The set of all $$\alpha \in R$$, for which $$w = \frac{1+(1-8\alpha)z}{1-z}$$ is a purely imaginary number, for all $$z \in C$$ satisfying $$|z| = 1$$ and Re(z) $$\neq$$ 1, is:

We have the transformation $$w=\dfrac{1+(1-8\alpha)z}{1-z}$$ and the condition $$|z|=1$$ with $$\operatorname{Re}(z)\neq1$$, that is, $$z\neq1$$ on the unit circle. Putting $$z=e^{i\theta}$$ (where $$\theta\in(0,2\pi)$$ and $$\theta\neq0$$) is the natural way to represent every such complex number.

Writing $$A=1-8\alpha$$ to shorten expressions, the formula becomes

$$w=\dfrac{1+A\,e^{i\theta}}{1-e^{i\theta}}.$$

To find the real part of $$w$$ we rationalise the denominator. The standard algebraic device is to multiply numerator and denominator by the conjugate of the denominator. Therefore

$$w=\dfrac{1+A\,e^{i\theta}}{1-e^{i\theta}}\;\dfrac{1-e^{-i\theta}}{1-e^{-i\theta}}=\dfrac{(1+A\,e^{i\theta})(1-e^{-i\theta})}{|\,1-e^{i\theta}\,|^{2}}.$$

The denominator is purely real because

$$|\,1-e^{i\theta}\,|^{2}=(1-e^{i\theta})(1-e^{-i\theta})=2-2\cos\theta =4\sin^{2}\dfrac{\theta}{2}.$$

Hence the real part of $$w$$ is simply the real part of the numerator divided by this real denominator. We expand the numerator completely:

$$\begin{aligned} (1+A\,e^{i\theta})(1-e^{-i\theta}) &=1-e^{-i\theta}+A\,e^{i\theta}-A\\[4pt] &=\bigl(1-\cos\theta\bigr)(1-A)\;+\;i\sin\theta\,(1+A). \\ \end{aligned}$$

Its real part is therefore

$$\operatorname{Re}\bigl[(1+A\,e^{i\theta})(1-e^{-i\theta})\bigr]=(1-\cos\theta)(1-A).$$

Dividing by the real denominator $$4\sin^{2}\dfrac{\theta}{2}=2(1-\cos\theta)$$ we get

$$\operatorname{Re}(w)=\dfrac{(1-\cos\theta)(1-A)}{2(1-\cos\theta)}=\dfrac{1-A}{2}.$$

The factor $$1-\cos\theta\neq0$$ because $$\theta\neq0$$, so it cancels cleanly. Substituting back $$A=1-8\alpha$$ we obtain

$$\operatorname{Re}(w)=\dfrac{1-(1-8\alpha)}{2}=\dfrac{8\alpha}{2}=4\alpha.$$

This remarkable simplification tells us that, for every point on the given unit circle (excluding $$z=1$$), the real part of $$w$$ is the same constant $$4\alpha$$. For $$w$$ to be purely imaginary we must have

$$\operatorname{Re}(w)=0\quad\Longrightarrow\quad4\alpha=0\quad\Longrightarrow\quad\alpha=0.$$

Thus the only real value of $$\alpha$$ that satisfies the requirement for all admissible $$z$$ is $$\alpha=0$$.

Hence, the correct answer is Option A.