The modulus of the complex number `z = (( 1 - i sqrt""3 ) ( cos theta + i sin theta))/( 2 ( 1 - i) ( cos theta - i sin theta))`

Options

A. `(1)/( sqrt""2)`
B. `(1)/(2 sqrt""2)`
C. `(1)/(sqrt""3)`
D. none of these

Answer

To find the modulus of the complex number

\[ z = \frac{(1 - i \sqrt{3})(\cos \theta + i \sin \theta)}{2(1 - i)(\cos \theta - i \sin \theta)} \]

we will follow these steps:

### Step 1: Calculate the modulus of the numerator
The numerator is \( (1 - i \sqrt{3})(\cos \theta + i \sin \theta) \).

1. **Find the modulus of \( 1 - i \sqrt{3} \)**:
\[ |1 - i \sqrt{3}| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

2. **Find the modulus of \( \cos \theta + i \sin \theta \)**:
\[ |\cos \theta + i \sin \theta| = \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt{1} = 1 \]

3. **Combine the moduli**:
\[ |(1 - i \sqrt{3})(\cos \theta + i \sin \theta)| = |1 - i \sqrt{3}| \cdot |\cos \theta + i \sin \theta| = 2 \cdot 1 = 2 \]

### Step 2: Calculate the modulus of the denominator
The denominator is \( 2(1 - i)(\cos \theta - i \sin \theta) \).

1. **Find the modulus of \( 1 - i \)**:
\[ |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Find the modulus of \( \cos \theta - i \sin \theta \)**:
\[ |\cos \theta - i \sin \theta| = \sqrt{\cos^2 \theta + (-\sin \theta)^2} = \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt{1} = 1 \]

3. **Combine the moduli**:
\[ |2(1 - i)(\cos \theta - i \sin \theta)| = 2 \cdot |1 - i| \cdot |\cos \theta - i \sin \theta| = 2 \cdot \sqrt{2} \cdot 1 = 2\sqrt{2} \]

### Step 3: Calculate the modulus of \( z \)
Now we can find the modulus of \( z \):
\[ |z| = \frac{|(1 - i \sqrt{3})(\cos \theta + i \sin \theta)|}{|2(1 - i)(\cos \theta - i \sin \theta)|} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
The modulus of the complex number \( z \) is
\[ \boxed{\frac{1}{\sqrt{2}}} \]