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Complex Numbers question

2021 · Shift 1 · Q35
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  5. /2021 · Shift 1 · Q35

Complex Numbers question

2021 · Shift 1 · Q35

JEE AdvancedMathematicsComplex NumbersMultiple correct+4 / −2
For any complex number w = c + id, let arg⁡(w)∈(−π,π]\arg (w) \in ( - \pi ,\pi ]arg(w)∈(−π,π], where i=−1i = \sqrt { - 1}i=−1​. Let α\alphaα and β\betaβ be real numbers such that for all complex numbers z = x + iy satisfying arg⁡(z+αz+β)=π4\arg \left( {{{z + \alpha } \over {z + \beta }}} \right) = {\pi \over 4}arg(z+βz+α​)=4π​, the ordered pair (x, y) lies on the circle x2+y2+5x−3y+4=0{x^2} + {y^2} + 5x - 3y + 4 = 0x2+y2+5x−3y+4=0, Then which of the following statements is (are) TRUE?
  1. A
    α\alphaα=−-− 1
  2. B
    αβ\alpha\betaαβ = 4
  3. C
    αβ\alpha\betaαβ=−-− 4
  4. D
    β\betaβ = 4
View written solutionFree

Correct answer: B, D

Step-by-Step Solution:

  1. Understand the Complex Number Condition: The given condition is arg((z + α) / (z + β)) = π / 4. Using the property of arguments, arg(w₁ / w₂) = arg(w₁) - arg(w₂), we can rewrite the condition as: arg(z+α)−arg(z+β)=π4arg(z + α) - arg(z + β) = {π \over 4}arg(z+α)−arg(z+β)=4π​

  2. Convert to Cartesian Coordinates: Let the complex number z be represented by the point (x, y), so z = x + iy. Since α and β are real numbers, we have: z + α = (x + α) + iy z + β = (x + β) + iy

    Let θ₁ = arg(z + α) and θ₂ = arg(z + β). The condition is θ₁ - θ₂ = π / 4. In terms of x and y: tan(θ₁) = y / (x + α) tan(θ₂) = y / (x + β)

  3. Derive the Locus Equation: We take the tangent of the argument difference: tan(θ1−θ2)=tan(π/4)=1tan(θ₁ - θ₂) = tan(π / 4) = 1tan(θ1​−θ2​)=tan(π/4)=1 Using the tangent subtraction formula, tan(A - B) = (tan A - tan B) / (1 + tan A tan B): tan(θ1)−tan(θ2)1+tan(θ1)tan(θ2)=1{{tan(θ₁) - tan(θ₂)} \over {1 + tan(θ₁)tan(θ₂)}} = 11+tan(θ1​)tan(θ2​)tan(θ1​)−tan(θ2​)​=1 {{y \over {x + α}} - {y \over {x + β}}} \over {1 + \left( {y \over {x + α}} \right)\left( {y \over {x + β}} \right)}} = 1 Simplifying the numerator and denominator: y(x+β)−y(x+α)(x+α)(x+β)(x+α)(x+β)+y2(x+α)(x+β)=1{{y(x + β) - y(x + α)} \over {(x + α)(x + β)}} \over {{(x + α)(x + β) + y^2} \over {(x + α)(x + β)}} = 1(x+α)(x+β)(x+α)(x+β)+y2​=1(x+α)(x+β)y(x+β)−y(x+α)​​ This simplifies to: y((x+β)−(x+α))=(x+α)(x+β)+y2y((x + β) - (x + α)) = (x + α)(x + β) + y^2y((x+β)−(x+α))=(x+α)(x+β)+y2 y(β−α)=x2+(α+β)x+αβ+y2y(β - α) = x^2 + (α + β)x + αβ + y^2y(β−α)=x2+(α+β)x+αβ+y2 Rearranging the terms, we get the equation of a circle: x2+y2+(α+β)x−(β−α)y+αβ=0x^2 + y^2 + (α + β)x - (β - α)y + αβ = 0x2+y2+(α+β)x−(β−α)y+αβ=0

  4. Compare with the Given Circle Equation: The problem states that the ordered pair (x, y) lies on the circle x² + y² + 5x - 3y + 4 = 0. This means the locus we derived is identical to the given circle. We can compare the coefficients of the two equations: x2+y2+(α+β)x−(β−α)y+αβ=0x^2 + y^2 + (α + β)x - (β - α)y + αβ = 0x2+y2+(α+β)x−(β−α)y+αβ=0 x2+y2+5x−3y+4=0x^2 + y^2 + 5x - 3y + 4 = 0x2+y2+5x−3y+4=0

    Comparing the coefficients, we get a system of equations:

    1. α + β = 5
    2. -(β - α) = -3 => β - α = 3
    3. αβ = 4
  5. Solve for α and β: We have a system of two linear equations for α and β: α + β = 5 β - α = 3 Adding these two equations gives: 2β = 8 => β = 4 Substituting β = 4 into the first equation: α + 4 = 5 => α = 1

  6. Verify the Solution: Let's check if these values satisfy the third equation αβ = 4: (1)(4) = 4. The values are consistent. So, we have uniquely determined α = 1 and β = 4.

  7. Evaluate the Options: Now we check each statement with the values α = 1 and β = 4.

    • A: α = -1: This is FALSE because α = 1.
    • B: αβ = 4: This is TRUE because (1)(4) = 4.
    • C: αβ = -4: This is FALSE.
    • D: β = 4: This is TRUE.

Conclusion:

The correct statements are B and D.

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