JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given equation is z^2 + 4z - (1 + 12i) = 0 Let the roots be z_1 and z_2. Sum of roots: z_1 + z_2 = -4 Product of roots: z_1 z_2 = -(1 + 12i) We know that (z_1 - z_2)^2 = (z_1 + z_2)^2 - 4z_1 z_2 (z_1 - z_2)^2 = (-4)^2 - 4(-(1 + 12i)) = 16 + 4 + 48i = 20 + 48i Taking modulus on both sides: |z_1 - z_2|^2 = |20 + 48i| = 20^2 + 48^2 = 400 + 2304 = 2704 = 52 Using the parallelogram law for complex numbers: |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2) |-4|^2 + 52 = 2(|z_1|^2 + |z_2|^2) 16 + 52 = 2(|z_1|^2 + |z_2|^2) 68 = 2(|z_1|^2 + |z_2|^2) |z_1|^2 + |z_2|^2 = 34 Answer: 34