If Z = cos ⁡ ϕ + i sin ⁡ ϕ ∀ ϕ ∈ π 3 , π , then the value of arg Z 2 - Z is equal to (where, arg ( Z ) represents the argument of the complex number Z lying in the interval - π , π and i 2 = - 1 )

Step-by-step solution

Z 2 - Z = Z Z - 1 = cos ⁡ ϕ + i sin ⁡ ϕ cos ⁡ ϕ - 1 + i sin ⁡ ϕ = cos ⁡ ϕ + i sin ⁡ ϕ - 2 s i n 2 ϕ 2 + i 2 s i n ϕ 2 c o s ϕ 2 = 2 i s i n ϕ 2 c o s 3 ϕ 2 + i s i n 3 ϕ 2 = 2 s i n ϕ 2 c o s 3 ϕ + π 2 + i s i n 3 ϕ + π 2 Now, 3 ϕ ∈ π , 3 π ⇒ 3 ϕ + π 2 ∈ π , 2 π But, the argument lies in - π , π , hence arg Z 2 - Z = 3 ϕ + π 2 - 2 π = 3 2 ϕ - π