Key Concepts and Formulas

• A complex number $Z = x + iy$ is purely imaginary if its real part, $x$, is zero. i.e., $Re(Z) = 0$.
• To rationalize a complex number of the form $\frac{a+bi}{c+di}$, multiply both the numerator and the denominator by the conjugate of the denominator, which is $c-di$.
• Remember that $i^2 = -1$.

Step-by-Step Solution

Step 1: Rationalize the denominator

We are given the complex number $Z = \frac{2 + 3i\sin\theta}{1 - 2i\sin\theta}$. To express it in the standard form $x+iy$, we first rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $1 + 2i\sin\theta$.

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

Step 2: Expand the numerator and denominator

Now we expand the numerator and the denominator separately.

• Numerator: $(2 + 3i\sin\theta)(1 + 2i\sin\theta) = 2(1) + 2(2i\sin\theta) + 3i\sin\theta(1) + 3i\sin\theta(2i\sin\theta)$ $= 2 + 4i\sin\theta + 3i\sin\theta + 6i^2\sin^2\theta$ Since $i^2 = -1$, we have: $= 2 + 7i\sin\theta - 6\sin^2\theta$ $= (2 - 6\sin^2\theta) + 7i\sin\theta$

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

Step 3: Express Z in the form x + iy

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

$Z = \frac{(2 - 6\sin^2\theta) + 7i\sin\theta}{1 + 4\sin^2\theta} = \frac{2 - 6\sin^2\theta}{1 + 4\sin^2\theta} + i\frac{7\sin\theta}{1 + 4\sin^2\theta}$

So, we have $Z = x + iy$, where $x = \frac{2 - 6\sin^2\theta}{1 + 4\sin^2\theta}$ and $y = \frac{7\sin\theta}{1 + 4\sin^2\theta}$.

Step 4: Apply the condition for a purely imaginary number

For $Z$ to be purely imaginary, its real part, $x$, must be zero. Thus, we need to solve:

$\frac{2 - 6\sin^2\theta}{1 + 4\sin^2\theta} = 0$

Step 5: Solve for sin θ

For the fraction to be zero, the numerator must be zero (and the denominator must be non-zero, which it always is in this case). So:

$2 - 6\sin^2\theta = 0$ $6\sin^2\theta = 2$ $\sin^2\theta = \frac{2}{6} = \frac{1}{3}$ $\sin\theta = \pm\sqrt{\frac{1}{3}} = \pm\frac{1}{\sqrt{3}}$

Step 6: Match with the options

We are looking for a value of $\theta$ such that $\sin\theta = \frac{1}{\sqrt{3}}$ or $\sin\theta = -\frac{1}{\sqrt{3}}$. The options are:

(A) ${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 4}} \right)$ (B) ${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)\,$ (C) ${\pi \over 3}$ (D) ${\pi \over 6}$

Option (B) is $\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$, which means $\sin\theta = \frac{1}{\sqrt{3}}$. This matches one of our solutions.

Common Mistakes & Tips:

• Remember to multiply both the numerator and the denominator by the conjugate when rationalizing.
• Carefully expand and simplify the expressions, especially when dealing with $i^2$.
• When solving for $\sin \theta$, remember to consider both positive and negative roots.

Summary:

To find the value of $\theta$ for which the given complex number is purely imaginary, we rationalized the denominator, separated the real and imaginary parts, set the real part equal to zero, and solved for $\sin \theta$. Comparing the result with the given options, we find that option (A) is the correct answer.

The final answer is \boxed{{\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 4}} \right)}, which corresponds to option (A).