Let S 1 = z 1 ∈ C : z 1 - 3 = 1 2 and S 2 = z 2 ∈ C : z 2 - z 2 + 1 = z 2 + z 2 - 1 . Then, for z 1 ∈ S 1 and z 2 ∈ S 2 , the least value of z 2 - z 1 is

Step-by-step solution

Here z 1 - 3 = 1 2 represents a circle on argand plane with centre 3 , 0 and radius 1 2 Given z 2 + z 2 − 1 2 = z 2 − z 2 + 1 2 ⇒ z 2 + z 2 - 1 z ¯ 2 + z 2 - 1 = z 2 - z 2 + 1 z ¯ 2 - z 2 + 1 ⇒ z 2 z 2 - 1 + z 2 + 1 + z ¯ 2 z 2 - 1 + z 2 + 1 = z 2 + 1 2 - z 2 - 1 2 ⇒ z 2 + z ¯ 2 z 2 + 1 + z 2 - 1 = 2 z 2 + z ¯ 2 ⇒ Either z 2 + z ¯ 2 = 0 or z 2 + 1 + z 2 - 1 = 2 i.e. z 2 lies on imaginary axis or it lies on the line segment joining - 1 , 0 and 1 , 0 S