Let A = {θ ∈ (−π/2, π) : 3 + 2i sin θ + i sin θ is purely imaginary}. Then the sum of the elements in A is 3π/4, 5π/6, π/2, π/3

Given: z = 3 + 2i sin θ + i sin θ is purely imaginary, so the real part becomes zero.
→ z = (3 + 2i sin θ + i sin θ) × (1 + 2i sin θ / 1 + 2i sin θ)
→ z = (3 − 8 sin²θ) + i(8 sin θ) / 1 + 4 sin²θ
Now Re(z) = 0 → 3 − 8 sin²θ / 1 + 4 sin²θ = 0 → sin²θ = 3/4
→ sin θ = ±√3/2 → θ = −π/3, π/3, 2π/3
∈ (−π/2, π) then sum of the elements in A is → −π/3 + π/3 + 2π/3 = 2π/3.