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 $Re\left(z\right)\ne 1$, is

• A. an empty set
• B. $\left\{0\right\}$
• C. $\{0,\frac{1}{4},-\frac{1}{4}\}$
• D. equal to $R$

Answer 1

The correct option is B $\left\{0\right\}$

$w=\frac{1+(1-8\alpha )z}{1-z}\overline{w}=\frac{1+(1-8\alpha )\overline{z}}{1-\overline{z}}$
Now, $z\overline{z}=1(\because |z|=1)$
$\Rightarrow \overline{z}=\frac{1}{z}$
So, $\overline{w}=\frac{1+(1-8\alpha )\frac{1}{z}}{1-\frac{1}{z}}$
$\Rightarrow \overline{w}=\frac{z+(1-8\alpha )}{z-1}$

$\because w$ is purely imaginary,
$w+\overline{w}=0\Rightarrow \frac{1+(1-8\alpha )z}{1-z}-\left(\frac{z+(1-8\alpha )}{1-z}\right)=0\Rightarrow \frac{8\alpha +z(-8\alpha )}{1-z}=0\Rightarrow \frac{(1-z)8\alpha }{(1-z)}=0$
As $|z|=1$ and $Re\left(z\right)\ne 1,$
$\Rightarrow z\ne 1$
$\therefore \alpha =0$