Official solution
As w = \[{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}\], w is purely imaginary
\[ \therefore w\] + \[\bar w\] = 0
\[ \Rightarrow \] \[{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}\] + \[{{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}\] = 0
\[ \Rightarrow \] [1 + (1 - 8\[\alpha \])][1 - \[\bar z\]] + [1 + ( 1 - 8\[\alpha \])][1 - z] = 0
\[ \Rightarrow \] 2 - (z + \[\bar z\]) + (1 - 8\[\alpha \])(z + \[\bar z\]) - 2(1 - 8\[\alpha \]) = 0
\[ \Rightarrow \] 2 - (z + \[\bar z\]) + (z + \[\bar z\]) - 8\[\alpha (z + \[\bar z\]) - 2 + 16\[\alpha \] = 0
\[ \Rightarrow \] 16\[\alpha \] = 8\[\alpha (z + \[\bar z)\]
z + \[\bar z\] = 2 or \[\alpha \] = 0
but z + \[\bar z\] = 2 is not possible as Re(Z) \[ \ne \] 1
\[ \therefore \] \[\alpha \] = 0
\[ \therefore \] \[\alpha \] \[ \in \] {0}