Let z 1 , z 2 and z 3 are the points on the argand plane which lie on the circle with equation z - z 0 = 4 3 (where z 0 is the centre of the circle). If z 1 = 0 , z 2 = - 4 and z 3 = 4 + 3 z 0 , then arg ⁡ z 0 is equal to (where arg Z ∈ - π , π )

Step-by-step solution

Here z 1 + z 2 + z 3 3 = z 0 ⇒ z 3 = 4 + 3 z 0 Therefore, center coincides with the circumcentre ⇒ Triangle is equilateral ⇒ z 1 - z 2 = 4 Clearly, z 3 either lie in the second or third quadrant So the centre z 0 also lies in the second or third quadrant. ∴ z 0 can be = - 2 + 2 3 i , - 2 - 2 3 i ⇒ arg z 0 = 5 π 6 , 7 π 6