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Step-by-Step Solution
Step 1: Express the Given Condition Mathematically
We are given the expression:
$$
w = \frac{1 + \bigl(1 - 8\alpha\bigr)z}{1 - z}
$$
and the condition that \(w\) is purely imaginary for all complex numbers \(z\) satisfying \(|z| = 1\) and \(\operatorname{Re}(z) \neq 1\). Being “purely imaginary” means that the real part of \(w\) must be zero, or equivalently \(w + \overline{w} = 0\). Here, \(\overline{w}\) denotes the complex conjugate of \(w\).
Step 2: Write the Conjugate Expression
First, note that if
$$
w = \frac{1 + \bigl(1 - 8\alpha\bigr)z}{1 - z},
$$
then its complex conjugate is
$$
\overline{w} = \frac{1 + \bigl(1 - 8\alpha\bigr)\overline{z}}{1 - \overline{z}}.
$$
Since \(|z| = 1\), we know \(\overline{z} = \frac{1}{z}\) if needed, but for now we only use that \(\overline{z}\) is just the conjugate of \(z\).
Step 3: Apply the Purely Imaginary Condition
Set \(w + \overline{w} = 0\):
$$
\frac{1 + \bigl(1 - 8\alpha\bigr)z}{1 - z}
\;+\;
\frac{1 + \bigl(1 - 8\alpha\bigr)\overline{z}}{1 - \overline{z}}
= 0.
$$
We aim to simplify this expression.
Step 4: Simplify the Numerator
Combine both fractions. A common strategy is to multiply numerator and denominator carefully, but here we can simplify term by term. We write:
$$
\bigl(1 + (1 - 8\alpha)z\bigr)\bigl(1 - \overline{z}\bigr)
+
\bigl(1 + (1 - 8\alpha)\overline{z}\bigr)\bigl(1 - z\bigr)
= 0 \times \bigl[(1-z)(1-\overline{z})\bigr].
$$
After expansion:
Step 5: Expand Each Part
Expand the first product:
\[
\bigl(1 + (1 - 8\alpha)z\bigr)\bigl(1 - \overline{z}\bigr)
= 1 \cdot 1
- 1 \cdot \overline{z}
+ (1 - 8\alpha)z \cdot 1
- (1 - 8\alpha)z \,\overline{z}.
\]
Similarly, expand the second product:
\[
\bigl(1 + (1 - 8\alpha)\overline{z}\bigr)\bigl(1 - z\bigr)
= 1 \cdot 1
- 1 \cdot z
+ (1 - 8\alpha)\overline{z} \cdot 1
- (1 - 8\alpha)\overline{z}z.
\]
Step 6: Combine Like Terms
Combining all expanded terms yields an expression involving \(z + \overline{z}\), and \(z \,\overline{z} = |z|^2 = 1\). In fact, one ultimately arrives at:
$$
2 - (z + \overline{z}) + (1 - 8\alpha)(z + \overline{z}) - 2(1 - 8\alpha) = 0.
$$
After rearranging:
$$
2 - (z + \overline{z}) + (z + \overline{z}) - 8\alpha(z + \overline{z}) - 2 + 16\alpha = 0,
$$
which simplifies to
$$
16\alpha = 8\alpha \,(z + \overline{z}).
$$
Step 7: Interpret the Simplified Result
The equation
$$
16\alpha = 8\alpha \,(z + \overline{z})
$$
can be reorganized as
$$
16\alpha - 8\alpha \,(z + \overline{z}) = 0,
$$
or
$$
8\alpha \bigl( 2 - (z + \overline{z})\bigr) = 0.
$$
This implies either \(8\alpha = 0\), which gives \(\alpha = 0\), or \(2 - (z + \overline{z}) = 0\), which implies \(z + \overline{z} = 2\). But note that \(z + \overline{z} = 2\) is equivalent to \(\operatorname{Re}(z) = 1\). However, by the problem statement, \(\operatorname{Re}(z) \neq 1\). Hence that possibility is disqualified.
Step 8: Conclude the Value of α
Since \(\operatorname{Re}(z) \neq 1\) prevents the condition \(z + \overline{z} = 2\), the only possibility left is \(\alpha = 0\). So the set of all real \(\alpha\) for which \(w\) remains purely imaginary under the given conditions is \(\{\,0\}\).
Final Answer
The correct answer is \(\{0\}\).
