Question

$If|z|=1and\omega =\frac{z-1}{z+1}(where,z\ne -1),thenRe\left(\omega \right)is$

• A. 0
• B. $\frac{1}{|z+1{|}^2}$
• C. $\left|\begin{array}{c}\frac{1}{z+1}\end{array}\right|.\frac{1}{|z+1{|}^2}$
• D. $\frac{\sqrt{2}}{|z+1{|}^2}$

Answer 1

Answer: (A)

0

Since,$|z|=1andomega=\frac{z-1}{z+1}\Rightarrow z-1=\omega z+\omega \Rightarrow z=\frac{1+\omega }{1-\omega }\Rightarrow |z|=\frac{|1+\omega |}{|1-\omega |}\Rightarrow |1-\omega |=|1+\omega |[\therefore |z|=1]$
On squaring both sides,we get
$1+|\omega {|}^2-2Re\left(\omega \right)=1+|\omega {|}^2+2Re\left(\omega \right)$
using $|z_1\pm z_2{|}^2=|z_1{|}^2+|z_2{|}^2\pm 2Re\left({\overline{z}}_1{z}_2\right)]\Rightarrow 4Re(\omega )=0\Rightarrow Re(\omega )=0$