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$z\ne 1$, is?

• A. $Anemptyset$
• B. $\left\{0\right\}$
• C. $\{0,\frac{1}{4},-\frac{1}{4}\}$
• D. $EqualtoR$

Answer 1

Answer: (B)

$\left\{0\right\}$

It is given that $|z|=1$ and $Rez\ne 1$

Let assume $z=x+iy⟹x^2+y^2=1$

$\omega =\frac{1+(1-8\alpha )z}{1-z}$

$\omega =\frac{(1+(1-8\alpha \left)\right)(x+iy)}{1-(x+iy)}$

On solving the above equation

$Re\omega =\frac{(1+x(1-8\alpha \left)\right)(1-x)}{(1-x^2)+y^2)}=0$ (since it is given that $\omega$ is purely imaginary)

$⟹(1-x)+x(1-8\alpha =(1-8\alpha )x^2+(1-8\alpha )y^2$

as given $x^2+y^2=1$

$(1-x)+x(1-8\alpha =(1-8\alpha )$

$⟹\alpha =0$

So, $\alpha ϵ\left\{0\right\}$