Question

The set of all $\alpha \epsilon R$, for which $w = \frac{1 + (1 - 8 \alpha)z}{1 - z}$ is a purely imaginary number, for all $z \neq 1$, is :

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

Answer 1

Answer: (B)

$\\{ 0 \\}$

Answer 2

As $\omega$ is purely imaginary $\omega+\bar{\omega}=0$ $\frac{1+(1-8 \alpha) z}{1-z}+\frac{1+(1-8 a) \bar{z}}{1-z}=0$ $\frac{1-\bar{z}+(1-8 a)(z-1)+a-z+(1-8 \alpha)(\bar{z}-1)}{(1-z)(1-\bar{z})}=0$ $1-\bar{z}+z-1-8 a z+8 a+1-z+\bar{z}-1-8 \bar{z}-1-8 \bar{z} \alpha+8 \alpha=0$ $-8 a(z+\bar{z})+16 \alpha=0$ $8 a[2-(2+2)]=0$ if $Re(z) \neq 1$