Question

For complex numbers \( z \), if \( |z| = 1 \), then the value of \( \frac{z-1}{z+1} \) is

• A. 0
• B. real
• C. purely imaginary
• D. complex numbers

Hint:

When \( |z| = 1 \), the ratio \( \frac{z-1}{z+1} \) will always be purely imaginary, as the real part cancels out.

Answer 1

The Correct Option is C

Step 1: Since \( |z|=1 \), \( \bar z = 1/z \).
Step 2: Let \( w = \dfrac{z-1}{z+1} \); then \( \bar w = \dfrac{1/z-1}{1/z+1} = \dfrac{1-z}{1+z} = -w \).
Step 3: \( \bar w = -w \) is exactly the condition for \( w \) to be purely imaginary.
Hence \( \dfrac{z-1}{z+1} \) is purely imaginary.

Answer 2

A third way is to test a couple of concrete points on the unit circle directly and observe the pattern, rather than working with the general conjugate identity.

At \( z = i \) (so \( |z|=1 \)): \( w = \dfrac{i-1}{i+1} \). Multiplying numerator and denominator by the conjugate of the denominator, \( \dfrac{(i-1)(1-i)}{(i+1)(1-i)} = \dfrac{i - i^2 -1 + i}{1 - i^2 + i - i}= \dfrac{2i}{2} = i \), a purely imaginary number. At \( z=-i \): \( w=\dfrac{-i-1}{-i+1} \), and a similar computation also yields a purely imaginary value. These sample points are consistent with a general purely-imaginary pattern for all \(|z|=1\), \(z\ne -1\).

1. 0: Neither sample point gave \(w=0\); this only happens at the single point \(z=1\), not generally.
2. real: Both sample computations gave purely imaginary results (\(i\) and a similar imaginary value), not real numbers, so this option is inconsistent with the computed samples.
3. purely imaginary: Both sample points computed above land exactly on the imaginary axis, consistent with this option.
4. complex numbers: Too broad a classification; the samples specifically land on the imaginary axis (zero real part), which "purely imaginary" captures precisely.

Sample evaluations on the unit circle confirm the purely imaginary classification.

Therefore, the correct answer is purely imaginary.