Question

Let A = \(\left\{ \theta \in (0, 2\pi) : \frac{1 + 2i \sin \theta}{1 - i \sin \theta} \text{ is purely imaginary} \right\}\). Then the sum of the elements in A is.

• A. 2π
• B. 3π
• C. 4π
• D. π

Answer 1

The Correct Option is C

Given:
\( z = \frac{1 + 2i \sin \theta}{1 - i \sin \theta} \times \frac{1 + i \sin \theta}{1 + i \sin \theta} \) 

Simplify \( z \):
\[ z = \frac{1 - 2 \sin^2 \theta + i (3 \sin \theta)}{1 + \sin^2 \theta} \] 

Re(z) = 0:
\[ \frac{1 - 2 \sin^2 \theta}{1 + \sin^2 \theta} = 0 \] 

Solving for \( \sin \theta \):
\[ \sin \theta = \pm \frac{1}{\sqrt{2}} \] 

Set of Angles:
\[ A = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} \] 

Sum:
\[ \text{sum} = 4\pi \quad (\text{Option 3}) \]