Let $\mathrm{A}=\left\{\mathrm{z} \in \mathrm{C}:\left|\frac{\mathrm{z}+1}{\mathrm{z}-1}<1\right|\right\}$ and $B=\left\{z \in C: \arg \left(\frac{z-1}{z+1}\right)=\frac{2 \pi}{3}\right\}$. Then $\mathrm{A} \cap \mathrm{B}$ is :

(B) a portion of a circle centred at $\left(0,-\frac{1}{\sqrt{3}}\right)$ that lies in the second quadrant only

(A) a portion of a circle centred at $\left(0,-\frac{1}{\sqrt{3}}\right)$ that lies in the second and third quadrants only

(D) a portion of a circle of radius $\frac{2}{\sqrt{3}}$ that lies in the third quadrant only

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Complex Numbers

Let

A = {z \in C : \frac{z + 1}{z - 1} < 1}

and

B = {z \in C : ar g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}}

.
Then

A \cap B

is :

(A)a portion of a circle centred at $(0, - \frac{1}{3})$ that lies in the second and third quadrants only
(B)a portion of a circle centred at $(0, - \frac{1}{3})$ that lies in the second quadrant only✓ Correct
(C)an empty set
(D)a portion of a circle of radius $\frac{2}{3}$ that lies in the third quadrant only

Answer: (B)

Set A:

\frac{z + 1}{z - 1} < 1

implies

∣ z + 1∣ < ∣ z - 1∣

, which gives

x < 0

(the left half-plane).

Set B:

ar g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}

represents a circular arc. Using the argument condition and simplifying, we get the circle

x^{2} + y^{2} + \frac{2 y}{3} - 1 = 0

with centre

(0, - \frac{1}{3})

.

The intersection

A \cap B

is the portion of this circle that lies in the left half-plane (

x < 0

). Since the centre has

y < 0

and the circle extends into the second quadrant only (not the third), the answer is (B).

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option