Let A = θ ∈ 0 , 2 π : 1 + 2 i sin θ 1 - i sin θ is purely imaginary Then the sum of the elements is in A is

Step-by-step solution

Let z = 1 + 2 i sin θ 1 - i sin θ ⇒ z = 1 + 2 i sin θ 1 - i sin θ × 1 + i sin θ 1 + i sin θ ⇒ z = 1 - 2 sin 2 θ + 3 i sin θ 1 + sin 2 θ ⇒ z = 1 - 2 sin 2 θ 1 + sin 2 θ + i 3 sin θ 1 + sin 2 θ Since, z is a purely imaginary number, so real part must be zero, hence 1 - 2 sin 2 θ 1 + sin 2 θ = 0 ⇒ 1 - 2 sin 2 θ = 0 ⇒ cos 2 θ = 0 ⇒ θ = π 4 , 3 π 4 , 5 π 4 , 7 π 4 , for &