A quadratic polynomial (z - a)(z - b) over complex numbers is said to be square invariant if (z - a)(z - b) = (z - a^2)(z - b^2). Suppose from the set of all square invariant quadratic polynomials we choose one at random. The probability that the roots of the chosen polynomial are equal is __________ (rounded off to one decimal place).

GATE 2025 · Engineering Mathematics · Square Invariant · medium

Answer: P(roots equal) = 2/4 = 0.5

1. Case A — each root is a fixed point of squaring: Both a and b independently in {0,1}. Distinct polynomials: roots (0,0) giving z^2; roots (0,1) giving z(z-1); roots (1,1) giving (z-1)^2.
2. Case B — roots swap under squaring: a=0 or a=1 lead to pairs already counted. New solution: a = omega, b = omega^2 where omega = e^(2*pi*i/3), a primitive cube root of unity. Polynomial: (z-omega)(z-omega^2) = z^2 + z + 1.
3. List all 4 distinct square invariant polynomials: 1) z^2 (equal roots: 0,0) 2) z(z-1) (distinct roots: 0,1) 3) (z-1)^2 (equal roots: 1,1) 4) z^2+z+1 (distinct roots: omega, omega^2).
4. Compute probability of equal roots: Polynomials with equal roots: z^2 and (z-1)^2. Count = 2. P(equal roots) = 2/4 = 0.5.