Understanding the Complex Number Problem

The problem asks us to find the angle \(\theta\) for which the given complex number \(z = \frac{1+\text{i}\sin \theta}{1−\text{i}\sin \theta}\) is purely real. A complex number is considered purely real if its imaginary part is equal to zero. We are given that i = \(\sqrt{-1}\).

Simplifying the Complex Number Expression

To find the real and imaginary parts of \(z\), we need to simplify the expression. We can do this by multiplying the numerator and the denominator by the conjugate of the denominator. The denominator is \(1-\text{i}\sin \theta\), and its conjugate is \(1+\text{i}\sin \theta\).

Let's perform the multiplication:

\[z = \frac{1+\text{i}\sin \theta}{1−\text{i}\sin \theta} \times \frac{1+\text{i}\sin \theta}{1+\text{i}\sin \theta}\]

Multiply the numerators and denominators:

Numerator: \((1+\text{i}\sin \theta)(1+\text{i}\sin \theta) = 1^2 + 2(1)(\text{i}\sin \theta) + (\text{i}\sin \theta)^2 = 1 + 2\text{i}\sin \theta + \text{i}^2\sin^2 \theta\)

Since \(\text{i}^2 = -1\), the numerator becomes: \(1 + 2\text{i}\sin \theta - \sin^2 \theta\)

Denominator: \((1-\text{i}\sin \theta)(1+\text{i}\sin \theta)\). This is in the form \((a-b)(a+b) = a^2 - b^2\).

Denominator: \(1^2 - (\text{i}\sin \theta)^2 = 1 - \text{i}^2\sin^2 \theta = 1 - (-\sin^2 \theta) = 1 + \sin^2 \theta\)

So, the simplified complex number \(z\) is:

\[z = \frac{1 - \sin^2 \theta + 2\text{i}\sin \theta}{1 + \sin^2 \theta}\]

Separating Real and Imaginary Parts

We can write \(z\) in the standard form \(a + \text{i}b\) by separating the real and imaginary parts:

\[z = \frac{1 - \sin^2 \theta}{1 + \sin^2 \theta} + \text{i}\frac{2\sin \theta}{1 + \sin^2 \theta}\]

Using the trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\), which means \(1 - \sin^2 \theta = \cos^2 \theta\), we can rewrite the real part:

\[z = \frac{\cos^2 \theta}{1 + \sin^2 \theta} + \text{i}\frac{2\sin \theta}{1 + \sin^2 \theta}\]

The real part of \(z\) is \(\text{Re}(z) = \frac{\cos^2 \theta}{1 + \sin^2 \theta}\).

The imaginary part of \(z\) is \(\text{Im}(z) = \frac{2\sin \theta}{1 + \sin^2 \theta}\).

Finding the Condition for z to be Purely Real

For \(z\) to be purely real, the imaginary part must be zero.

\[\text{Im}(z) = 0\]

\[\frac{2\sin \theta}{1 + \sin^2 \theta} = 0\]

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

Therefore, we only need the numerator to be zero:

\[2\sin \theta = 0\]

\[\sin \theta = 0\]

Solving the Trigonometric Equation for Angle θ

We need to find the general solution for the equation \(\sin \theta = 0\). The sine function is zero at integer multiples of \(\pi\).

So, the general solution is:

\[\theta = n\pi\]

where \(n\) is an integer (\(n \in \mathbb{Z}\)).

Comparing with Given Options

Let's examine the options provided:

• \(\frac{n\pi}{2}\): This includes values like \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, ...\). \(\sin(\pi/2) = 1 \neq 0\). So this is not correct.
• \(\frac{(2n+1)\pi}{2}\): This includes values like \(\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...\) (odd multiples of \(\pi/2\)). \(\sin(\theta) = \pm 1\) for these values. So this is not correct.
• \(n\pi\): This includes values like \(..., -2\pi, -\pi, 0, \pi, 2\pi, ...\) (integer multiples of \(\pi\)). For all these values, \(\sin(n\pi) = 0\). This matches our solution.
• \(2n\pi\) only: This includes values like \(..., -4\pi, -2\pi, 0, 2\pi, 4\pi, ...\) (even multiples of \(\pi\)). While \(\sin(2n\pi) = 0\), this does not include odd multiples of \(\pi\) (like \(\pi, 3\pi\)), for which \(\sin \theta\) is also 0. Thus, this is not the complete solution.

The angle \(\theta\) such that \(z\) is purely real is \(n\pi\), where \(n\) is an integer.

Revision Table: Complex Numbers and Purely Real Condition

Additional Information: Trigonometric Equations

Solving trigonometric equations is crucial in complex numbers and other areas of mathematics. Here are some basic general solutions:

• \(\sin x = 0 \implies x = n\pi\), where \(n \in \mathbb{Z}\).
• \(\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}\), where \(n \in \mathbb{Z}\).
• \(\tan x = 0 \implies x = n\pi\), where \(n \in \mathbb{Z}\).
• \(\sin x = \sin \alpha \implies x = n\pi + (-1)^n \alpha\), where \(n \in \mathbb{Z}\).
• \(\cos x = \cos \alpha \implies x = 2n\pi \pm \alpha\), where \(n \in \mathbb{Z}\).
• \(\tan x = \tan \alpha \implies x = n\pi + \alpha\), where \(n \in \mathbb{Z}\).

In our problem, we solved \(\sin \theta = 0\), which directly corresponds to the first case, giving \(\theta = n\pi\).