Let $$w(Im\ w \neq 0)$$ be a complex number. Then, the set of all complex numbers $$z$$ satisfying the equation $$w - \bar{w}z = k(1 - z)$$, for some real number $$k$$, is:

We are given a complex number $$ w $$ with non-zero imaginary part, meaning $$ \text{Im}(w) \neq 0 $$, and we need to find the set of all complex numbers $$ z $$ such that the equation $$ w - \bar{w} z = k(1 - z) $$ holds for some real number $$ k $$.

First, rearrange the equation to solve for $$ z $$. Starting with:

$$ w - \bar{w} z = k(1 - z) $$

Expand the right side:

$$ w - \bar{w} z = k - k z $$

Bring all terms to the left side:

$$ w - \bar{w} z - k + k z = 0 $$

Group terms containing $$ z $$:

$$ w - k + z(k - \bar{w}) = 0 $$

Isolate the term with $$ z $$:

$$ z(k - \bar{w}) = k - w $$

Solve for $$ z $$:

$$ z = \frac{k - w}{k - \bar{w}} $$

Note that $$ k $$ is real, so $$ \bar{k} = k $$. The denominator $$ k - \bar{w} $$ must not be zero. Since $$ \text{Im}(w) \neq 0 $$, $$ w $$ is not real, and thus $$ \bar{w} $$ is also not real. Therefore, for any real $$ k $$, $$ k - \bar{w} \neq 0 $$ and similarly $$ k - w \neq 0 $$, ensuring $$ z $$ is defined.

To find the set of $$ z $$, consider the conjugate of $$ z $$. Given:

$$ z = \frac{k - w}{k - \bar{w}} $$

Take the conjugate:

$$ \bar{z} = \overline{ \left( \frac{k - w}{k - \bar{w}} \right) } = \frac{ \overline{k - w} }{ \overline{k - \bar{w}} } = \frac{ \bar{k} - \bar{w} }{ \bar{k} - \overline{\bar{w}} } = \frac{ k - \bar{w} }{ k - w } $$

Now, multiply $$ z $$ and $$ \bar{z} $$:

$$ z \bar{z} = \left( \frac{k - w}{k - \bar{w}} \right) \left( \frac{k - \bar{w}}{k - w} \right) = 1 $$

Since $$ z \bar{z} = |z|^2 $$, we have:

$$ |z|^2 = 1 $$

Thus, $$ |z| = 1 $$ (as magnitude is non-negative).

Now, check if $$ z = 1 $$ is possible. Substitute $$ z = 1 $$ into the original equation:

$$ w - \bar{w} \cdot 1 = k(1 - 1) \implies w - \bar{w} = 0 \implies w = \bar{w} $$

This implies $$ w $$ is real, but $$ \text{Im}(w) \neq 0 $$, so $$ w $$ is not real. Thus, $$ z = 1 $$ is not possible and must be excluded.

So far, we have that $$ z $$ must satisfy $$ |z| = 1 $$ and $$ z \neq 1 $$. Now, verify that every such $$ z $$ can be achieved with some real $$ k $$. Let $$ z $$ be any complex number with $$ |z| = 1 $$ and $$ z \neq 1 $$. Solve for $$ k $$ in:

$$ z = \frac{k - w}{k - \bar{w}} $$

Rearrange:

$$ z(k - \bar{w}) = k - w $$

$$ z k - z \bar{w} = k - w $$

Bring terms with $$ k $$ to one side:

$$ z k - k = z \bar{w} - w $$

$$ k(z - 1) = z \bar{w} - w $$

Solve for $$ k $$:

$$ k = \frac{z \bar{w} - w}{z - 1} $$

Since $$ z \neq 1 $$, the denominator is non-zero. Now, show that $$ k $$ is real by proving $$ k = \bar{k} $$. Compute the conjugate of $$ k $$:

$$ \bar{k} = \overline{ \left( \frac{z \bar{w} - w}{z - 1} \right) } = \frac{ \overline{z \bar{w} - w} }{ \overline{z - 1} } = \frac{ \bar{z} \overline{\bar{w}} - \bar{w} }{ \bar{z} - 1 } = \frac{ \bar{z} w - \bar{w} }{ \bar{z} - 1 } $$

Since $$ |z| = 1 $$, $$ \bar{z} = \frac{1}{z} $$. Substitute:

$$ \bar{k} = \frac{ \frac{1}{z} w - \bar{w} }{ \frac{1}{z} - 1 } = \frac{ \frac{w - z \bar{w}}{z} }{ \frac{1 - z}{z} } = \frac{w - z \bar{w}}{z} \cdot \frac{z}{1 - z} = \frac{w - z \bar{w}}{1 - z} $$

Note that $$ 1 - z = -(z - 1) $$, so:

$$ \bar{k} = \frac{w - z \bar{w}}{-(z - 1)} = -\frac{w - z \bar{w}}{z - 1} $$

Compare to $$ k $$:

$$ k = \frac{z \bar{w} - w}{z - 1} = -\frac{w - z \bar{w}}{z - 1} $$

Thus, $$ \bar{k} = k $$, confirming $$ k $$ is real. Therefore, for every $$ z $$ with $$ |z| = 1 $$ and $$ z \neq 1 $$, there exists a real $$ k $$ satisfying the equation.

Now, evaluate the options:

A. $$ \{z : z \neq 1\} $$ — This includes points not on the unit circle, but we require $$ |z| = 1 $$, so incorrect.

B. $$ \{z : |z| = 1, z \neq 1\} $$ — Matches our derived set.

C. $$ \{z : z = \bar{z}\} $$ — This is the real line, but $$ z $$ on the unit circle may not be real, so incorrect.

D. $$ \{z : |z| = 1\} $$ — Includes $$ z = 1 $$, which is excluded, so incorrect.

Hence, the correct answer is Option B.