Let α and β be the sum and the product of all the non-zero solutions of the equation $$(\bar{z})^2 + |z| = 0,\ z \in \mathbb{C}$$. Then $$4(\alpha^2 + \beta^2)$$ is equal to:

Let z = x + iy. Then z̄ = x - iy, |z| = √(x² + y²).

(z̄)² + |z| = 0 → (x-iy)² + √(x²+y²) = 0

x² - y² - 2ixy + √(x²+y²) = 0

Real: x² - y² + √(x²+y²) = 0, Imaginary: -2xy = 0

From imaginary part: x = 0 or y = 0.

If y = 0: x² + |x| = 0 → only x = 0 (trivial).

If x = 0: -y² + |y| = 0 → |y|(1-|y|) = 0 → |y| = 1, so y = ±1.

Non-zero solutions: z = i and z = -i.

α = sum = i + (-i) = 0, β = product = i(-i) = 1

4(α² + β²) = 4(0 + 1) = 4

The correct answer is Option 4: 4.