Let S = z ∈ C : z − 1 = 1 and 2 − 1 z + z ¯ - i z - z ¯ = 2 2 . Let z 1 , z 2 ∈ S be such that z 1 = max z ∈ s z and z 2 = min z ∈ s z . Then 2 z 1 − z 2 2 equals:

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Correct answer: D. 2

Given, z − 1 = 1 Now, let z = x + i y ⇒ x − 1 2 + y 2 = 1 . . . i And 2 − 1 z + z ¯ - i z - z ¯ = 2 2 ⇒ 2 − 1 2 x − i 2 i y = 2 2 ⇒ 2 − 1 x + y = 2 . . . i i Solving i and i i , ⇒ x - 1 2 + 2 - 2 - 1 x 2 = 1 ⇒ x 2 + 1 - 2 x + 2 + 2 - 1 2 x 2 - 2 2 2 - 1 x = 1 ⇒ x 2 + 1 - 2 x + 2 + 2 + 1 - 2 2 x 2 - 4 x + 2 2 x = 1 ⇒ x 2 - 2 x + 2 + 3 x 2 - 2 x 2 2 - 4 x + 2 2 x = 0 ⇒ 4 x 2 - 2 x 2 2 - 6 x + 2 2 x + 2 = 0 ⇒ 4 - 2 2 x 2 + - 6 + 2 2 x + 2 = 0 ⇒ 4 - 2 2 x 2 + - 4 + 2 2 x - 2 x + 2 = 0 ⇒ 4 - 2 2 x x - 1 - 2 x - 1 = 0 ⇒

Let S = z ∈ C : z − 1 = 1 and 2 − 1 z + z ¯ - i z - z ¯ = 2 2 . Let z 1 , z 2 ∈ S be such that z 1 = max z ∈ s z and z 2 = min z ∈ s z . Then 2 z 1 − z 2 2 equals:

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