Question

If \( z = 1 + \cos\theta - i\sin\theta \) and \( 0<\theta<\pi \), then \[ \left[ |z - 1|^2 \cdot \frac{|z|^2}{4} \right]^{1/2} = \]

• A. \( \sqrt{2} \cos\theta \)
• B. \( \sqrt{2} \sin\theta \)
• C. \( \cos\left( \frac{\theta}{2} \right) \)
• D. \( \sin\left( \frac{\theta}{2} \right) \)

Hint:

Use the identity \( \cos^2\left( \frac{\theta}{2} \right) = \frac{1 + \cos\theta}{2} \) to simplify modulus-based expressions.

Answer 1

The Correct Option is C

Step 1: Compute \( z - 1 \)
\[ z - 1 = (1 + \cos\theta - i\sin\theta) - 1 = \cos\theta - i\sin\theta \]
Step 2: Find \( |z - 1| \) and \( |z - 1|^2 \)
\[ |z - 1| = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1 \] \[ |z - 1|^2 = 1^2 = 1 \]
Step 3: Find \( |z| \) and \( |z|^2 \)
\[ |z| = \sqrt{(1 + \cos\theta)^2 + \sin^2\theta} = \sqrt{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta} \] Using the identity \( \cos^2\theta + \sin^2\theta = 1 \): \[ |z| = \sqrt{2 + 2\cos\theta} = \sqrt{2(1 + \cos\theta)} \] \[ |z|^2 = 2(1 + \cos\theta) \]
Step 4: Substitute into the expression
\[ \left[ |z - 1|^2 \cdot \frac{|z|^2}{4} \right]^{1/2} = \left[ 1 \cdot \frac{2(1 + \cos\theta)}{4} \right]^{1/2} = \sqrt{\frac{1 + \cos\theta}{2}} \]
Step 5: Simplify using trigonometric identity
Recall the half-angle identity: \[ \cos\left( \frac{\theta}{2} \right) = \sqrt{\frac{1 + \cos\theta}{2}} \] Since \( 0<\theta<\pi \), \( \cos\left( \frac{\theta}{2} \right) \) is positive.
Step 6: Match with the given options
The simplified form matches option 3. Verification:
Let \( \theta = \frac{\pi}{2} \): \[ z = 1 - i \implies |z - 1| = 1, \quad |z| = \sqrt{2} \] \[ \left[ 1 \cdot \frac{2}{4} \right]^{1/2} = \frac{\sqrt{2}}{2} = \cos\left( \frac{\pi}{4} \right) \] This confirms option 3 is correct. Conclusion: The correct answer is \(\boxed{3}\).