JEE Mains · Maths · STD 11 - 4.1 complex nubers
Lets \(S=\{z \in C:|z-1|=1\) and \((\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}\). Let \(\mathrm{z}_1, \mathrm{z}_2\) \(\in S\) be such that \(\left|z_1\right|=\max _{z \in S}|z|\) and \(\left|z_2\right|=\min _{z \in S}|z|\). Then \(\left|\sqrt{2} z_1-z_2\right|^2\) equals :
- A \(1\)
- B \(4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
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…
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