JEE Main 2026 — Complex Number Question with Solution

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

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: $x+2y+3k=0\Rightarrow x=-2y-3k$

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

$\Delta =(12k-5{)}^2-4(5)(9k^2-4k)\ge 0$

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

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

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

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

The sum of the roots is given by:

$\alpha +\beta =-\frac{40}{36}=-\frac{10}{9}$

Therefore, $9(\alpha +\beta )=9\left(-\frac{10}{9}\right)=-10$.

Answer: $-10$