Let A=z∈C: z+1z−1<1 and B=z∈C: argz−1z+1=2π3 . Then A ∩ B is

⇒z+1z−1<1⇒|z+1|<|z−1|⇒(x + 1)2 + y2 < (x – 1)2 + y2 ⇒ x < 0 Set : B ⇒arg(z−1z+1)=2π3⇒tan−1(yx−1)−tan−1(yx+1)=2π3 ⇒x2+y2+2y3−1=0A∩B⇒centre(0,−13)

Let A=z∈C: z+1z−1<1 and B=z∈C: argz−1z+1=2π3 . Then A ∩ B is

⇒z+1z−1<1⇒|z+1|<|z−1|⇒(x + 1)2 + y2 < (x – 1)2 + y2 ⇒ x < 0 Set : B ⇒arg(z−1z+1)=2π3⇒tan−1(yx−1)−tan−1(yx+1)=2π3 ⇒x2+y2+2y3−1=0A∩B⇒centre(0,−13)

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

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

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

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

answer is B.

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$\Rightarrow |\frac{z + 1}{z - 1}| < 1 \Rightarrow | z + 1 | < | z - 1 |$

$\Rightarrow {(x + 1)}^{2} + y^{2} < {(x - 1)}^{2} + y^{2}$
$\Rightarrow$ x < 0
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Set : B

$\Rightarrow \arg (\frac{z - 1}{z + 1}) = \frac{2 π}{3} \Rightarrow \tan^{- 1} (\frac{y}{x - 1}) - \tan^{- 1} (\frac{y}{x + 1}) = \frac{2 π}{3} \Rightarrow x^{2} + y^{2} + \frac{2 y}{\sqrt{3}} - 1 = 0$

A∩B

$\Rightarrow c e n t r e (0, - \frac{1}{\sqrt{3}})$

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Let A=z∈C: z+1z−1<1 and B=z∈C: argz−1z+1=2π3 . Then A ∩ B is

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