Let $A = \left\{\theta \in \left(-\dfrac{\pi}{2}, \pi\right) : \dfrac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \text{ is purely imaginary}\right\}$. Then the sum of the elements in $A$ is
Step-by-Step Solution
Key Concept: A complex number is purely imaginary iff its real part is zero AND its imaginary part is non-zero; after rationalising, set only the real part to zero.
**Step 1: Extract real part**
Multiply numerator and denominator by the conjugate $1 + 2i\sin\theta$:
$$\text{Re}\!\left(\dfrac{3+2i\sin\theta}{1-2i\sin\theta}\right) = \dfrac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta}.$$
**Step 2: Set real part to 0**
$3 - 4\sin^2\theta = 0 \Rightarrow \sin^2\theta = \dfrac{3}{4} \Rightarrow \sin\theta = \pm\dfrac{\sqrt{3}}{2}$, giving $\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, -\dfrac{\pi}{3}$ (within the given interval; $-\dfrac{2\pi}{3} < -\dfrac{\pi}{2}$ is excluded).
**Step 3: Sum the valid elements**
$\dfrac{\pi}{3} + \dfrac{2\pi}{3} + \left(-\dfrac{\pi}{3}\right) = \dfrac{2\pi}{3}$.
Correct Answer: 4