Suppose$z$ is a complex number such that $z\ne -1$, $\left|z\right|=1,$and$\arg \left(z\right)=$$\theta$Let$\omega$$=\frac{z(1-\overline{z})}{\overline{z}(1+z)}$,than $\mathrm{Re}\left(\omega \right)$ ) is equal to

$\begin{array}{l}\omega =\frac{z-z\overline{z}}{\overline{z}+\overline{z}z}=\frac{z-1}{1/z+1}=\frac{(z-1)z}{z+1}[\because |z|=1]\\ =\frac{z^2-1}{z+1}-\frac{z-1}{z+1}=z-1+\frac{1-z}{1+z}\\ \mathrm{Re}\left(\omega \right)=\mathrm{Re}\left(z\right)-1+\mathrm{Re}\left(\frac{1-z}{1+z}\right)\end{array}$

But $\mathrm{Re}\left(\frac{1-z}{1+z}\right)=\frac{1}{2}\left(\frac{1-z}{1+z}+\frac{1-\overline{z}}{1+\overline{z}}\right)$

$=\frac{1}{2}\left(\frac{1-z}{1+z}+\frac{\overline{z}-1}{\overline{z}+1}\right)=0$

$\therefore \mathrm{Re}\left(\omega \right)=\cos \theta -1=-2{\sin }^2(\theta /2)$