JEE Main 2024 — 1 February, Shift 1 — Question 6

Let $\mathrm{Z}=\mathrm{x}+\mathrm{iy}$

Then $(\mathrm{x}-1)^{2}+\mathrm{y}^{2}=1 \quad \rightarrow(1)$

$\begin{aligned} \& & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} & \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)\end{aligned}$

Solving (1) & (2) we get Either $x=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow(3)$

On solving (3) with (2) we get For

$\mathrm{x}=1 \Rightarrow \mathrm{y}=1 \Rightarrow Z_{2}=1+i$ $

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 $\left|\sqrt{2} z_{1}-z_{2}\right|^{2}$

$=\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^{2}$

$=(\sqrt{2})^{2}$

$=2$