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 simplify a complex fraction $\frac{z_1}{z_2}$, multiply the numerator and denominator by the conjugate of the denominator, $\bar{z_2}$. If $z_2 = a + bi$, then $\bar{z_2} = a - bi$.
• The property $(a+bi)(a-bi) = a^2 + b^2$ is used to rationalize the denominator.

Step-by-Step Solution

Step 1: Simplify the given complex expression

We are given the complex number $Z = \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta}$. To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator, which is $1 + 2i\sin\theta$.

$Z = \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \times \frac{1 + 2i\sin\theta}{1 + 2i\sin\theta}$

Why this step? Multiplying by the conjugate eliminates the imaginary part from the denominator, making it easier to separate the real and imaginary parts of the complex number.

Now, we expand the numerator and the denominator:

Numerator: $(3 + 2i\sin\theta)(1 + 2i\sin\theta) = 3 + 6i\sin\theta + 2i\sin\theta + 4i^2\sin^2\theta$ Since $i^2 = -1$: $= 3 + 8i\sin\theta - 4\sin^2\theta = (3 - 4\sin^2\theta) + i(8\sin\theta)$

Denominator: $(1 - 2i\sin\theta)(1 + 2i\sin\theta) = 1 - (2i\sin\theta)^2 = 1 - 4i^2\sin^2\theta$ Since $i^2 = -1$: $= 1 + 4\sin^2\theta$

Now, substitute these back into the expression for $Z$:

$Z = \frac{(3 - 4\sin^2\theta) + i(8\sin\theta)}{1 + 4\sin^2\theta}$

Step 2: Identify the Real Part of the Complex Expression

We rewrite $Z$ by separating the real and imaginary components:

$Z = \frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta} + i\frac{8\sin\theta}{1 + 4\sin^2\theta}$

The real part of $Z$ is $\text{Re}(Z) = \frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta}$. The imaginary part of $Z$ is $\text{Im}(Z) = \frac{8\sin\theta}{1 + 4\sin^2\theta}$.

Step 3: Apply the condition for a purely imaginary number

The problem states that the given complex number is purely imaginary. This means its real part must be zero.

Why this step? This is the crucial condition that allows us to form an equation and solve for $\theta$.

Therefore, we set the real part equal to zero:

$\frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta} = 0$

For a fraction to be zero, its numerator must be zero (provided the denominator is not zero). The denominator $1 + 4\sin^2\theta$ is always greater than or equal to 1 since $\sin^2\theta \ge 0$, so it can never be zero.

Thus, we only need to set the numerator to zero: $3 - 4\sin^2\theta = 0$

Step 4: Solve the trigonometric equation for $\sin\theta$

From the equation $3 - 4\sin^2\theta = 0$, we isolate $\sin^2\theta$:

$4\sin^2\theta = 3$ $\sin^2\theta = \frac{3}{4}$

Now, take the square root of both sides to find $\sin\theta$:

$\sin\theta = \pm\sqrt{\frac{3}{4}}$ $\sin\theta = \pm\frac{\sqrt{3}}{2}$

Why both positive and negative roots? We must consider both positive and negative possibilities for $\sin\theta$ because squaring either a positive or a negative number yields a positive result.

Step 5: Find the values of $\theta$ in the given interval

We need to find values of $\theta$ in the interval $\left( -\frac{\pi}{2}, \pi \right)$ such that $\sin\theta = \frac{\sqrt{3}}{2}$ or $\sin\theta = -\frac{\sqrt{3}}{2}$.

• Case 1: $\sin\theta = \frac{\sqrt{3}}{2}$ The solutions in the interval $[0, 2\pi)$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{2\pi}{3}$. Both are within the given interval $\left( -\frac{\pi}{2}, \pi \right)$.

• Case 2: $\sin\theta = -\frac{\sqrt{3}}{2}$ The solutions in the interval $[0, 2\pi)$ are $\theta = \frac{4\pi}{3}$ and $\theta = \frac{5\pi}{3}$. However, these are outside the given interval $\left( -\frac{\pi}{2}, \pi \right)$. Since the sine function is negative in the fourth quadrant, we can find a solution in the interval $\left( -\frac{\pi}{2}, 0 \right)$ by taking $\theta = -\frac{\pi}{3}$.

Combining all valid solutions within the interval $\left( -\frac{\pi}{2}, \pi \right)$, the set $A$ is: $A = \left\{ -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} \right\}$

Step 6: Calculate the sum of the elements in A

Finally, we sum the elements found in set $A$:

Sum $= -\frac{\pi}{3} + \frac{\pi}{3} + \frac{2\pi}{3} = \frac{2\pi}{3}$

Common Mistakes & Tips

• Interval Check: Always check if the solutions obtained lie within the specified interval.
• Conjugate Multiplication: Double-check the algebraic expansion when multiplying by the conjugate to avoid errors with signs or $i^2$.
• Trigonometric Solutions: Remember to consider all possible solutions for $\sin\theta$ and $\cos\theta$ within the given interval.

Summary

This problem involves simplifying a complex number, applying the condition for it to be purely imaginary, solving the resulting trigonometric equation, and finding the sum of the solutions within the given interval. The key is to carefully simplify the expression, apply the condition correctly, and check the solutions against the interval.

The final answer is $\boxed{\frac{2\pi}{3}}$, which corresponds to option (D).