Let α and β be the sum and the product of all the non-zero solutions of the equation (z̄)²+|z|=0, z ∈ C. Then 4(α²+β²) is equal to :

Solution

Maths sir · 4 steps

Conjugate the whole equation: z² becomes a negative real, so z must be purely imaginary.

1. Conjugate both sides. Since |z| is real, the conjugate of (z̄)² + |z| = 0 is z² + |z| = 0. So z² = -|z|, which is a real number that is negative for z ≠ 0.
2. z is purely imaginary. A complex number whose square is a negative real must lie on the imaginary axis. Put z = iy with y real and y ≠ 0. Then z² = -y² and |z| = |y|.

   -y² + |y| = 0 ⇒ |y|(1 - |y|) = 0 ⇒ |y| = 1

3. Non-zero solutions. So z = i or z = -i. Check z = i: z̄ = -i, (z̄)² = -1, and -1 + 1 = 0. The same check works for z = -i.
4. So. α = i + (-i) = 0 and β = i × (-i) = 1. Then 4(α² + β²) = 4(0 + 1) = 4, option (a).

The idea to keep. When an equation mixes z̄ and |z|, conjugating it often turns it into a condition on z itself. If z² comes out real and negative, z must be purely imaginary.

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