Let z₁, z₂ and z₃ be three complex numbers on the circle |z|=1 with arg (z₁)=-π/4, arg (z₂)=0 and arg (z₃)=π/4. If |z₁ z̄₂+z₂ z̄₃+z₃ z̄₁|²=α+β √2, α, β ∈ Z, then the value of α²+β² is :

Solution

Maths sir · 4 steps

On the unit circle, zw̄ is just e^(i(θ - φ)), so subtract the arguments and add three points.

1. Use polar form. Each number has modulus 1, so z₁ = e^(-iπ/4), z₂ = 1 and z₃ = e^(iπ/4). For such numbers zw̄ = eⁱ⁽ᵃʳᵍᶻ⁻ᵃʳᵍʷ⁾.
2. Each product. z₁z̄₂ = e^(-iπ/4), z₂z̄₃ = e^(-iπ/4) and z₃z̄₁ = e^(iπ/2) = i. Since e^(-iπ/4) = 1/√2 - i/√2, the first two together give √2 - i√2.

   z₁z̄₂ + z₂z̄₃ + z₃z̄₁ = √2 + i(1 - √2)

3. Square the modulus. | · |² = (√2)² + (1-√2)² = 2 + (3 - 2√2) = 5 - 2√2. So α = 5 and β = -2.
4. So. α² + β² = 25 + 4 = 29, which is option (b).

The idea to keep. When every number sits on |z| = 1, write it as e^(iθ). Products with conjugates only subtract angles, so the algebra becomes adding a few known points.

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