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

Answer: (B)

$3\pi$

Answer 2

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: $3\pi$.