JEE Challenger
More from Complex Numbers and Quadratic Equations
MathHardAdv 2021 • Paper 1

Complex Argument Locus and Circle Parameter Analysis

Given a complex number w=c+idw = c + id, let arg(w)∈(−π,π]\text{arg}(w) \in (-\pi, \pi], where i=−1i = \sqrt{-1}. Suppose α\alpha and β\beta are real numbers such that for all complex numbers z=x+iyz = x + iy satisfying arg(z+αz+β)=π4\text{arg}\left(\frac{z+\alpha}{z+\beta}\right) = \frac{\pi}{4} the ordered pair (x,y)(x, y) lies on the circle x2+y2+5x−3y+4=0x^2 + y^2 + 5x - 3y + 4 = 0 Then which of the following statements is (are) TRUE ?

Options

A

α=−1\alpha = -1

B

αβ=4\alpha\beta = 4

Correct
C

αβ=−4\alpha\beta = -4

D

β=4\beta = 4

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex NumbersCircles

Step-by-Step Solution

To determine the correct statements, we analyze the given complex locus equation: arg(z+αz+β)=π4\text{arg}\left(\frac{z+\alpha}{z+\beta}\right) = \frac{\pi}{4}

Step 1: Finding the intersection points with the real axis

The given locus equation represents the set of points z=x+iyz = x + iy from which the directed segment from z=−βz = -\beta to z=−αz = -\alpha subtends an angle of π4\frac{\pi}{4}.

The locus of such complex numbers zz lies on a major arc of a circle passing through the points A(−α,0)A(-\alpha, 0) and B(−β,0)B(-\beta, 0) on the real axis (x-axis).

Since the given circle is: x2+y2+5x−3y+4=0x^2 + y^2 + 5x - 3y + 4 = 0

The points A(−α,0)A(-\alpha, 0) and B(−β,0)B(-\beta, 0) must lie on this circle. Setting y=0y = 0 in the circle's equation gives: x2+5x+4=0x^2 + 5x + 4 = 0 (x+1)(x+4)=0  ⟹  x=−1orx=−4(x + 1)(x + 4) = 0 \implies x = -1 \quad \text{or} \quad x = -4

Thus, the set of values for {−α,−β}\{-\alpha, -\beta\} is {−1,−4}\{-1, -4\}, which implies: {α,β}={1,4}\{\alpha, \beta\} = \{1, 4\}

Step 2: Identifying the values of α\alpha and β\beta

We now determine which of the two cases holds:

  1. Case 1: α=1\alpha = 1 and β=4\beta = 4
  2. Case 2: α=4\alpha = 4 and β=1\beta = 1

To test these cases, we select a point z=x+iyz = x + iy on the circle. Substitute x=−1x = -1 into the circle equation: (−1)2+y2+5(−1)−3y+4=0  ⟹  y2−3y=0  ⟹  y=3(y≠0)(-1)^2 + y^2 + 5(-1) - 3y + 4 = 0 \implies y^2 - 3y = 0 \implies y = 3 \quad (y \neq 0)

So, the point z=−1+3iz = -1 + 3i lies on the circle.

Testing Case 1 (α=1,β=4\alpha = 1, \beta = 4):

z+αz+β=(−1+3i)+1(−1+3i)+4=3i3+3i=i1+i\frac{z + \alpha}{z + \beta} = \frac{(-1 + 3i) + 1}{(-1 + 3i) + 4} = \frac{3i}{3 + 3i} = \frac{i}{1 + i} Multiplying the numerator and denominator by the conjugate (1−i)(1 - i): i(1−i)(1+i)(1−i)=i−i22=1+i2\frac{i(1 - i)}{(1 + i)(1 - i)} = \frac{i - i^2}{2} = \frac{1 + i}{2} The argument is: arg(1+i2)=π4\text{arg}\left(\frac{1 + i}{2}\right) = \frac{\pi}{4} This matches the given condition arg(z+αz+β)=π4\text{arg}\left(\frac{z+\alpha}{z+\beta}\right) = \frac{\pi}{4}.

Testing Case 2 (α=4,β=1\alpha = 4, \beta = 1):

z+αz+β=(−1+3i)+4(−1+3i)+1=3+3i3i=1+ii=1−i\frac{z + \alpha}{z + \beta} = \frac{(-1 + 3i) + 4}{(-1 + 3i) + 1} = \frac{3 + 3i}{3i} = \frac{1 + i}{i} = 1 - i The argument is: arg(1−i)=−π4≠π4\text{arg}(1 - i) = -\frac{\pi}{4} \neq \frac{\pi}{4} Thus, Case 2 is invalid.

Hence, we uniquely get: α=1andβ=4\alpha = 1 \quad \text{and} \quad \beta = 4

Step 3: Evaluating the Options

  • Option (A): α=−1\alpha = -1 (False, since α=1\alpha = 1)
  • Option (B): αβ=(1)(4)=4\alpha\beta = (1)(4) = 4 (True)
  • Option (C): αβ=−4\alpha\beta = -4 (False, since αβ=4\alpha\beta = 4)
  • Option (D): β=4\beta = 4 (True)

Conclusion

The correct options are (B) and (D).

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