Let S be the set of all α , β , π < α , β < 2 π , for which the complex number 1 - i sin α 1 + 2 i sin α is purely imaginary and 1 + i cos β 1 - 2 i cos β is purely real. Let Z α β = sin 2 α + i cos 2 β , α , β ∈ S . Then ∑ α , β ∈ S i Z α β + 1 i Z ¯ α β is equal to

Step-by-step solution

Given π < α , β < 2 π 1 - i sin α 1 + i 2 sin α is purely imaginary ⇒ 1 - i sin α 1 - i 2 sin α 1 + 4 sin 2 α is purely imaginary ⇒ 1 - 2 sin 2 α 1 + 4 sin 2 α = 0 ⇒ sin 2 α = 1 2 ⇒ α = 5 π 4 , 7 π 4 Also 1 + i cos β 1 + i - 2 cos β is purely real ⇒ 1 + i cos β 1 + 2 i cos β 1 + 4 cos 2 β is purely real ⇒ 3 cos β = 0 ⇒ β = 3 π 2 Now Z α β = si