If z is a complex number of unit modulus and argument `theta`, then `arg((1+z)/(1+bar(z)))` equals to

Options

A. `pi/2-theta`
B. `theta`
C. `pi-theta`
D. `-theta`

Answer

To solve the problem, we need to find the argument of the expression \(\frac{1 + z}{1 + \bar{z}}\) where \(z\) is a complex number of unit modulus and argument \(\theta\).

### Step-by-Step Solution:

1. **Express \(z\)**:
Since \(z\) has unit modulus and argument \(\theta\), we can express \(z\) in the polar form:
\[ z = e^{i\theta} = \cos(\theta) + i\sin(\theta) \]

2. **Find \(\bar{z}\)**:
The conjugate of \(z\) is given by:
\[ \bar{z} = e^{-i\theta} = \cos(\theta) - i\sin(\theta) \]

3. **Substitute \(z\) and \(\bar{z}\) into the expression**:
We need to evaluate:
\[ \frac{1 + z}{1 + \bar{z}} = \frac{1 + (\cos(\theta) + i\sin(\theta))}{1 + (\cos(\theta) - i\sin(\theta))} \]
This simplifies to:
\[ \frac{1 + \cos(\theta) + i\sin(\theta)}{1 + \cos(\theta) - i\sin(\theta)} \]

4. **Multiply numerator and denominator by the conjugate of the denominator**:
To simplify the expression, we multiply both the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(1 + \cos(\theta) + i\sin(\theta))(1 + \cos(\theta) + i\sin(\theta))}{(1 + \cos(\theta))^2 + \sin^2(\theta)} \]

5. **Simplify the denominator**:
The denominator simplifies as follows:
\[ (1 + \cos(\theta))^2 + \sin^2(\theta) = 1 + 2\cos(\theta) + \cos^2(\theta) + \sin^2(\theta) = 2 + 2\cos(\theta) = 2(1 + \cos(\theta)) \]

6. **Simplify the numerator**:
The numerator becomes:
\[ (1 + \cos(\theta))^2 + i(1 + \cos(\theta))\sin(\theta) = (1 + \cos(\theta))^2 + i(1 + \cos(\theta))\sin(\theta) \]

7. **Combine the results**:
Thus, we have:
\[ \frac{(1 + \cos(\theta))^2 + i(1 + \cos(\theta))\sin(\theta)}{2(1 + \cos(\theta))} \]
This simplifies to:
\[ \frac{1 + \cos(\theta)}{2} + i\frac{\sin(\theta)}{2} \]

8. **Find the argument**:
The argument of a complex number \(x + iy\) is given by \(\tan^{-1}(\frac{y}{x})\). Here:
\[ x = \frac{1 + \cos(\theta)}{2}, \quad y = \frac{\sin(\theta)}{2} \]
Therefore, the argument is:
\[ \arg\left(\frac{1 + z}{1 + \bar{z}}\right) = \tan^{-1}\left(\frac{\frac{\sin(\theta)}{2}}{\frac{1 + \cos(\theta)}{2}}\right) = \tan^{-1}\left(\frac{\sin(\theta)}{1 + \cos(\theta)}\right) \]

9. **Use trigonometric identities**:
We know that:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \]
Thus:
\[ \arg\left(\frac{1 + z}{1 + \bar{z}}\right) = \frac{\theta}{2} \]

### Final Result:
The argument \(\arg\left(\frac{1 + z}{1 + \bar{z}}\right) = \frac{\theta}{2}\).