Let Z be a complex number satisfying the relation Z 3 + 4 Z ¯ 2 Z = 0 . If the least possible argument of Z is – k π , then k is equal to (here, arg ⁡ Z ∈ - π , π )

Step-by-step solution

Let, Z = r e i θ ⇒ r 3 e i θ + 4 r e - i 2 θ = 0 ⇒ r 2 e i 5 θ = - 4 ⇒ r 2 = 4 and e i 5 θ = - 1 ⇒ r = 2 and θ = - 3 π 5 , - π 5 , π 5 , 3 π 5 , π ⇒ least arg ⁡ Z is - 3 π 5 ⇒ k = 0 . 6