If z is a complex number of unit modulus and argument `theta`, then the real part of `(z(1-barz))/(barz(1+z))`, is

Options

A. `2cos^(2)(theta/2)`
B. `1-cos(theta/2)`
C. `1+sin(pi/2)`
D. `-2sin^(2)(theta/2)`

Explanation

To find the real part of the expression \(\frac{z(1 - \bar{z})}{\bar{z}(1 + z)}\) where \(z\) is a complex number of unit modulus and argument \(\theta\), we can follow these steps:

### Step 1: Express \(z\) in terms of \(\theta\)
Since \(z\) is a complex number of unit modulus, we can express it as:
\[ z = e^{i\theta} = \cos \theta + i \sin \theta \]
The conjugate of \(z\) is:
\[ \bar{z} = e^{-i\theta} = \cos \theta - i \sin \theta \]

### Step 2: Substitute \(z\) and \(\bar{z}\) into the expression
Substituting \(z\) and \(\bar{z}\) into the expression, we have:
\[ \frac{z(1 - \bar{z})}{\bar{z}(1 + z)} = \frac{e^{i\theta}(1 - e^{-i\theta})}{e^{-i\theta}(1 + e^{i\theta})} \]

### Step 3: Simplify the numerator and denominator
Calculating the numerator:
\[ 1 - \bar{z} = 1 - (\cos \theta - i \sin \theta) = 1 - \cos \theta + i \sin \theta \]
Thus, the numerator becomes:
\[ z(1 - \bar{z}) = e^{i\theta}(1 - \cos \theta + i \sin \theta) = (\cos \theta + i \sin \theta)(1 - \cos \theta + i \sin \theta) \]

Calculating the denominator:
\[ 1 + z = 1 + (\cos \theta + i \sin \theta) = 1 + \cos \theta + i \sin \theta \]
Thus, the denominator becomes:
\[ \bar{z}(1 + z) = e^{-i\theta}(1 + \cos \theta + i \sin \theta) = (\cos \theta - i \sin \theta)(1 + \cos \theta + i \sin \theta) \]

### Step 4: Expand both the numerator and denominator
Expanding the numerator:
\[ (\cos \theta + i \sin \theta)(1 - \cos \theta + i \sin \theta) = \cos \theta(1 - \cos \theta) + i \cos \theta \sin \theta + i \sin \theta(1 - \cos \theta) - \sin^2 \theta \]
Combining the real and imaginary parts:
\[ = \cos \theta(1 - \cos \theta) - \sin^2 \theta + i(\cos \theta \sin \theta + \sin \theta(1 - \cos \theta)) \]

Expanding the denominator:
\[ (\cos \theta - i \sin \theta)(1 + \cos \theta + i \sin \theta) = \cos \theta(1 + \cos \theta) + i \cos \theta \sin \theta - i \sin \theta(1 + \cos \theta) - \sin^2 \theta \]
Combining the real and imaginary parts:
\[ = \cos \theta(1 + \cos \theta) - \sin^2 \theta + i(\cos \theta \sin \theta - \sin \theta(1 + \cos \theta)) \]

### Step 5: Formulate the final expression
Now we have:
\[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{\text{Real part} + i \text{Imaginary part}}{\text{Real part} + i \text{Imaginary part}} \]

### Step 6: Find the real part
To find the real part of the expression, we can use the fact that the real part of a complex fraction \(\frac{a + bi}{c + di}\) can be computed using:
\[ \text{Re}\left(\frac{a + bi}{c + di}\right) = \frac{ac + bd}{c^2 + d^2} \]
where \(a\), \(b\), \(c\), and \(d\) are the respective real and imaginary parts of the numerator and denominator.

### Final Result
After simplification, we find that the real part of the expression is:
\[ -2 \sin^2\left(\frac{\theta}{2}\right) \]