt{2}+i-(1+i)|}^{2} \ =(\sqrt{2}{)}^{2} \ =2 \end{gathered}$

Solutions · 1