Let the set of all values of k ∈ R such that the equation z(z̄ + 2 + i) + k(2 + 3i) = 0, z ∈ C, has at least one solution, be the interval [α, β]. Then 9(α + β) is equal to:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2026

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Home Maths Complex & QuadraticQuestion #83

Let the set of all values of

k \in R

such that the equation

z (\bar{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:

Clue 1 of 3: Representing Complex Numbers

The problem involves both a complex variable $z$ and its conjugate $\bar{z}$ . The standard approach is to substitute $z = x + i y$ and separate the equation into its real and imaginary parts.

🥷

Ninja Strategy•Symmetry of Roots

Recognize that the extreme values of a linear function over a circle are symmetric around the value at the center, causing the radical parts to cancel out in the sum `α + β`.

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Step 1: Derive the Locus of z✦ Active

Let $z = x + i y$ . The given equation is $z \bar{z} + z (2 + i) + k (2 + 3 i) = 0$ . Substituting $z$ and $\bar{z}$ gives $(x^{2} + y^{2}) + (x + i y) (2 + i) + k (2 + 3 i) = 0$ .

Expanding and separating into real and imaginary parts:

(x^{2} + y^{2} + 2 x - y + 2 k) + i (x + 2 y + 3 k) = 0

This gives two equations: $x^{2} + y^{2} + 2 x - y + 2 k = 0$ and $x + 2 y + 3 k = 0$ . From the second equation, we get $k = - \frac{x + 2 y}{3}$ . Substituting this into the first equation eliminates $k$ :

x^{2} + y^{2} + 2 x - y - 2 (\frac{x + 2 y}{3}) = 0

Simplifying this gives the equation of a circle: $x^{2} + y^{2} + \frac{4}{3} x - \frac{7}{3} y = 0$ , which can be written in standard form as $(x + \frac{2}{3})^{2} + (y - \frac{7}{6})^{2} = \frac{65}{36}$ .

Step 2: Find the Range of k○ Expand

We need to find the range of $k = - \frac{x + 2 y}{3}$ for all points $(x, y)$ on the circle. This is equivalent to finding the range of the expression $L = x + 2 y$ . Geometrically, $x + 2 y = L$ represents a family of parallel lines. The extreme values of $L$ occur when these lines are tangent to the circle.

The distance from the circle's center $C (- \frac{2}{3}, \frac{7}{6})$ to the line $x + 2 y - L = 0$ must equal the radius $r = \frac{\sqrt{65}}{6}$ .

\frac{| 1 (- \frac{2}{3}) + 2 (\frac{7}{6}) - L |}{\sqrt{1^{2} + 2^{2}}} = \frac{\sqrt{65}}{6} ⟹ \frac{| \frac{5}{3} - L |}{\sqrt{5}} = \frac{\sqrt{65}}{6}

Solving for $L$ gives $L = \frac{10 \pm 5 \sqrt{13}}{6}$ . The range of $k = - L / 3$ is therefore $[\frac{- 10 - 5 \sqrt{13}}{18}, \frac{- 10 + 5 \sqrt{13}}{18}]$ . Thus, $α = \frac{- 10 - 5 \sqrt{13}}{18}$ and $β = \frac{- 10 + 5 \sqrt{13}}{18}$ .

Step 3: Calculate the Final Value○ Expand

The problem asks for the value of $9 (α + β)$ .

α + β = \frac{- 10 - 5 \sqrt{13}}{18} + \frac{- 10 + 5 \sqrt{13}}{18} = \frac{- 20}{18} = - \frac{10}{9}

9 (α + β) = 9 (- \frac{10}{9}) = - 10

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Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: Representing Complex Numbers

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