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 $Rez\ne 1$, 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 $\left\{0\right\}$

Assuming $z=\cos \theta +i\sin \theta$
$w=\frac{1+(1-8\alpha )z}{1-z}\Rightarrow w=\frac{1+(1-8\alpha )z}{1-z}\times \frac{1-\overline{z}}{1-\overline{z}}\Rightarrow w=\frac{1-\overline{z}+(1-8\alpha )z-(1-8\alpha )z\overline{z}}{1-z-\overline{z}+z\overline{z}}\Rightarrow w=\frac{1-\overline{z}+z-8\alpha z-1+8\alpha }{2-2\cos \theta }$
$\Rightarrow w=\frac{8\alpha (1-\cos \theta )+2i\sin \theta (1-4\alpha )}{2-2\cos \theta };$
As $w$ is purely imaginary so,
$Re\left(w\right)=0\Rightarrow \frac{2\alpha (1-\cos \theta )}{1-\cos \theta }=0\Rightarrow \alpha =0[\because Rez\ne 1\Rightarrow \cos \theta \ne 1]$