If $S=\left\{z \in C: \frac{z-i}{z+2i} \in R\right\}$, then

(B) S is a straight line in the complex plane

(A) S is a circle in the complex plane

(C) S contains only one element

(D) S contains exactly two elements

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Complex Numbers

S = {z \in C : \frac{z - i}{z + 2 i} \in R}

, then

Answer: (B)

For

\frac{z - i}{z + 2 i}

to be real, it must equal its conjugate:

\frac{z - i}{z + 2 i} = \frac{z - i}{z + 2 i}

\frac{z - i}{z + 2 i} = \frac{z ˉ + i}{z ˉ - 2 i}

Cross-multiplying:

(z - i) (\overset{z}{ˉ} - 2 i) = (z + 2 i) (\overset{z}{ˉ} + i)

z \overset{z}{ˉ} - 2 i z - i \overset{z}{ˉ} + 2 i^{2} = z \overset{z}{ˉ} + i z + 2 i \overset{z}{ˉ} + 2 i^{2}

- 2 i z - i \overset{z}{ˉ} - 2 = i z + 2 i \overset{z}{ˉ} - 2

- 3 i z - 3 i \overset{z}{ˉ} = 0

z + \overset{z}{ˉ} = 0

Let

z = x + i y

. Then

z + \overset{z}{ˉ} = 2 x = 0

, so

x = 0

.

This means

z

is purely imaginary, i.e.,

z = i y

for any real

y

.

Therefore,

S

is a straight line (the imaginary axis) in the complex plane.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option