Consider the complex number `z = (1 - isin theta)/(1+ icos theta)`.
If agrument of z is `pi/4`, then (a) `theta = npi, n in I` only (b) `theta= (2n + 1), n in I `only (c) both `theta= npi and theta = (2n + 1)(pi)/(2), n in I` (d) none of these

Options

A. `theta = npi, n in I` only
B. `theta= (2n + 1), n in I `only
C. both `theta= npi and theta = (2n + 1)(pi)/(2), n in I`
D. none of these

Explanation

To solve the problem, we need to analyze the complex number \( z = \frac{1 - i \sin \theta}{1 + i \cos \theta} \) and find the values of \( \theta \) such that the argument of \( z \) is \( \frac{\pi}{4} \).

### Step 1: Rationalize the denominator
We start by multiplying the numerator and denominator by the conjugate of the denominator:

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

### Step 2: Simplify the denominator
The denominator simplifies as follows:

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

### Step 3: Expand the numerator
Now expand the numerator:

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

### Step 4: Combine the results
Putting it all together, we have:

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

### Step 5: Identify the real and imaginary parts
The real part \( \text{Re}(z) \) and imaginary part \( \text{Im}(z) \) are:

\[ \text{Re}(z) = \frac{1 + \sin \theta \cos \theta}{1 + \cos^2 \theta} \]
\[ \text{Im}(z) = \frac{-(\sin \theta + \cos \theta)}{1 + \cos^2 \theta} \]

### Step 6: Set up the argument condition
Given that the argument of \( z \) is \( \frac{\pi}{4} \), we have:

\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

Thus, we set up the equation:

\[ \frac{\text{Im}(z)}{\text{Re}(z)} = 1 \]

This leads to:

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

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:

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

### Step 8: Expand and simplify the equation
Expanding both sides and simplifying leads to:

\[ -\sin \theta - \cos \theta - \sin \theta \cos^2 \theta - \cos \theta \cos^2 \theta = 1 + 2 \sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta \]

### Step 9: Set up a quadratic equation
Rearranging gives a quadratic equation in terms of \( \sin \theta \) and \( \cos \theta \).

### Step 10: Solve for \( \theta \)
From the simplifications, we find that:

1. Either \( \sin \theta = 0 \) which gives \( \theta = n\pi \) for \( n \in \mathbb{Z} \).
2. Or \( \cos \theta = 0 \) which gives \( \theta = (2n + 1)\frac{\pi}{2} \) for \( n \in \mathbb{Z} \).

### Conclusion
Thus, the values of \( \theta \) that satisfy the condition are:

\[ \theta = n\pi \quad \text{or} \quad \theta = (2n + 1)\frac{\pi}{2}, \quad n \in \mathbb{Z} \]

The correct answer is option (c): both \( \theta = n\pi \) and \( \theta = (2n + 1)\frac{\pi}{2} \).