Let S=|z ∈ C:| z-1 |=1 and (√2-1)(z+z̄)-i(z-z̄)=2 √2 |. Let z₁, z₂ ∈ S be such that |z₁|=max(z ∈ s)|z| and |z₂|=min _(z ∈ S)|z|. Then |√2 z₁-z₂|² equals :

Solution

Maths sir · 4 steps

Turn both conditions into x and y: a circle meets a line, so S is just two points to compare.

1. Use x and y. Put z = x + iy. Then z + z̄ = 2x and -i(z - z̄) = -i(2iy) = 2y. So S is where the circle (x-1)² + y² = 1 meets the line (√2-1)x + y = √2.
2. Solve together. Substitute y = √2 - (√2-1)x into the circle and divide by 2. This gives (2-√2)x² - (3-√2)x + 1 = 0. Here x = 1 is a root, and the product of the roots is 1/(2-√2) = 1 + 1/√2, so the other root is x = 1 + 1/√2.

   Point	z	|z|²
   x = 1, y = 1	1 + i	2, the smaller, so this is z₂
   x = 1 + 1/√2, y = 1/√2	(1 + 1/√2) + i/√2	2 + √2, the larger, so this is z₁

3. Combine them. √2z₁ = (√2 + 1) + i. Subtracting z₂ = 1 + i leaves √2z₁ - z₂ = √2, a purely real number.
4. So. |√2z₁ - z₂|² = (√2)² = 2, which is option (d).

The idea to keep. When z and z̄ appear together, switch to x and y at once. A circle meeting a line gives at most two points, so max and min of |z| just means comparing them.

Answer from the NTA final answer key. Solution by PrepWiser.