JEE Main 2026 — Mathematics Algebra

Let $z = x + iy$z=x+iy, then $\bar{z} = x - iy$zˉ=x−iy.

Substituting $z$z into the given equation:

$$(x + iy)(x - iy + 2 + i) + k(2 + 3i) = 0$$(x+iy)(x−iy+2+i)+k(2+3i)=0

$$x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0$$x2+y2+2x+ix+2iy−y+2k+3ki=0

Separating the real and imaginary parts, we get:

Real part: $x^2 + y^2 + 2x - y + 2k = 0$x2+y2+2x−y+2k=0

Imaginary part: $x + 2y + 3k = 0 \Rightarrow x = -2y - 3k$x+2y+3k=0⇒x=−2y−3k

Substituting $x$x into the real part equation:

$$(-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0$$(−2y−3k)2+y2+2(−2y−3k)−y+2k=0

$$4y^2 + 12ky + 9k^2 + y^2 - 4y - 6k - y + 2k = 0$$4y2+12ky+9k2+y2−4y−6k−y+2k=0

$$5y^2 + (12k - 5)y + 9k^2 - 4k = 0$$5y2+(12k−5)y+9k2−4k=0

For the equation to have at least one solution $z \in \mathbb{C}$z∈C, there must be at least one real value of $y$y. Thus, the discriminant of this quadratic equation in $y$y must be non-negative ($\Delta \ge 0$Δ≥0):

$$\Delta = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$$Δ=(12k−5)2−4(5)(9k2−4k)≥0

$$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$$144k2−120k+25−180k2+80k≥0

$$-36k^2 - 40k + 25 \ge 0$$−36k2−40k+25≥0

$$36k^2 + 40k - 25 \le 0$$36k2+40k−25≤0

The values of $k$k lie in the interval $[\alpha, \beta]$[α,β], where $\alpha$α and $\beta$β are the roots of the equation $36k^2 + 40k - 25 = 0$36k2+40k−25=0.

The sum of the roots is given by:

$$\alpha + \beta = -\dfrac{40}{36} = -\dfrac{10}{9}$$α+β=−3640​=−910​

Therefore, $9(\alpha + \beta) = 9 \left( -\dfrac{10}{9} \right) = -10$9(α+β)=9(−910​)=−10.

Answer: $-10$−10