Let ℂ be the set of all complex numbers. Let S 1 = z ∈ ℂ : z - 2 ≤ 1 and S 2 = z ∈ ℂ : z 1 + i + z ¯ 1 - i ≥ 4 . Then, the maximum value of z - 5 2 2 for z ∈ S 1 ∩ S 2 is equal to :

Step-by-step solution

Here, z - 2 ≤ 1 Put z = x + i y x - 2 2 + y 2 ≤ 1 Also, z 1 + i + z ¯ 1 - i ≥ 4 Gives x - y ≥ 2 Let point on circle be A 2 + cos θ , sin θ θ ∈ - 3 π 4 , π 4 Let P be 5 2 , 0 So, z - 5 2 2 = AP 2 = 2 + cos θ - 5 2 2 + sin 2 θ = cos 2 θ - cos θ + 1 4 + sin 2 θ = 5 4 - cos θ For AP 2 to be maximum θ = - 3 π 4 AP 2 = 5 4 + 1 2 = 5 2 + 4 4 2 = 5 + 2 2 4