The set of all alpha in R, fo | vedclass

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Question

$\because|z|=1$ and $	ext { Re } z 
eq 1$

Suppose $z=x+i y$ $ \Rightarrow x^{2}+y^{2}=1$

Now, $w = \frac{{1 + (1 - 8\alpha )z}}{{1 - z}}$

$

Vedclass · Accepted Answer

$\left\{ 0 \right\}$

Vedclass · Answer

$\because|z|=1$ and $	ext { Re } z 
eq 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 $w = \frac{{[(1 + x(1 - 8\alpha ))(1 - \alpha ) - (1 - 8\alpha ){y^2}]}}{{{{(1 - x)}^2} + {y^2}}}$ $=0$

$\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y^{2}$

$\Rightarrow(1-x)+x(1-8 \alpha)-x^{2}(1-8 \alpha)=(1-8 x) y^{2}$

$\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha$

$\Rightarrow 1-8 \alpha=1$

$\Rightarrow \alpha=0$

$	herefore \mathrm{c} \in\{0\}$

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

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