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 \((\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\} \). Let \(z_{1}, z_{2}\in S\) 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 :

(1) 1

(2) 4

(3) 3

(4) 2

Answer 1

Correct option is (4) 2

Let \(Z=x+ iy\)

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

\(\&\ (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}\)

\( \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2}\quad\quad ....(2)\)

Solving (1) & (2) we get

Either \(\mathrm{x}=1\ or \ x=\frac{1}{2-\sqrt{2}} \quad\quad ....(3)\)

On solving (3) with (2) we get

For \(\mathrm{x}=1 \Rightarrow \mathrm{y}=1 \Rightarrow \mathrm{Z}_{2}=1+\mathrm{i}\)

& 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

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

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

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

\(=2\)