Let A = z ∈ C : z + 1 z - 1 < 1 and B = z ∈ C : arg z - 1 z + 1 = 2 π 3 . Then A ∩ B is

Step-by-step solution

Set A is z ∈ C : z + 1 z - 1 < 1 ⇒ z ¯ + 1 z ¯ - 1 < 1 ⇒ z ¯ + 1 < z ¯ - 1 ⇒ x + 1 2 + y 2 < x - 1 2 + y 2 ⇒   x < 0 Set B is z ∈ C : arg z - 1 z + 1 = 2 π 3 ⇒ arg z - 1 z + 1 = 2 π 3 ⇒ tan - 1 y x - 1 - tan - 1 y x + 1 = 2 π 3 ⇒ x 2 + y 2 + 2 y 3 - 1 = 0 ⇒ Centre 0 , - 1 3 Hence, A ∩ B will represent a portion of a circle centred at 0 , - 1 3 that lies in the second quadrant only