Question

Let \( S = \{ z \in \mathbb{C} : |z - 1| = 1 \} \) and \((\sqrt{2} - 1)(z + \overline{z}) - i(z - \overline{z}) = 2\sqrt{2}.\)
Let \( z_1, z_2 \in S \) be such that \( |z_1| = \max_{z \in S} |z| \) and \( |z_2| = \min_{z \in S} |z| \).  
Then \( \sqrt{2}|z_1 - z_2|^2 \) equals:

• A. 1
• B. 4
• C. 3
• D. 2

Answer 1

The Correct Option is D

Let \( Z = x + iy \).

\(\text{Then } (x - 1)^2 + y^2 = 1 \quad \cdots (1)\) \(\text{and } (\sqrt{2} - 1)(2x) + i(2y) = 2\sqrt{2}\)

\(\implies (\sqrt{2} - 1)x + y = \sqrt{2} \quad \cdots (2)\)

Solving (1) and (2), we get:
\(x = 1 \quad \text{or} \quad x = -\frac{1}{\sqrt{2}} \quad \cdots (3)\)

On solving (3) with (2), we get:
\(\text{For } x = 1 \implies y = 1 \implies Z_1 = 1 + i\)
and for
\(x = -\frac{1}{\sqrt{2}} \implies y = \sqrt{2} - \frac{1}{\sqrt{2}} \implies Z_2 = \left( -\frac{1}{\sqrt{2}} \right) + i \left( \sqrt{2} - \frac{1}{\sqrt{2}} \right).\)

Now:
\(\sqrt{2} |Z_1 - Z_2|^2\)

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

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