QJEEM24Ma170d: JEE Main 2024 Mathematics - Class 11 JEEM24Ma170d Solution: Let and . Let be such that and . Then equals :

Question

Verified Solution • 2025–26 Session moderate

Let

\mathrm{S}=|\mathrm{z} \in \mathrm{C}:| z-1 \mid=1

and

(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \mid

. Let

z_1, z_2 \in \mathrm{S}

be such that

\left|z_1\right|=\max\limits_{z \in s}|z|

and

\left|z_2\right|=\min\limits _{z \in S}|z|

. Then

\left|\sqrt{2} z_1-z_2\right|^2

equals :

Accepted Answer

Let

z=x+i y

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

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

Solving (1) and (2), we get

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

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

Now,

\begin{aligned}&amp; \left|\sqrt{2} z_1-z_2\right|^2 \\\\&amp; =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\\\&amp; =(\sqrt{2})^2 \\\\&amp; =2\end{aligned}

Step-by-Step Verified Solution