JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let z = x + iy, then z = x - iy. Substituting z into the given equation: (x + iy)(x - iy + 2 + i) + k(2 + 3i) = 0 x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0 Separating the real and imaginary parts, we get: Real part: x^2 + y^2 + 2x - y + 2k = 0 Imaginary part: x + 2y + 3k = 0 x = -2y - 3k Substituting x into the real part equation: (-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0 4y^2 + 12ky + 9k^2 + y^2 - 4y - 6k - y + 2k = 0 5y^2 + (12k - 5)y + 9k^2 - 4k = 0 For the equation to have at least one solution z C , there must be at least one real value of y. Thus, the discriminant of this quadratic equation in y must be non-negative ( 0): = (12k - 5)^2 - 4(5)(9k^2 - 4k) 0 144k^2 - 120k + 25 - 180k^2 + 80k 0 -36k^2 - 40k + 25 0 36k^2 + 40k - 25 0 The values of k lie in the interval [ , ], where and are the roots of the equation 36k^2 + 40k - 25 = 0. The sum of the roots is given by: + = - 40 36 = - 10 9 Therefore, 9( + ) = 9 ( - 10 9 ) = -10. Answer: -10