$\because|z|=1$ and $\text { Re } z \neq 1$

Suppose $z=x+i y$ $ \Rightarrow x^{2}+y^{2}=1$

Now, $w = \frac{{1 + (1 – 8\alpha )z}}{{1 – z}}$

$ \Rightarrow w = \frac{{1 + (1 – 8\alpha )(x + iy)}}{{1 – (x + iy)}}$

$\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y))((1-x)+i y)}{1-(x+i y))((1-x)+i y)}$

$ \Rightarrow w = \frac{{[1 + x(1 – 8\alpha )(1 – x) – (1 – 8){y^2}]}}{{{{(1 – x)}^2} + {y^2}}}$  $+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^{2}+y^{2}}$

If, $w$ is purely imaginary. So,

Re $w = \frac{{[(1 + x(1 – 8\alpha ))(1 – \alpha ) – (1 – 8\alpha ){y^2}]}}{{{{(1 – x)}^2} + {y^2}}}$ $=0$

$\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y^{2}$

$\Rightarrow(1-x)+x(1-8 \alpha)-x^{2}(1-8 \alpha)=(1-8 x) y^{2}$

$\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha$

$\Rightarrow 1-8 \alpha=1$

$\Rightarrow \alpha=0$

$\therefore \mathrm{c} \in\{0\}$