Mathematics - Complex Number Question with Solution | TestHub

$$\begin{array}{rl}& \because |z|=1\text{ \& }\mathrm{Re}z\ne 1\\ & \text{ Suppose }z=x+iy\Rightarrow x^2+y^2=1\dots \dots (\mathrm{i})\\ & \text{ Now, }w=\frac{1+(1-8\alpha )z}{1-z}\\ & \Rightarrow w=\frac{1+(1-8\alpha )(x+iy)}{1-(x+iy)}\\ & \Rightarrow w=\frac{1+(1-8\alpha )(x+iy))((1-\mathrm{x})+\mathrm{iy})}{1-(x+iy))((1-\mathrm{x})+\mathrm{iy})}\\ & \Rightarrow w=\frac{[\left(1+x(1-8\alpha )(1-x)-(1-8)y^2\right]}{(1-x{)}^2+y^2}\\ & +i\frac{[(1+x(1-8\alpha ))y-(1-8\alpha )y(1-x)]}{(1-x{)}^2+y^2}\end{array}$$ If, $w$ is purely imaginary. So, $$\begin{array}{rl}& \mathrm{Re}w=\frac{\left[(1+x(1-8\alpha ))(1-\alpha )-(1-8\alpha )y^2\right]}{(1-x{)}^2+y^2}\\ & =0\\ & \Rightarrow (1-x)+x(1-8\alpha )(1-x)=(1-8)y^2\\ & \Rightarrow (1-x)+x(1-8\alpha )-x^2(1-8\alpha )=(1-8x)y^2\\ & \Rightarrow (1-x)+x(1-8\alpha )=1-8\alpha \\ & \left[\text{ From }(i),x^2+y^2=1\right]\end{array}$$ $\Rightarrow 1-8\alpha =1$ $\Rightarrow \alpha =0$ $\therefore \alpha \in \{0\}$