The question asks for the angle \(\theta\) such that the given complex number $z$ is purely imaginary. A complex number is purely imaginary if its real part is equal to zero and its imaginary part is not equal to zero.

The given complex number is \(z = \frac{1+\text{i}\sin \theta}{1−\text{i}\sin \theta}\). To find the real and imaginary parts, we need to express $z$ in the standard form $x + iy$. We can do this by multiplying the numerator and the denominator by the conjugate of the denominator.

The conjugate of the denominator \(1 - i \sin \theta\) is \(1 + i \sin \theta\).

Let's simplify $z$:

\(\qquad z = \frac{1+\text{i}\sin \theta}{1−\text{i}\sin \theta} \times \frac{1+\text{i}\sin \theta}{1+\text{i}\sin \theta}\)

The numerator is \((1 + i \sin \theta)^2\). Expanding this, we get:

\(\qquad (1 + i \sin \theta)^2 = 1^2 + 2(1)(i \sin \theta) + (i \sin \theta)^2\)

\(\qquad = 1 + 2i \sin \theta + i^2 \sin^2 \theta\)

Since \(i^2 = -1\), this becomes:

\(\qquad = 1 + 2i \sin \theta - \sin^2 \theta\)

The denominator is \((1 - i \sin \theta)(1 + i \sin \theta)\). This is of the form \((a-b)(a+b) = a^2 - b^2\).

\(\qquad (1 - i \sin \theta)(1 + i \sin \theta) = 1^2 - (i \sin \theta)^2\)

\(\qquad = 1 - i^2 \sin^2 \theta\)

Since \(i^2 = -1\), this becomes:

\(\qquad = 1 - (-\sin^2 \theta) = 1 + \sin^2 \theta\)

Now, substitute the simplified numerator and denominator back into the expression for $z$:

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

We can separate this into real and imaginary parts:

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

For $z$ to be purely imaginary, the real part must be zero.

Real part of \(z = \frac{1 - \sin^2 \theta}{1 + \sin^2 \theta}\).

Set the real part to zero:

\(\qquad \frac{1 - \sin^2 \theta}{1 + \sin^2 \theta} = 0\)

For a fraction to be zero, the numerator must be zero, provided the denominator is non-zero.

Numerator: \(1 - \sin^2 \theta = 0\)

\(\qquad \sin^2 \theta = 1\)

This implies \(\sin \theta = 1\) or \(\sin \theta = -1\).

Denominator: \(1 + \sin^2 \theta\). Since \(\sin^2 \theta \ge 0\), \(1 + \sin^2 \theta \ge 1\). Thus, the denominator is never zero.

So, the condition for $z$ to be purely imaginary is \(\sin \theta = 1\) or \(\sin \theta = -1\). These values occur when \(\theta\) is an odd multiple of \(\frac{\pi}{2}\).

The general solution for \(\sin \theta = 1\) is \(\theta = 2n\pi + \frac{\pi}{2}\), where $n$ is an integer.

The general solution for \(\sin \theta = -1\) is \(\theta = 2n\pi - \frac{\pi}{2}\), where $n$ is an integer.

These two sets of solutions can be combined into a single expression representing all odd multiples of \(\frac{\pi}{2}\). An odd integer can be represented as $2n+1$ for any integer $n$. Thus, the general solution is \(\theta = (2n+1)\frac{\pi}{2} = \frac{(2n+1)\pi}{2}\), where $n$ is an integer.

We also need to ensure the imaginary part is non-zero when the real part is zero. The imaginary part is \(\frac{2 \sin \theta}{1 + \sin^2 \theta}\). When \(\sin \theta = \pm 1\), the imaginary part is \(\frac{2(\pm 1)}{1 + (\pm 1)^2} = \frac{\pm 2}{1+1} = \frac{\pm 2}{2} = \pm 1\). Since \(\pm 1\) is non-zero, $z$ is indeed purely imaginary for these values of \(\theta\).

Let's compare this result with the given options:

• Option 1: \(\frac{n\pi}{2}\). This includes angles like \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \dots\). For \(\theta = 0\) or \(\theta = \pi\) (where \(\sin \theta = 0\)), \(z = \frac{1+0}{1-0} = 1\), which is purely real. So, this option is incorrect.
• Option 2: \(\frac{(2n+1)\pi}{2}\). This represents angles like \(\frac{\pi}{2}, \frac{3\pi}{2}, -\frac{\pi}{2}, \dots\) where \(\sin \theta = \pm 1\). For these values, the real part is zero and the imaginary part is non-zero, making $z$ purely imaginary. This matches our result.
• Option 3: \(n\pi\). This represents angles like \(0, \pi, 2\pi, \dots\) where \(\sin \theta = 0\). As shown above, $z=1$, which is purely real. Incorrect.
• Option 4: \(2n\pi\). This represents angles like \(0, 2\pi, 4\pi, \dots\) where \(\sin \theta = 0\). As shown above, $z=1$, which is purely real. Incorrect.

Therefore, the angle \(\theta\) for which $z$ is purely imaginary is \(\frac{(2n+1)\pi}{2}\), where $n$ is an integer.

Complex Number Analysis Summary

To summarize the analysis of the complex number \(z = \frac{1+\text{i}\sin \theta}{1−\text{i}\sin \theta}\), we performed the following steps:

• Expressed the complex number in the form $x+iy$.
• Identified the real part \(x = \frac{1 - \sin^2 \theta}{1 + \sin^2 \theta}\).
• Set the real part to zero: \(1 - \sin^2 \theta = 0\).
• Solved the trigonometric equation \(\sin^2 \theta = 1\), yielding \(\sin \theta = \pm 1\).
• Determined the general solution for \(\theta\) corresponding to \(\sin \theta = \pm 1\).
• Verified that the imaginary part is non-zero for these values of \(\theta\).

Revision Table - Complex Numbers and Trigonometry

Additional Information - Properties of Purely Imaginary Numbers

A complex number $z$ is said to be purely imaginary if it can be written in the form $z = iy$, where $y$ is a real number and \(y \ne 0\). The set of purely imaginary numbers is the set \(\{iy \mid y \in \mathbb{R}, y \ne 0\}\).

• The real part of a purely imaginary number is 0.
• The imaginary part of a purely imaginary number is non-zero.
• Geometrically, purely imaginary numbers lie on the imaginary axis in the complex plane, excluding the origin $(0,0)$. The origin $0 = 0 + i0$ has both real and imaginary parts equal to zero, and is considered purely real and purely imaginary simultaneously. However, typically "purely imaginary" implies the imaginary part is non-zero. In the context of an MCQ, the requirement that the real part is zero is the primary condition to isolate these cases from purely real numbers. The question implies $z$ is non-zero as it's a ratio. If $z=0$, then \(1+i\sin\theta = 0\), which is impossible as the real part is 1.
• If $z$ is purely imaginary, then \(z = - \bar{z}\) (where \(\bar{z}\) is the complex conjugate). If $z = iy$, then \(\bar{z} = -iy\). So, $z = -(-iy) = iy$, which holds. If $z = x+iy$, \(\bar{z} = x-iy\). The condition \(z = -\bar{z}\) gives $x+iy = -(x-iy) = -x+iy$, which implies $x = -x$, so $2x=0$, meaning $x=0$. This confirms that having a zero real part is key.
• Another property: If $z$ is purely imaginary, then \(z^2\) is a non-positive real number. For example, \((3i)^2 = 9i^2 = -9\), \((-5i)^2 = 25i^2 = -25\). If $z=iy$ with \(y\ne 0\), \(z^2 = (iy)^2 = i^2 y^2 = -y^2\). Since $y$ is real and \(y \ne 0\), \(y^2 > 0\), so \(-y^2 < 0\).

These properties help in understanding the nature of complex numbers and their classification.