Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\overline{z}+2+i)+k(2+3 i)=0, z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha+\beta)$ is equal to:

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Complex Numbers

Let the set of all values of

k \in R

such that the equation

z (\overline{z} + 2 + i) + k (2 + 3 i) = 0, z \in C

, has at least one solution, be the interval

[α, β]

. Then

9 (α + β)

is equal to:

Answer: (A)

Put

z = x + i y

; then

z (\overset{z}{ˉ} + 2 + i) = x^{2} + y^{2} + 2 x - y + i (x + 2 y)

, so the equation

z (\overset{z}{ˉ} + 2 + i) + k (2 + 3 i) = 0

splits into

x^{2} + y^{2} + 2 x - y + 2 k = 0

and

x + 2 y + 3 k = 0

. Substituting

x = - 2 y - 3 k

into the first gives

5 y^{2} + (12 k - 5) y + 9 k^{2} - 4 k = 0

. A real

y

requires

(12 k - 5)^{2} - 20 (9 k^{2} - 4 k) \geq 0

, i.e.

36 k^{2} + 40 k - 25 \leq 0

. The endpoints

α, β

are the roots of

36 k^{2} + 40 k - 25 = 0

, with

α + β = - \frac{40}{36} = - \frac{10}{9}

, so

9 (α + β) = - 10

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option