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The set of all $\alpha \in R$, for which $w = \frac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and ${\mathop{\rm Re}\nolimits} \,z \ne 1$, is
$\left\{ 0 \right\}$
an empty set
$\left\{ {0,\frac{1}{4}, - \frac{1}{4}} \right\}$
equal to $R$
Solution
$\because|z|=1$ and $\text { Re } z \neq 1$
Suppose $z=x+i y$ $ \Rightarrow x^{2}+y^{2}=1$
Now, $w = \frac{{1 + (1 – 8\alpha )z}}{{1 – z}}$
$ \Rightarrow w = \frac{{1 + (1 – 8\alpha )(x + iy)}}{{1 – (x + iy)}}$
$\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y))((1-x)+i y)}{1-(x+i y))((1-x)+i y)}$
$ \Rightarrow w = \frac{{[1 + x(1 – 8\alpha )(1 – x) – (1 – 8){y^2}]}}{{{{(1 – x)}^2} + {y^2}}}$ $+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^{2}+y^{2}}$
If, $w$ is purely imaginary. So,
Re $w = \frac{{[(1 + x(1 – 8\alpha ))(1 – \alpha ) – (1 – 8\alpha ){y^2}]}}{{{{(1 – x)}^2} + {y^2}}}$ $=0$
$\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y^{2}$
$\Rightarrow(1-x)+x(1-8 \alpha)-x^{2}(1-8 \alpha)=(1-8 x) y^{2}$
$\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha$
$\Rightarrow 1-8 \alpha=1$
$\Rightarrow \alpha=0$
$\therefore \mathrm{c} \in\{0\}$
