Key Concepts and Formulas

• A complex number $z = x + iy$ is purely imaginary if its real part $x$ is zero, i.e., $\text{Re}(z) = 0$.
• To rationalize a complex number of the form $\frac{a+bi}{c+di}$, multiply both the numerator and denominator by the conjugate of the denominator, $c-di$.
• The trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ and the fact that $\sin(\pi - \theta) = \sin(\theta)$.

Step-by-Step Solution

Step 1: Rationalizing the Complex Expression

We are given the complex number $z = \frac{1+2 i \sin \theta}{1-i \sin \theta}$. To express it in the standard form $x+iy$, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1+i \sin \theta)$. This eliminates the imaginary component from the denominator.

$$z = \frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta}$$

Step 2: Expanding the Numerator

We expand the numerator by multiplying the two complex numbers:

$$\begin{aligned} (1+2 i \sin \theta)(1+i \sin \theta) &= 1(1) + 1(i \sin \theta) + (2 i \sin \theta)(1) + (2 i \sin \theta)(i \sin \theta) \\ &= 1 + i \sin \theta + 2 i \sin \theta + 2 i^2 \sin^2 \theta \\ &= 1 + 3 i \sin \theta - 2 \sin^2 \theta \quad (\text{since } i^2 = -1) \\ &= (1 - 2 \sin^2 \theta) + i(3 \sin \theta) \end{aligned}$$

Step 3: Expanding the Denominator

We expand the denominator, which is the product of a complex number and its conjugate:

$$\begin{aligned} (1-i \sin \theta)(1+i \sin \theta) &= 1^2 - (i \sin \theta)^2 \\ &= 1 - i^2 \sin^2 \theta \\ &= 1 - (-1) \sin^2 \theta \\ &= 1 + \sin^2 \theta \end{aligned}$$

Step 4: Expressing z in Standard Form

Substitute the expanded numerator and denominator back into the expression for $z$:

$$z = \frac{(1 - 2 \sin^2 \theta) + i(3 \sin \theta)}{1+\sin^2 \theta}$$

Separate the real and imaginary parts:

$$z = \frac{1 - 2 \sin^2 \theta}{1+\sin^2 \theta} + i \frac{3 \sin \theta}{1+\sin^2 \theta}$$

Step 5: Applying the Purely Imaginary Condition

For $z$ to be purely imaginary, its real part must be zero:

$$\text{Re}(z) = 0 \Rightarrow \frac{1 - 2 \sin^2 \theta}{1+\sin^2 \theta} = 0$$

Since the denominator $1 + \sin^2 \theta$ is always positive and therefore non-zero, we only need to set the numerator to zero:

$$1 - 2 \sin^2 \theta = 0 \\ 2 \sin^2 \theta = 1 \\ \sin^2 \theta = \frac{1}{2}$$

Taking the square root of both sides gives:

$$\sin \theta = \pm \frac{1}{\sqrt{2}}$$

Step 6: Finding Solutions for $\theta$ in the Given Interval

We need to find all values of $\theta$ in the interval $(0, 2\pi)$ such that $\sin \theta = \frac{1}{\sqrt{2}}$ or $\sin \theta = -\frac{1}{\sqrt{2}}$.

Case 1: $\sin \theta = \frac{1}{\sqrt{2}}$ The reference angle is $\frac{\pi}{4}$. Since $\sin \theta$ is positive, $\theta$ lies in the first or second quadrant.

• First Quadrant: $\theta = \frac{\pi}{4}$
• Second Quadrant: $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

Case 2: $\sin \theta = -\frac{1}{\sqrt{2}}$ Since $\sin \theta$ is negative, $\theta$ lies in the third or fourth quadrant.

• Third Quadrant: $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
• Fourth Quadrant: $\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$

Thus, the set $A$ is:

$$A = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$$

Step 7: Calculating the Sum of Elements in A

Finally, we need to find the sum of all elements in set $A$:

$$\text{Sum} = \frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$$

Common Mistakes & Tips

• Remember to consider both positive and negative roots when solving $\sin^2 \theta = k$.
• Pay close attention to the given interval for $\theta$ and ensure all solutions lie within it.
• Double-check the multiplication when rationalizing the denominator.

Summary

This problem involves converting a complex number to standard form, applying the condition for a purely imaginary number (real part equals zero), and solving the resulting trigonometric equation within a specified domain. The sum of the $\theta$ values that satisfy the condition in the given interval $(0, 2\pi)$ is $4\pi$.

The final answer is \boxed{4\pi}, which corresponds to option (D).