Question

Let \( w \neq \pm 1 \) be a complex number. If \( |w| = 1 \) and \( z = \frac{w - 1}{w + 1} \), then \( \text{Re}(z) \) is equal to:

• A. \( 1 \)
• B. \( \frac{1}{|w + 1|} \)
• C. \( \text{Re}(w) \)
• D. \( 0 \)
• E. \( w + \bar{w} \)

Hint:

Geometrically, if $|w|=1$, then $z = \frac{w-1}{w+1}$ maps a point on the unit circle to the imaginary axis. This is a common transformation in complex analysis.

Answer 1

The Correct Option is D

Concept: The real part of a complex number \( z \) is zero if \( z \) is purely imaginary. A complex number is purely imaginary if \( z + \bar{z} = 0 \) or \( \bar{z} = -z \). We also use the property that if \( |w| = 1 \), then \( w\bar{w} = 1 \) or \( \bar{w} = 1/w \).

Step 1: Find the conjugate of \( z \).
Given \( z = \frac{w - 1}{w + 1} \). \[ \bar{z} = \frac{\bar{w} - 1}{\bar{w} + 1} \]

Step 2: Substitute \( \bar{w} = 1/w \).
\[ \bar{z} = \frac{\frac{1}{w} - 1}{\frac{1}{w} + 1} = \frac{\frac{1 - w}{w}}{\frac{1 + w}{w}} \] \[ \bar{z} = \frac{1 - w}{1 + w} \]

Step 3: Compare \( z \) and \( \bar{z} \).
\[ \bar{z} = -\left( \frac{w - 1}{w + 1} \right) = -z \] Since \( \bar{z} = -z \), the complex number \( z \) is purely imaginary. Therefore, \( \text{Re}(z) = 0 \).