The real vaule of `theta` for which the expression `(1+icostheta)/(1-2icos theta)` is a purely imaginary number is

Options

A. `npi pm (pi)/(6)`
B. `npi pm (pi)/(3)`
C. `(2n+1) (pi)/(2)`
D. ` npi pm (pi)/(4)`

Answer

To find the real value of \( \theta \) for which the expression

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

is purely imaginary, we will follow these steps:

### Step 1: Define the expression
Let

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

### Step 2: Multiply numerator and denominator by the conjugate of the denominator
To simplify \( z \), we multiply both the numerator and the denominator by the conjugate of the denominator, which is \( 1 + 2 i \cos \theta \):

\[ z = \frac{(1 + i \cos \theta)(1 + 2 i \cos \theta)}{(1 - 2 i \cos \theta)(1 + 2 i \cos \theta)} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:

\[ (1 - 2 i \cos \theta)(1 + 2 i \cos \theta) = 1^2 - (2 i \cos \theta)^2 = 1 - 4(-\cos^2 \theta) = 1 + 4 \cos^2 \theta \]

### Step 4: Simplify the numerator
Now, simplify the numerator:

\[ (1 + i \cos \theta)(1 + 2 i \cos \theta) = 1 + 2 i \cos \theta + i \cos \theta + 2 i^2 \cos^2 \theta = 1 + 3 i \cos \theta - 2 \cos^2 \theta \]

### Step 5: Combine the results
Now we can write \( z \) as:

\[ z = \frac{(1 - 2 \cos^2 \theta) + 3 i \cos \theta}{1 + 4 \cos^2 \theta} \]

### Step 6: Separate real and imaginary parts
This gives us:

\[ z = \frac{1 - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta} + i \frac{3 \cos \theta}{1 + 4 \cos^2 \theta} \]

### Step 7: Set the real part to zero
For \( z \) to be purely imaginary, the real part must be zero:

\[ \frac{1 - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta} = 0 \]

This implies:

\[ 1 - 2 \cos^2 \theta = 0 \]

### Step 8: Solve for \( \cos^2 \theta \)
Solving for \( \cos^2 \theta \):

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

### Step 9: Find \( \cos \theta \)
Taking the square root:

\[ \cos \theta = \pm \frac{1}{\sqrt{2}} = \pm \cos \frac{\pi}{4} \]

### Step 10: Determine \( \theta \)
Thus, the general solutions for \( \theta \) are:

\[ \theta = n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Conclusion
The values of \( \theta \) for which the expression is purely imaginary are given by:

\[ \theta = n\pi \pm \frac{\pi}{4} \]