The set of all `alpha in R` for which `w = (1+(1-8 alpha)z)/(1-z)`is a purely imaginary number, for all `z in C`, satisfying |z| = 1 and `Re(z) ne 1`, is

Options

A. an empty set
B. {0}
C. `{0, (1)/(4),-(1)/(4)}`
D. equal to R

Answer

To solve the problem, we need to determine the set of all real values of \( \alpha \) for which the expression

\[ w = \frac{1 + (1 - 8\alpha)z}{1 - z} \]

is purely imaginary for all \( z \in \mathbb{C} \) such that \( |z| = 1 \) and \( \text{Re}(z) \neq 1 \).

### Step-by-step Solution:

1. **Understanding the Condition**:
We want \( w \) to be purely imaginary. This means that the real part of \( w \) must be zero for all valid \( z \).

2. **Substituting \( z \)**:
Since \( z \) lies on the unit circle, we can express \( z \) as \( z = e^{i\theta} \) where \( \theta \) is a real number.

3. **Calculating \( w \)**:
Substitute \( z = e^{i\theta} \) into the expression for \( w \):

\[ w = \frac{1 + (1 - 8\alpha)e^{i\theta}}{1 - e^{i\theta}} \]

4. **Separating Real and Imaginary Parts**:
To analyze this expression, we can multiply the numerator and denominator by the conjugate of the denominator:

\[ w = \frac{(1 + (1 - 8\alpha)e^{i\theta})(1 - e^{-i\theta})}{(1 - e^{i\theta})(1 - e^{-i\theta})} \]

The denominator simplifies to:

\[ |1 - e^{i\theta}|^2 = 2(1 - \cos(\theta)) = 4\sin^2\left(\frac{\theta}{2}\right) \]

5. **Expanding the Numerator**:
The numerator becomes:

\[ (1 + (1 - 8\alpha)e^{i\theta})(1 - e^{-i\theta}) = (1 - e^{-i\theta}) + (1 - 8\alpha)(e^{i\theta} - 1) \]

This can be simplified further.

6. **Setting the Real Part to Zero**:
For \( w \) to be purely imaginary, the real part of the numerator must equal zero. This leads to a condition involving \( \alpha \).

7. **Finding the Condition**:
After simplifying, we find that the real part of the numerator can be expressed in terms of \( \alpha \). Setting this equal to zero gives us a condition on \( \alpha \).

8. **Solving for \( \alpha \)**:
After manipulating the equations, we find that the only solution that satisfies the condition for all \( z \) is:

\[ 8\alpha = 0 \implies \alpha = 0 \]

### Conclusion:
Thus, the set of all \( \alpha \in \mathbb{R} \) for which \( w \) is purely imaginary for all \( z \) satisfying the given conditions is:

\[ \{0\} \]