The set of all α ∈ R , for which w = (1 + (1 - 8α)z)/(1 - z) is a...

For $w$ to be purely imaginary, $\dfrac{1 + \left(1 - 8\alpha\right)z}{1 - z} + \dfrac{1 + \left(1 - 8\alpha\right)\overset{¯}{z}}{1 - \overset{¯}{z}} = 0$

$⇒ 1 - \overset{¯}{z} + \left(1 - 8\alpha\right)z - \left(1 - 8\alpha\right) + 1 - z + \left(1 - 8\alpha\right)\overset{¯}{z} - \left(1 - 8\alpha\right) = 0$

$⇒ 2 - \left(z + \overset{¯}{z}\right) + \left(1 - 8\alpha\right)\left(z + \overset{¯}{z}\right) - 2 + 16\alpha = 0$

$⇒ 8\alpha\left(z + \overset{¯}{z}\right) = 16\alpha$

$⇒ z + \overset{¯}{z} = 2$ or $\alpha = 0$

For all $z \in C$, we have $\alpha = 0$.