Complex NumbershardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Complex Numbers: Let Set Values Equation Least Solution Interval Equal (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let the set of all values of k∈Rk\in\mathbb{R} such that the equation z(zˉ+2+i)+k(2+3i)=0z(\bar z+2+i)+k(2+3i)=0, z∈Cz\in\mathbb{C}, has at least one solution, be the interval [α,β][\alpha,\beta]. Then 9(α+β)9(\alpha+\beta) is equal to
A−10-10correct
B−8-8
C101310\sqrt{13}
D8138\sqrt{13}
Solution
Step 1: Let z=x+iy⇒zˉ=x−iyz=x+iy\Rightarrow\bar z=x-iy.
zˉ+2+i=(x+2)+i(1−y).\bar z+2+i=(x+2)+i(1-y).
z(zˉ+2+i)=(x+iy)[(x+2)+i(1−y)].z(\bar z+2+i)=(x+iy)\big[(x+2)+i(1-y)\big].
 Re =x(x+2)−y(1−y)=x(x+2)-y(1-y); Im =x(1−y)+y(x+2)=x(1-y)+y(x+2). Step 2: z(zˉ+2+i)+k(2+3i)=0z(\bar z+2+i)+k(2+3i)=0 gives
Re: x(x+2)−y(1−y)+2k=0⇒x2+2x−y+y2+2k=0,\text{Re: }x(x+2)-y(1-y)+2k=0\Rightarrow x^2+2x-y+y^2+2k=0,
Im: x(1−y)+y(x+2)+3k=0⇒x−xy+xy+2y+3k=0⇒x+2y+3k=0.\text{Im: }x(1-y)+y(x+2)+3k=0\Rightarrow x-xy+xy+2y+3k=0\Rightarrow x+2y+3k=0.
 Step 3: x=−(2y+3k)x=-(2y+3k). Substitute into Re:
(2y+3k)2−2(2y+3k)−y+y2+2k=0.(2y+3k)^2-2(2y+3k)-y+y^2+2k=0.
(2y+3k)2=4y2+12ky+9k2,−2(2y+3k)=−4y−6k.(2y+3k)^2=4y^2+12ky+9k^2,\quad-2(2y+3k)=-4y-6k.
⇒4y2+12ky+9k2−4y−6k−y+y2+2k=0⇒5y2+(12k−5)y+(9k2−4k)=0.\Rightarrow4y^2+12ky+9k^2-4y-6k-y+y^2+2k=0\Rightarrow5y^2+(12k-5)y+(9k^2-4k)=0.
 Step 4: Real yy ∴ discriminant ≥0\ge0. With A=5,B=12k−5,C=9k2−4kA=5,B=12k-5,C=9k^2-4k:
(12k−5)2−20(9k2−4k)≥0.(12k-5)^2-20(9k^2-4k)\ge0.
(12k−5)2=144k2−120k+25,20(9k2−4k)=180k2−80k.(12k-5)^2=144k^2-120k+25,\quad20(9k^2-4k)=180k^2-80k.
⇒144k2−120k+25−180k2+80k≥0⇒−36k2−40k+25≥0⇒36k2+40k−25≤0.\Rightarrow144k^2-120k+25-180k^2+80k\ge0\Rightarrow-36k^2-40k+25\ge0\Rightarrow36k^2+40k-25\le0.
 Step 5: Roots of 36k2+40k−25=036k^2+40k-25=0 bound [α,β][\alpha,\beta].
α+β=−4036=−109⇒9(α+β)=9⋅(−109)=−10.\alpha+\beta=-\dfrac{40}{36}=-\dfrac{10}{9}\Rightarrow9(\alpha+\beta)=9\cdot\left(-\dfrac{10}{9}\right)=-10.
 Correct answer: (1)
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