All the points in the set $\mathrm{S}=\left\{\frac{\alpha+\mathrm{i}}{\alpha-\mathrm{i}}: \alpha \in \mathrm{R}\right\}(\mathrm{i}=\sqrt{-1})$ lie on a :

(A) straight line whose slope is 1

(B) circle whose radius is $\sqrt{2}$

(C) straight line whose slope is -1

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Complex Numbers

All the points in the set

S = {\frac{α + i}{α - i} : α \in R} (i = - 1)

lie on a :

Answer: (D)

For

z = \frac{α + i}{α - i}

with real

α

, the numerator and denominator are complex conjugates, so

∣ z ∣ = \frac{∣ α + i ∣}{∣ α - i ∣} = \frac{α ^{2} + 1}{α ^{2} + 1} = 1

. Since this holds for every real

α

, all the points sit on the circle of radius

1

about the origin.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option