JEE Advanced Mathematics Question (2019) — Solution

Step-by-step solution

If z is a complex number Then | z | 2 = z z ¯ Now, a + b ω + c ω 2 2 = a + b ω + c ω 2 a + b ω + c ω 2 ¯ a + b ω + c ω 2 2 = a + b ω + c ω 2 a + b ω - + c ω - 2 i f   a ∈ R ⇒ a - = a z 1 z 2 ¯ = z - 1 z - 2 ω - = ω 2 ,    ω - 2 = ω a + b ω + c ω 2 2 = a + b ω + c ω 2 a + b ω 2 + c ω = a 2 + b 2 + c 2 - a b - b c - c a ∵   ω 3 = 1 ,    1 + ω + ω 2 = 0 | a + b ω + c ω 2 | 2 = 1 2 [ a - b 2 + b - c 2 + c - a 2 ] Now for this to be minimum Take a = 1 ,   b = 2 ,   c = 3 , as a ,   b ,   c are distinct and non-zero integers Minimum of  a + b ω + c ω 2 2 = 3