The set of all $\alpha \in R$, for which $w=\frac{1+(1-8 \alpha) z}{1-z}$ is a purely imaginary number, for all $z \in C$ satisfying $|z|=1$ and $\operatorname{Re} z
eq 1$, is

Question

The set of all $\alpha \in R$, for which $w=\frac{1+(1-8 \alpha) z}{1-z}$ is a purely imaginary number, for all $z \in C$ satisfying $|z|=1$ and $\operatorname{Re} z 
eq 1$, is

Aicharya · Accepted Answer

$ \begin{aligned} &\because|z|=1 	ext { \& } \operatorname{Re} z 
eq 1 \ &	ext { Suppose } z=x+i y \Rightarrow x^2+y^2=1 \ldots \ldots(\mathrm{i}) \ &	ext { 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ight]ight.}{(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} $ If, $w$ is purely imaginary. So, $ \begin{aligned} &\operatorname{Re} w=\frac{\left[(1+x(1-8 \alpha))(1-\alpha)-(1-8 \alpha) y^2ight]}{(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[	ext { From }(i), x^2+y^2=1ight]} \end{aligned} $ $\Rightarrow 1-8 \alpha=1$ $\Rightarrow \alpha=0$ $	herefore \alpha \in\{0\}$

The set of all $\alpha \in R$, for which $w=\frac{1+(1-8 \alpha) z}{1-z}$ is a purely imaginary number, for…

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