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Complex Numbers question

2025 · 28 Jan · Shift 2 · Q39
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  5. /2025 · 28 Jan · Shift 2 · Q39

Complex Numbers question

2025 · 28 Jan · Shift 2 · Q39

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If α+iβ\alpha + i\betaα+iβ and γ+iδ\gamma + i\deltaγ+iδ are the roots of x2−(3−2i)x−(2i−2)=0x^2 - (3 - 2i)x - (2i - 2) = 0x2−(3−2i)x−(2i−2)=0, i=−1i = \sqrt{-1}i=−1​, then αγ+βδ\alpha \gamma + \beta \deltaαγ+βδ is equal to:
  1. A
    2
  2. B
    -6
  3. C
    6
  4. D
    -2
View written solutionFree

Correct answer: A

  1. Let the roots of x2−(3−2i)x−(2i−2)=0x^2-(3-2i)x-(2i-2)=0x2−(3−2i)x−(2i−2)=0 be z1=α+iβ,z2=γ+iδ.z_1=\alpha+i\beta, \qquad z_2=\gamma+i\delta.z1​=α+iβ,z2​=γ+iδ.

  2. By Vieta’s formulas, the product of the roots is the constant term: z1z2=−(2i−2)=2−2i.z_1z_2=-(2i-2)=2-2i.z1​z2​=−(2i−2)=2−2i.

  3. Now compute z1z2z_1z_2z1​z2​ in terms of α,β,γ,δ\alpha,\beta,\gamma,\deltaα,β,γ,δ: (α+iβ)(γ+iδ)=αγ−βδ+i(αδ+βγ).(\alpha+i\beta)(\gamma+i\delta)=\alpha\gamma-\beta\delta+i(\alpha\delta+\beta\gamma).(α+iβ)(γ+iδ)=αγ−βδ+i(αδ+βγ). So the real part of z1z2z_1z_2z1​z2​ is αγ−βδ=2.\alpha\gamma-\beta\delta=2.αγ−βδ=2.

This does not directly give αγ+βδ\alpha\gamma+\beta\deltaαγ+βδ.

  1. Since the quadratic has non-real coefficients, the roots need not be conjugates. So we find the roots explicitly.

The equation is x2−(3−2i)x+(2−2i)=0.x^2-(3-2i)x+(2-2i)=0.x2−(3−2i)x+(2−2i)=0. Discriminant: D=(3−2i)2−4(2−2i).D=(3-2i)^2-4(2-2i).D=(3−2i)2−4(2−2i). Now, (3−2i)2=9−12i+4i2=9−12i−4=5−12i,(3-2i)^2=9-12i+4i^2=9-12i-4=5-12i,(3−2i)2=9−12i+4i2=9−12i−4=5−12i, and 4(2−2i)=8−8i.4(2-2i)=8-8i.4(2−2i)=8−8i. Hence, D=(5−12i)−(8−8i)=−3−4i.D=(5-12i)-(8-8i)=-3-4i.D=(5−12i)−(8−8i)=−3−4i.

  1. We need −3−4i\sqrt{-3-4i}−3−4i​. Observe that (1−2i)2=1−4i+4i2=1−4i−4=−3−4i.(1-2i)^2=1-4i+4i^2=1-4i-4=-3-4i.(1−2i)2=1−4i+4i2=1−4i−4=−3−4i. So, D=±(1−2i).\sqrt{D}=\pm(1-2i).D​=±(1−2i).

  2. Therefore the roots are x=(3−2i)±(1−2i)2.x=\frac{(3-2i)\pm(1-2i)}{2}.x=2(3−2i)±(1−2i)​. So the two roots are: z1=(3−2i)+(1−2i)2=4−4i2=2−2i,z_1=\frac{(3-2i)+(1-2i)}{2}=\frac{4-4i}{2}=2-2i,z1​=2(3−2i)+(1−2i)​=24−4i​=2−2i, z2=(3−2i)−(1−2i)2=22=1.z_2=\frac{(3-2i)-(1-2i)}{2}=\frac{2}{2}=1.z2​=2(3−2i)−(1−2i)​=22​=1.

Thus we may take α=2, β=−2, γ=1, δ=0.\alpha=2,\ \beta=-2,\ \gamma=1,\ \delta=0.α=2, β=−2, γ=1, δ=0.

  1. Now compute αγ+βδ=(2)(1)+(−2)(0)=2.\alpha\gamma+\beta\delta=(2)(1)+(-2)(0)=2.αγ+βδ=(2)(1)+(−2)(0)=2.

  2. Hence the correct option is 2.\boxed{2}.2​.

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