If `Re ((z-1)/(z+1)) =0` where `z= x+iy` is a complex number, then which one of the following is correct ?

Options

A. A. `z= 1+i`
B. B. `|z| =2`
C. C. `z = 1-i`
D. D. `|z|=1`

Answer

To solve the problem, we need to analyze the condition given: \( \text{Re} \left( \frac{z-1}{z+1} \right) = 0 \), where \( z = x + iy \) is a complex number.

### Step-by-Step Solution:

1. **Substituting \( z \)**:
We start by substituting \( z = x + iy \) into the expression:
\[ \frac{z - 1}{z + 1} = \frac{(x + iy) - 1}{(x + iy) + 1} = \frac{(x - 1) + iy}{(x + 1) + iy} \]

2. **Rationalizing the Denominator**:
To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)} \]

3. **Calculating the Denominator**:
The denominator simplifies as follows:
\[ (x + 1)^2 + y^2 \]

4. **Calculating the Numerator**:
The numerator expands to:
\[ (x - 1)(x + 1) - i(y(x + 1) - y(x - 1)) = (x^2 - 1) + i(2y) \]

5. **Combining the Results**:
Thus, we have:
\[ \frac{(x^2 - 1) + 2yi}{(x + 1)^2 + y^2} \]

6. **Finding the Real Part**:
The real part of this expression is:
\[ \text{Re} \left( \frac{(x^2 - 1) + 2yi}{(x + 1)^2 + y^2} \right) = \frac{x^2 - 1}{(x + 1)^2 + y^2} \]
Setting this equal to zero gives:
\[ x^2 - 1 = 0 \]

7. **Solving for \( x \)**:
Solving \( x^2 - 1 = 0 \) gives:
\[ x^2 = 1 \implies x = \pm 1 \]

8. **Using the Condition**:
We also know that \( z \) is a complex number, so \( x^2 + y^2 = 1 \) must hold true. This means:
\[ y^2 = 1 - x^2 \]

9. **Finding \( y \)**:
If \( x = 1 \) or \( x = -1 \), then:
- If \( x = 1 \), \( y^2 = 0 \implies y = 0 \).
- If \( x = -1 \), \( y^2 = 0 \implies y = 0 \).

10. **Finding the Modulus**:
Thus, the modulus of \( z \) is given by:
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{1} = 1 \]

### Conclusion:
The correct answer is that \( |z| = 1 \).