Question

The sum of all possible values of \(\theta \in [-\pi, 2\pi]\), for which \[ \frac{1 + i \cos\theta}{1 - 2i \cos\theta} \] is purely imaginary, is equal to:

• A. \(2\pi\)
• B. \(3\pi\)
• C. \(5\pi\)
• D. \(4\pi\)

Answer 1

The Correct Option is B

Let:

\[ Z = \frac{1 + i \cos \theta}{1 - 2i \cos \theta}. \]

For \( Z \) to be purely imaginary:

\[ Z + \overline{Z} = 0, \]

where \( \overline{Z} \) is the complex conjugate of \( Z \). Thus:

\[ Z + \overline{Z} = \frac{1 + i \cos \theta}{1 - 2i \cos \theta} + \frac{1 - i \cos \theta}{1 + 2i \cos \theta} = 0. \]

Simplify:

\[ (1 + i \cos \theta)(1 - 2i \cos \theta) + (1 - i \cos \theta)(1 + 2i \cos \theta) = 0. \]

Expand both terms:

\[ (1 + i \cos \theta)(1 - 2i \cos \theta) = 1 - 2i \cos \theta + i \cos \theta - 2 \cos^2 \theta, \]

\[ (1 - i \cos \theta)(1 + 2i \cos \theta) = 1 + 2i \cos \theta - i \cos \theta - 2 \cos^2 \theta. \]

Combine:

\[ (1 - 2i \cos \theta + i \cos \theta - 2 \cos^2 \theta) + (1 + 2i \cos \theta - i \cos \theta - 2 \cos^2 \theta) = 0. \]

Simplify further:

\[ 2 - 4 \cos^2 \theta = 0. \]

Solve for \( \cos^2 \theta \):

\[ \cos^2 \theta = \frac{1}{2}. \]

Step 2: Find all possible values of \( \theta \). If \( \cos^2 \theta = \frac{1}{2} \), then:

\[ \cos \theta = \pm \frac{1}{\sqrt{2}}. \]

The possible values of \( \theta \in [-\pi, \pi] \) are:

\[ \theta = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}. \]

Step 3: Sum of all possible values

\[ \text{Sum} = \frac{\pi}{4} + \left( -\frac{\pi}{4} \right) + \frac{3\pi}{4} + \left( -\frac{3\pi}{4} \right) = 3\pi. \]
Final Answer: \( 3\pi \).