Let `z(ne -1)` be any complex number such that |z| = 1. Then the imaginary part of `(bar(z)(1-z))/(z(1+bar(z)))`is : (Here `theta` = Arg(z))

Options

A. `-tan((theta)/(2))sin theta`
B. `tan((theta)/(2))cos theta`
C. `-tan((theta)/(2))cos theta`
D. `tan((theta)/(2))sin theta`

Answer

To find the imaginary part of the expression \(\frac{\bar{z}(1-z)}{z(1+\bar{z})}\) where \(|z| = 1\), we can follow these steps:

### Step 1: Express \(z\) in terms of its components
Let \(z = x + iy\) where \(x\) and \(y\) are real numbers. Since \(|z| = 1\), we have:
\[ x^2 + y^2 = 1 \]

### Step 2: Write \(\bar{z}\)
The conjugate of \(z\) is given by:
\[ \bar{z} = x - iy \]

### Step 3: Substitute into the expression
Now substitute \(\bar{z}\) into the expression:
\[ \frac{\bar{z}(1-z)}{z(1+\bar{z})} = \frac{(x - iy)(1 - (x + iy))}{(x + iy)(1 + (x - iy))} \]

### Step 4: Simplify the numerator
The numerator becomes:
\[ (x - iy)(1 - x - iy) = (x - iy)(1 - x - iy) = (x(1 - x) + y^2) + i(y(1 - x) - x) \]
Thus, the numerator simplifies to:
\[ (1 - x^2 + y^2) + i(y(1 - x) - x) = (1 - 2x) + i(y(1 - x) - x) \]

### Step 5: Simplify the denominator
The denominator becomes:
\[ (x + iy)(1 + x - iy) = (x(1 + x) + y^2) + i(y(1 + x) - x) \]
Thus, the denominator simplifies to:
\[ (1 + x^2 + y^2) + i(y(1 + x) - x) = (2 + 2x) + i(y(1 + x) - x) \]

### Step 6: Combine the results
Now we have:
\[ \frac{(1 - 2x) + i(y(1 - x) - x)}{(2 + 2x) + i(y(1 + x) - x)} \]

### Step 7: Multiply by the conjugate of the denominator
To find the imaginary part, multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{[(1 - 2x) + i(y(1 - x) - x)] \cdot [(2 + 2x) - i(y(1 + x) - x)]}{(2 + 2x)^2 + (y(1 + x) - x)^2} \]

### Step 8: Extract the imaginary part
The imaginary part will come from the product of the real part of the numerator and the imaginary part of the conjugate of the denominator, minus the product of the imaginary part of the numerator and the real part of the denominator.

### Step 9: Simplify the imaginary part
After simplification, we will find that the imaginary part is:
\[ -\frac{2xy}{2 + 2x} \]

### Step 10: Substitute \(x = \cos \theta\) and \(y = \sin \theta\)
Using the relationships \(x = \cos \theta\) and \(y = \sin \theta\) (since \(|z| = 1\)):
\[ -\frac{2 \cos \theta \sin \theta}{2(1 + \cos \theta)} = -\frac{\sin(2\theta)}{1 + \cos \theta} \]

### Final Result
Thus, the imaginary part of \(\frac{\bar{z}(1-z)}{z(1+\bar{z})}\) is:
\[ -\cos \theta \tan\left(\frac{\theta}{2}\right) \]