Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`.
The value of `theta` for which z is purely imaginary are

Options

A. `npi-(pi)/(4), n in I`
B. `pin +(pi)/(4) , n in I`
C. `npi, n in I`
D. no real values of `theta`

Explanation

To find the values of \( \theta \) for which the complex number

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

is purely imaginary, we will follow these steps:

### Step 1: Multiply by the Conjugate
To simplify \( z \), we multiply the numerator and the 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 becomes:

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

### Step 3: Expand the Numerator
Now, we expand the numerator:

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

Since \( i^2 = -1 \), this simplifies to:

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

### Step 4: Combine the Results
Now we can write \( z \) as:

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

### Step 5: Identify Real and Imaginary Parts
For \( z \) to be purely imaginary, the real part must be zero. Thus, we set the real part to zero:

\[ 1 - \sin \theta \cos \theta = 0 \]

### Step 6: Solve the Equation
This gives us:

\[ \sin \theta \cos \theta = 1 \]

Using the double angle identity, we can rewrite this as:

\[ \frac{1}{2} \sin(2\theta) = 1 \]

Thus,

\[ \sin(2\theta) = 2 \]

### Step 7: Analyze the Result
The sine function has a maximum value of 1, so \( \sin(2\theta) = 2 \) is impossible. Therefore, there are no values of \( \theta \) for which \( z \) is purely imaginary.

### Final Answer
The conclusion is that there are no values of \( \theta \) for which \( z \) is purely imaginary.

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