Let A = \ z ∈ C: |(z + 1)/(z - 1)| < 1\ and B = \ z ∈ C: arg((z -...

Let

A = \left\{z \in C: \left|\dfrac{z + 1}{z - 1}\right| < 1\right\}

and

B = \left\{z \in C: arg\left(\dfrac{z - 1}{z + 1}\right) = \dfrac{2\pi}{3}\right\}

. Then

A \cap B

is $Let A = \ z ∈ C: |(z + 1)/(z - 1)| < 1\ and B = \ z ∈ C: arg((z - 1)/(z + 1)) = (2π)/(3)\ . Then A B is$ $Let A = \ z ∈ C: |(z + 1)/(z - 1)| < 1\ and B = \ z ∈ C: arg((z - 1)/(z + 1)) = (2π)/(3)\ . Then A B is$

Options

Aa portion of a circle centred at $\left(0, - \dfrac{1}{\sqrt{3}}\right)$ that lies in the second and third quadrants only
Ba portion of a circle centred at $\left(0, - \dfrac{1}{\sqrt{3}}\right)$ that lies in the second quadrant only
Can empty set
Da portion of a circle of radius $\dfrac{2}{\sqrt{3}}$ that lies in the third quadrant only

Correct answer

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

Step-by-step solution

Set $A$ is \left\{z \in C: \left|\dfrac{z + 1}{z - 1}\right| < 1\right\} $$ ⇒ \left|\dfrac{\overset{¯}{z} + 1}{\overset{¯}{z} - 1}\right| < 1 $$ ⇒ \left|\overset{¯}{z} + 1\right| < \left|\overset{¯}{z} - 1\right| $$ ⇒ {\left(x + 1\right)}^{2} + {y}^{2} < {\left(x - 1\right)}^{2} + {y}^{2} $$ ⇒ x < 0 $<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b3b59eaf-6dc4-4bf7-bf71-ca2a024e2a32-image.png" style="width: 180px; height: 145px;" />Set$ B $is \left\{z \in C: arg\left(\dfrac{z - 1}{z + 1}\right) = \dfrac{2\pi}{3}\right\}$

$⇒ arg\left(\dfrac{z - 1}{z + 1}\right) = \dfrac{2\pi}{3}$

$⇒ \tan^{- 1}\left(\dfrac{y}{x - 1}\right) - \tan^{- 1}\left(\dfrac{y}{x + 1}\right) = \dfrac{2\pi}{3}$

$⇒ {x}^{2} + {y}^{2} + \dfrac{2y}{\sqrt{3}} - 1 = 0$

$⇒$ Centre \left(0, - \dfrac{1}{\sqrt{3}}\right) $<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/f5846ea3-a740-46f2-911d-2b98b3cb95a8-image.png" style="width: 280px; height: 208px;" />Hence,$ A \cap B $will represent a portion of a circle centred at \left(0, - \dfrac{1}{\sqrt{3}}\right)$ that lies in the second quadrant only

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