Multiple choice

If roots of the equation $z^2 + \alpha z + \beta = 0 (\alpha, \beta \in C)$ are real, then value of $(Im(\beta))^2 + (Im(\alpha))(Im(\alpha\bar\beta))$ is Note: $Im(z)$ denotes imaginary part of $z$, where $z$ is a complex number.

  1. $0$
  2. $1$
  3. $-1$
  4. $i$
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A Correct answer
Explanation

If roots are real, then the discriminant D = alpha^2 - 4*beta >= 0. Since alpha, beta are complex, this implies specific relations. However, for a quadratic with real roots, the coefficients must be real if the roots are real. If alpha and beta are real, then Im(alpha) = 0 and Im(beta) = 0. Thus, (Im(beta))^2 + Im(alpha)*Im(alpha*conj(beta)) = 0 + 0 = 0.

AI explanation

Let alpha = a + ib and beta = c + id where a, b, c, d are real numbers, so Im(alpha) = b and Im(beta) = d. The given expression expands to d^2 + b(Im((a + ib)(c - id))) = d^2 + b(bc + ad). If the roots of z^2 + alpha z + beta = 0 are real, the discriminant alpha^2 - 4beta must be non-negative, forcing the imaginary parts to satisfy b^2 - 4d = 0, so d = b^2 / 4. Because the real roots are given by the quadratic formula, the imaginary part of the roots (-alpha/2) must be zero, implying b = 0. Substituting b = 0 into the expression yields d = 0, making the sum evaluate to 0.

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