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Complex Numbers Part 2

Section: Mathematics

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Question : 3 of 106

Marks: +1, -0

Let the set of all values of

k \in R

such that the equation

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

, has at least one solution, be the interval [α, β]. Then

9(\alpha+\beta)

[JEE Main 6 Apr 2026 shift 1]

-10
-8
$10\sqrt{13}$
$8\sqrt{13}$

Solution:

Put

z=x+iy, \overline{z}=x-iy

\Rightarrow \overline{z}+2+i=x-iy+2+i

\Rightarrow z(\overline{z}+2+i)=(x+iy)(x+2+i(1-y))

\Rightarrow

Real part

=x(x+2)-y(1-y)

Imaginary part

=x(1-y)+y(x+2)

\Rightarrow x^{2}+y^{2}+2x-y+2k=0 \qquad ...(1)

and

x+2y+3k=0 \qquad ...(2)

eliminating x from (1) and (2)

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

since

y \in R

use

D \geq 0

(12k-5)^{2}-4.5(9k^{2}-4k) \geq 0

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

\Rightarrow \alpha+\beta=\frac{-10}{9}

\Rightarrow 9(\alpha+\beta)=-10

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