The set of all α∈R, for which ω=1+(1−8α)z1−z is a purely imaginary number, for all z∈C satisfying |z|=1 and Re⁡(z)≠1, is

∵|z|=1&Re⁡z≠1Suppose z=x+iy⇒x2+y2=1Now, ω=1+(1−8α)z1−z⇒ω=1+(1−8α)(x+iy)1−(x+iy)⇒ω=(1+(1−8α)(x+iy))((1−x)+iy)(1−(x+iy))((1−x)+iy)⇒ω=1+x(1−8α)(1−x)−(1−8α)y2(1−x)2+y2+[(1+x(1−8α))y−(1−8α)y(1−x)](1−x)2+y2As, ω is purely imaginary. So,Re⁡(ω)=(1+x(1−8α))(1−x)−(1−8α)y2(1−x)2+y2=0⇒(1−x)+x(1−8α)(1−x)=(1−8α)y2⇒(1−x)+x(1−8α)−x2(1−8α)=(1−8α)y2⇒(1−x)+x(1−8α)=1−8α[From (i) x2 + y2 =1]⇒1−8α=1 since re(Z)≠1⇒α=0∴α∈{0}

The set of all α∈R, for which ω=1+(1−8α)z1−z is a purely imaginary number, for all z∈C satisfying |z|=1 and Re⁡(z)≠1, is

∵|z|=1&Re⁡z≠1Suppose z=x+iy⇒x2+y2=1Now, ω=1+(1−8α)z1−z⇒ω=1+(1−8α)(x+iy)1−(x+iy)⇒ω=(1+(1−8α)(x+iy))((1−x)+iy)(1−(x+iy))((1−x)+iy)⇒ω=1+x(1−8α)(1−x)−(1−8α)y2(1−x)2+y2+[(1+x(1−8α))y−(1−8α)y(1−x)](1−x)2+y2As, ω is purely imaginary. So,Re⁡(ω)=(1+x(1−8α))(1−x)−(1−8α)y2(1−x)2+y2=0⇒(1−x)+x(1−8α)(1−x)=(1−8α)y2⇒(1−x)+x(1−8α)−x2(1−8α)=(1−8α)y2⇒(1−x)+x(1−8α)=1−8α[From (i) x2 + y2 =1]⇒1−8α=1 since re(Z)≠1⇒α=0∴α∈{0}

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

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An empty set

{0}

$\{0, \frac{1}{4}, - \frac{1}{4}\}$

Equal to R

answer is A.

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Detailed Solution

$∵ | z | = 1 & Re z \neq 1$
Suppose $z = x + iy \Rightarrow x^{2} + y^{2} = 1$
Now, $ω = \frac{1 + (1 - 8 α) z}{1 - z}$
$\begin{array}{l} \Rightarrow ω = \frac{1 + (1 - 8 α) (x + iy)}{1 - (x + i y)} \\ \Rightarrow ω = \frac{(1 + (1 - 8 α) (x + iy)) ((1 - x) + iy)}{(1 - (x + iy)) ((1 - x) + iy)} \\ \Rightarrow ω = \frac{[1 + x (1 - 8 α) (1 - x) - (1 - 8 α) y^{2}]}{(1 - x)^{2} + y^{2}} \\ + \frac{[(1 + x (1 - 8 α)) y - (1 - 8 α) y (1 - x)]}{(1 - x)^{2} + y^{2}} \end{array}$
As, $ω$ is purely imaginary. So,
$Re (ω) = \frac{[(1 + x (1 - 8 α)) (1 - x) - (1 - 8 α) y^{2}]}{(1 - x)^{2} + y^{2}} = 0$
$\begin{array}{l} \Rightarrow (1 - x) + x (1 - 8 α) (1 - x) = (1 - 8 α) y^{2} \\ \Rightarrow (1 - x) + x (1 - 8 α) - x^{2} (1 - 8 α) = (1 - 8 α) y^{2} \\ \Rightarrow (1 - x) + x (1 - 8 α) = 1 - 8 α \end{array}$
[From (i) x² + y² =1]
$\begin{array}{l} \Rightarrow 1 - 8 α = 1 \sin c e r e (Z) \neq 1 \\ \Rightarrow α = 0 \\ ∴ α \in {0} \end{array}$