Let `z=(costheta+isintheta)/(costheta-isintheta), pi/4 lt theta lt pi/2`. Then arg(z) =

Options

A. `2theta`
B. `2theta-pi`
C. `pi+2theta`
D. none of these

Explanation

To find the argument of the complex number \( z = \frac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta} \) for \( \frac{\pi}{4} < \theta < \frac{\pi}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Write down the expression for \( z \)**:
\[ z = \frac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta} \]

2. **Rationalize the denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(\cos \theta + i \sin \theta)(\cos \theta + i \sin \theta)}{(\cos \theta - i \sin \theta)(\cos \theta + i \sin \theta)} \]
This simplifies to:
\[ z = \frac{(\cos \theta + i \sin \theta)^2}{\cos^2 \theta + \sin^2 \theta} \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:
\[ z = (\cos \theta + i \sin \theta)^2 \]

3. **Use the identity for complex exponentials**:
The expression \( \cos \theta + i \sin \theta \) can be represented as \( e^{i\theta} \). Therefore:
\[ z = (e^{i\theta})^2 = e^{2i\theta} \]

4. **Find the argument of \( z \)**:
The argument of \( e^{2i\theta} \) is simply \( 2\theta \):
\[ \arg(z) = 2\theta \]

5. **Determine the range of \( 2\theta \)**:
Since \( \frac{\pi}{4} < \theta < \frac{\pi}{2} \), multiplying by 2 gives:
\[ \frac{\pi}{2} < 2\theta < \pi \]
This means \( 2\theta \) lies in the interval \( \left(\frac{\pi}{2}, \pi\right) \).

### Final Answer:
Thus, the argument of \( z \) is:
\[ \arg(z) = 2\theta \]