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

Options

A. `- theta`
B. `(pi)/(2)+ theta`
C. `theta`
D. ` pi- theta`

Answer

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

### Step-by-Step Solution:

1. **Express \(z\) in terms of its modulus and argument**:
Since \(z\) is a complex number of unit modulus, we can express it as:
\[ z = e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
where \(\bar{z}\) (the conjugate of \(z\)) is:
\[ \bar{z} = \cos(\theta) - i\sin(\theta) \]

2. **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:
\[ \frac{1 + \cos(\theta) + i\sin(\theta)}{1 - \cos(\theta) + i\sin(\theta)} \]

3. **Multiply numerator and denominator by the conjugate of the denominator**:
To make the denominator real, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(1 + \cos(\theta) + i\sin(\theta))(1 - \cos(\theta) - i\sin(\theta))}{(1 - \cos(\theta) + i\sin(\theta))(1 - \cos(\theta) - i\sin(\theta))} \]

4. **Simplify the denominator**:
The denominator becomes:
\[ (1 - \cos(\theta))^2 + \sin^2(\theta) = 1 - 2\cos(\theta) + \cos^2(\theta) + \sin^2(\theta) = 2(1 - \cos(\theta)) \]

5. **Simplify the numerator**:
The numerator expands to:
\[ (1 + \cos(\theta))(1 - \cos(\theta)) - \sin^2(\theta) + i\sin(\theta)(1 - \cos(\theta) + 1 + \cos(\theta)) \]
Simplifying gives:
\[ (1 - \cos^2(\theta) - \sin^2(\theta)) + i\sin(\theta)(2) = 0 + 2i\sin(\theta) \]

6. **Combine the results**:
Thus, we have:
\[ \frac{2i\sin(\theta)}{2(1 - \cos(\theta))} = \frac{i\sin(\theta)}{1 - \cos(\theta)} \]

7. **Find the argument**:
The argument of a complex number \(\frac{a + bi}{c}\) is given by:
\[ \arg\left(\frac{a + bi}{c}\right) = \arg(a + bi) + \arg(c) \]
Here, since \(c\) is real, we have:
\[ \arg\left(\frac{i\sin(\theta)}{1 - \cos(\theta)}\right) = \arg(i\sin(\theta)) = \frac{\pi}{2} \text{ (since } \sin(\theta) > 0\text{)} \]

8. **Final result**:
Therefore, we conclude:
\[ \arg\left(\frac{1 + z}{1 - \bar{z}}\right) = \frac{\pi}{2} + \theta \]