The set of all alpha in R,for which w = frac{ | vedclass

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(A) Given $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$w + \bar{w} = \frac{1 + (1 - 8\alpha)z}{1 - z} + \f

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$\left\{ 0 \right\}$

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(A) Given $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$w + \bar{w} = \frac{1 + (1 - 8\alpha)z}{1 - z} + \frac{1 + (1 - 8\alpha)\bar{z}}{1 - \bar{z}} = 0$$\Rightarrow (1 + (1 - 8\alpha)z)(1 - \bar{z}) + (1 + (1 - 8\alpha)\bar{z})(1 - z) = 0$$\Rightarrow 1 - \bar{z} + (1 - 8\alpha)z - (1 - 8\alpha)z\bar{z} + 1 - z + (1 - 8\alpha)\bar{z} - (1 - 8\alpha)z\bar{z} = 0$Since $|z| = 1$,we have $z\bar{z} = 1$.$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(\bar{z} + z) - 2(1 - 8\alpha) = 0$$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(z + \bar{z}) - 2 + 16\alpha = 0$$\Rightarrow (z + \bar{z})(1 - 8\alpha - 1) + 16\alpha = 0$$\Rightarrow (z + \bar{z})(-8\alpha) + 16\alpha = 0$$\Rightarrow -8\alpha(z + \bar{z} - 2) = 0$Since $	ext{Re}(z) 
eq 1$,$z + \bar{z} 
eq 2$,so $z + \bar{z} - 2 
eq 0$.Therefore,$-8\alpha = 0$,which implies $\alpha = 0$.

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

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