Key Concepts and Formulas

• Purely Imaginary Numbers: A complex number $w$ is purely imaginary if and only if its real part is zero, i.e., $w + \bar{w} = 0$.
• Complex Conjugate Properties: $\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}$, $\overline{z_1 z_2} = \bar{z_1} \bar{z_2}$, $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}$, and $\overline{kz} = k\bar{z}$ for real $k$.
• Unit Circle: If $|z|=1$, then $z\bar{z} = 1$.
• Real Part: $z + \bar{z} = 2 \text{Re}(z)$.

Step 1: Apply the purely imaginary condition.

We are given that $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$ is purely imaginary. Therefore, $w + \bar{w} = 0$. First, find the conjugate of $w$: $\bar{w} = \overline{\left(\frac{1 + (1 - 8\alpha)z}{1 - z}\right)} = \frac{\overline{1 + (1 - 8\alpha)z}}{\overline{1 - z}} = \frac{1 + (1 - 8\alpha)\bar{z}}{1 - \bar{z}}$ Now, substitute $w$ and $\bar{w}$ into the equation $w + \bar{w} = 0$: $\frac{1 + (1 - 8\alpha)z}{1 - z} + \frac{1 + (1 - 8\alpha)\bar{z}}{1 - \bar{z}} = 0$

Step 2: Combine fractions and simplify the numerator.

Combine the fractions by finding a common denominator: $\frac{[1 + (1 - 8\alpha)z](1 - \bar{z}) + [1 + (1 - 8\alpha)\bar{z}](1 - z)}{(1 - z)(1 - \bar{z})} = 0$ For this equation to hold, the numerator must be zero, provided the denominator is non-zero, which is true since Re(z) != 1 implies z != 1. Expand the numerator: $(1 - \bar{z} + (1 - 8\alpha)z - (1 - 8\alpha)z\bar{z}) + (1 - z + (1 - 8\alpha)\bar{z} - (1 - 8\alpha)z\bar{z}) = 0$

Step 3: Use the condition $|z|=1$ to further simplify.

Since $|z|=1$, we have $z\bar{z} = 1$. Substitute this into the expanded numerator: $(1 - \bar{z} + (1 - 8\alpha)z - (1 - 8\alpha)) + (1 - z + (1 - 8\alpha)\bar{z} - (1 - 8\alpha)) = 0$ $1 - \bar{z} + z - 8\alpha z - 1 + 8\alpha + 1 - z + \bar{z} - 8\alpha \bar{z} - 1 + 8\alpha = 0$ Combine like terms: $16\alpha - 8\alpha z - 8\alpha \bar{z} = 0$ $16\alpha - 8\alpha(z + \bar{z}) = 0$

Step 4: Solve the simplified equation for $\alpha$.

Factor out $8\alpha$: $8\alpha(2 - (z + \bar{z})) = 0$ Substitute $z + \bar{z} = 2\text{Re}(z)$: $8\alpha(2 - 2\text{Re}(z)) = 0$ $16\alpha(1 - \text{Re}(z)) = 0$

Step 5: Analyze the result based on the given conditions.

We have $16\alpha(1 - \text{Re}(z)) = 0$. The problem states this must hold for all $z$ such that $|z|=1$ and $\text{Re}(z) \ne 1$. Since $\text{Re}(z) \ne 1$, then $1 - \text{Re}(z) \ne 0$. Therefore, the only way for the equation to hold true for all such $z$ is if $\alpha = 0$.

Common Mistakes & Tips

• Remember the definition of purely imaginary numbers: $w + \bar{w} = 0$.
• Always use the condition $|z|=1$ by substituting $z\bar{z} = 1$.
• Pay close attention to "for all $z$". This means the equation must hold for every valid $z$.

Summary

By applying the definition of purely imaginary numbers, the properties of complex conjugates, and the condition $|z|=1$, we simplified the given expression to $16\alpha(1 - \text{Re}(z)) = 0$. Since this must hold for all $z$ where $\text{Re}(z) \ne 1$, we conclude that $\alpha = 0$.

The final answer is \boxed{0}, which corresponds to option (B).