The set of all α ∈ R, for which w= 1+(1-8 α) z 1-z is a purely imaginary number, for all z ∈ C satisfying |z|=1 and Re z ≠ 1, is

$$\begin{aligned} &\because|z|=1 \text { \& } \operatorname{Re} z \neq 1 \\ &\text { Suppose } z=x+i y \Rightarrow x^2+y^2=1 \ldots \ldots(\mathrm{i}) \\ &\text { Now, } w=\frac{1+(1-8 \alpha) z}{1-z} \\ &\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y)}{1-(x+i y)} \\ &\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y))((1-\mathrm{x})+\mathrm{iy})}{1-(x+i y))((1-\mathrm{x})+\mathrm{iy})} \\ &\Rightarrow \quad w=\frac{\left[\left(1+x(1-8 \alpha)(1-x)-(1-8) y^2\right]\right.}{(1-x)^2+y^2} \\ &+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^2+y^2} \end{aligned}$$​∵∣z∣=1 & Rez=1 Suppose z=x+iy⇒x2+y2=1……(i) Now, w=1−z1+(1−8α)z​⇒w=1−(x+iy)1+(1−8α)(x+iy)​⇒w=1−(x+iy))((1−x)+iy)1+(1−8α)(x+iy))((1−x)+iy)​⇒w=(1−x)2+y2[(1+x(1−8α)(1−x)−(1−8)y2]​+i(1−x)2+y2[(1+x(1−8α))y−(1−8α)y(1−x)]​​ If, $w$w is purely imaginary. So, $$\begin{aligned} &\operatorname{Re} w=\frac{\left[(1+x(1-8 \alpha))(1-\alpha)-(1-8 \alpha) y^2\right]}{(1-x)^2+y^2} \\ &=0 \\ &\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y^2 \\ &\Rightarrow(1-x)+x(1-8 \alpha)-x^2(1-8 \alpha)=(1-8 x) y^2 \\ &\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha \\ &{\left[\text { From }(i), x^2+y^2=1\right]} \end{aligned}$$​Rew=(1−x)2+y2[(1+x(1−8α))(1−α)−(1−8α)y2]​=0⇒(1−x)+x(1−8α)(1−x)=(1−8)y2⇒(1−x)+x(1−8α)−x2(1−8α)=(1−8x)y2⇒(1−x)+x(1−8α)=1−8α[ From (i),x2+y2=1]​ $\Rightarrow 1-8 \alpha=1$⇒1−8α=1 $\Rightarrow \alpha=0$⇒α=0 $\therefore \alpha \in\{0\}$∴α∈{0}