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Complex Numbers question

2019 · 9 Apr · Shift 1 · Q29
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  5. /2019 · 9 Apr · Shift 1 · Q29

Complex Numbers question

2019 · 9 Apr · Shift 1 · Q29

JEE MainMathematicsComplex NumbersMCQ+4 / −1
All the points in the set S={α+iα−i:α∈R}(i=−1)S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )S={α−iα+i​:α∈R}(i=−1​) lie on a :
  1. A
    straight line whose slope is –1
  2. B
    straight line whose slope is 1.
  3. C
    circle whose radius is 1.
  4. D
    circle whose radius is 2\sqrt 22​ .
View written solutionFree

Correct answer: C

  1. Let z=α+iα−i,α∈R.z=\frac{\alpha+i}{\alpha-i},\qquad \alpha\in\mathbb R.z=α−iα+i​,α∈R. We need to find the locus of all possible points represented by zzz in the complex plane.

  2. Compute the modulus of zzz: ∣z∣=∣α+iα−i∣=∣α+i∣∣α−i∣.|z|=\left|\frac{\alpha+i}{\alpha-i}\right|=\frac{|\alpha+i|}{|\alpha-i|}.∣z∣=​α−iα+i​​=∣α−i∣∣α+i∣​. Now, ∣α+i∣=α2+1,∣α−i∣=α2+1.|\alpha+i|=\sqrt{\alpha^2+1},\qquad |\alpha-i|=\sqrt{\alpha^2+1}.∣α+i∣=α2+1​,∣α−i∣=α2+1​. Hence, ∣z∣=α2+1α2+1=1.|z|=\frac{\sqrt{\alpha^2+1}}{\sqrt{\alpha^2+1}}=1.∣z∣=α2+1​α2+1​​=1.

  3. Therefore every point zzz satisfies ∣z∣=1,|z|=1,∣z∣=1, which is the equation of a circle of radius 111 centered at the origin.

  4. So the set SSS lies on the unit circle.

  5. Check options:

  • A: straight line with slope −1-1−1 — false
  • B: straight line with slope 111 — false
  • C: circle with radius 111 — true
  • D: circle with radius 2\sqrt22​ — false

Therefore, the correct option is C.

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