Let S = { z ? C : | z - 1 | = 1 and ( 2 - 1 ) ( z + z ¯ ) - i ( z - z ¯ ) = 2 2 } . Let z 1 , z 2 ? S be such that | z 1 | = max z ? S | z | and | z 2 | = min z ? S | z | . Then | 2 z 1 - z 2 | 2 equals: [2024]

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Solution

Q.
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 $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

1 3

2 2

3 1

4 4

Ans.
(2)

We have, $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}}$

Let $z = x + i y$ and $\bar{z} = x - i y$

${(x - 1)}^{2} + y^{2} = 1$ ...(i)

$(\sqrt{2} - 1) x + y = \sqrt{2}$ ...(ii)

Equation (i) and (ii) intersect at (1, 1) and $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

   $∴$ $S = {(1, 1), (1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})}$

   $| z_{1} |$ $= \max_{z \in S}$ $| z |$ at $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

   $∴$    $z_{1} = 1 + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

   $z_{2} = 1 + i$

$∴$    $\sqrt{2} z + 1 + i - 1 - i = \sqrt{2}$

$∴$    $| \sqrt{2} z_{1} - z_{2} |^{2} = 2$

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