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Mathematics Question on complex numbers

Let S=z∈C:∣z−1∣=1S = \\{ z \in \mathbb{C} : |z - 1| = 1 \\} and (2−1)(z+z‾)−i(z−z‾)=22.(\sqrt{2} - 1)(z + \overline{z}) - i(z - \overline{z}) = 2\sqrt{2}.
Let z1,z2∈Sz_1, z_2 \in S be such that ∣z1∣=max⁡z∈S∣z∣|z_1| = \max_{z \in S} |z| and ∣z2∣=min⁡z∈S∣z∣|z_2| = \min_{z \in S} |z|.
Then 2∣z1−z2∣2\sqrt{2}|z_1 - z_2|^2 equals:

A

1

B

4

C

3

D

2

Answer

2

Explanation

Solution

Let Z=x+iyZ = x + iy.

Then (x−1)2+y2=1⋯(1)\text{Then } (x - 1)^2 + y^2 = 1 \quad \cdots (1) and (2−1)(2x)+i(2y)=22\text{and } (\sqrt{2} - 1)(2x) + i(2y) = 2\sqrt{2}

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

Solving (1) and (2), we get:
x=1orx=−12⋯(3)x = 1 \quad \text{or} \quad x = -\frac{1}{\sqrt{2}} \quad \cdots (3)

On solving (3) with (2), we get:
For x=1  ⟹  y=1  ⟹  Z1=1+i\text{For } x = 1 \implies y = 1 \implies Z_1 = 1 + i
and for
x=−12  ⟹  y=2−12  ⟹  Z2=(−12)+i(2−12).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:
2∣Z1−Z2∣2\sqrt{2} |Z_1 - Z_2|^2

=∣(1+12)2+i(1−(2−1))∣2= \left| \left( 1 + \frac{1}{\sqrt{2}} \right) \sqrt{2} + i\left( 1 - (\sqrt{2} - 1) \right) \right|^2

=∣(2)2∣=2= |(\sqrt{2})^2| = 2