Question

Let \( z \neq 1 \) be a complex number and let \( \omega = x + iy, y \neq 0 \). If \[ \frac{\omega -\overline{\omega}z}{1 -z} \] is purely real, then \( |z| \) is equal to

• A. \( |\omega| \)
• B. \( |\omega|^2 \)
• C. \( \frac{1}{|\omega|^2} \)
• D. 1

Hint:

For a complex number to be purely real, its imaginary part must be zero. This condition can be used to solve for unknown variables in complex expressions.

Answer 1

The Correct Option is D

Step 1: Understand the condition for a purely real number. 
A complex number is purely real if its imaginary part is zero. 
Step 2: Express the given expression in terms of real and imaginary parts. 
Let \( \omega = x + iy \) and \( \overline{\omega} = x -iy \). 
The expression becomes: \[ \frac{\omega -\overline{\omega}z}{1 -z} = \frac{(x + iy) -(x -iy)z}{1 -z} \] 
Step 3: Simplify the expression. \[ \frac{(x + iy) -(x -iy)z}{1 -z} = \frac{x(1 -z) + iy(1 + z)}{1 -z} \] \[ = x + iy \cdot \frac{1 + z}{1 -z} \] 
Step 4: Set the imaginary part to zero. 
For the expression to be purely real, the imaginary part must be zero: \[ y \cdot \frac{1 + z}{1 -z} = 0 \] 
Since \( y \neq 0 \), we have: \[ \frac{1 + z}{1 -z} = 0 \quad \Rightarrow \quad 1 + z = 0 \quad \Rightarrow \quad z = -1 \] 
Step 5: Determine \( |z| \). 
If \( z = -1 \), then \( |z| = 1 \).