If `w=alpha+ibeta` where `beta ne 0 ` and `z ne 1` satisfies the condition that `((w- bar wz)/(1-z))` is purely real then the set of values of z is

Options

A. `|z|=1,z ne 2`
B. `|z|=1 and z ne 1`
C. `z= barz `
D. None of these

Explanation

To solve the problem, we start with the given condition that \( w = \alpha + i\beta \) where \( \beta \neq 0 \) and \( z \neq 1 \). We need to find the set of values of \( z \) such that the expression

\[ \frac{w - \overline{w}z}{1 - z} \]

is purely real.

### Step-by-Step Solution:

1. **Express \( w \) and its conjugate**:
\[ w = \alpha + i\beta \quad \text{and} \quad \overline{w} = \alpha - i\beta \]

2. **Substitute \( w \) and \( \overline{w} \) into the expression**:
\[ \frac{(\alpha + i\beta) - (\alpha - i\beta)z}{1 - z} = \frac{\alpha + i\beta - \alpha z + i\beta z}{1 - z} \]
This simplifies to:
\[ \frac{(1 - z)\alpha + i\beta(1 + z)}{1 - z} \]

3. **Separate real and imaginary parts**:
The expression can be rewritten as:
\[ \frac{(1 - z)\alpha}{1 - z} + i\frac{\beta(1 + z)}{1 - z} \]
Which simplifies to:
\[ \alpha + i\frac{\beta(1 + z)}{1 - z} \]

4. **Condition for being purely real**:
For the expression to be purely real, the imaginary part must be zero:
\[ \frac{\beta(1 + z)}{1 - z} = 0 \]
Since \( \beta \neq 0 \), we can conclude that:
\[ 1 + z = 0 \quad \Rightarrow \quad z = -1 \]

5. **Check the condition \( z \neq 1 \)**:
The value \( z = -1 \) satisfies the condition \( z \neq 1 \).

6. **Conclusion**:
The only value of \( z \) that satisfies the given condition is:
\[ z = -1 \]

### Final Answer:
The set of values of \( z \) is \( \{-1\} \).