Question

If $\left|z\right|=1$ and $\omega =(z-1)/(z+1)$ (where $z\ne -1$), then $Re\left(\omega \right)$ is

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

Answer 1

The correct option is A $0$

Given,

$\left|z\right|=1\&w=\frac{(z-1)}{(z+1)}$

To find $Re\left(w\right)$

solution,

$Let,z=a+ib\therefore \overline{z}=a-ib\left|z\right|=\sqrt{a^2+b^2}\left|\overline{z}\right|=\sqrt{a^2+b^2}$

We know that

$Re\left(w\right)=\frac{w+\overline{w}}{2}Re\left(w\right)=\frac{\frac{(z-1)}{(z+1)}+\frac{\overline{z}-1}{\overline{z}+1}}{2}Re\left(w\right)=\frac{(z-1)(\overline{z}+1)+(\overline{z}-1)(z+1)}{2(z+1)(\overline{z}+1)}Re\left(w\right)=\frac{z\overline{z}-\overline{z}+z-1+z\overline{z}+\overline{z}-z-1}{2(z+1)(\overline{z}+1)}Re\left(w\right)=\frac{2z\overline{z}-2}{2(z+1)(\overline{z}+1)}Re\left(w\right)=\frac{z\overline{z}-1}{(z+1)(\overline{z}+1)}Re\left(w\right)=\frac{a^2+b^2-1}{(z+1)(\overline{z}+1)}$

Given, $a^2+b^2=1$

Hence, $Re\left(w\right)=0$