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

 

$\left|z\right|=1,\arg \left(z\right)=\theta$

$\Rightarrow z=\cos \theta +i\sin \theta$

Now, $w=\frac{z(1-\overline{z})}{\overline{z}(1+z)}=\frac{z-z\overline{z}}{z+z\overline{z}}$

$=\frac{z-1}{\overline{z}+1}$

$\overline{w}=\frac{\overline{z}-1}{z+1}$

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

$\begin{array}{r}=\frac{z^2-1+{\overline{z}}^2-1}{(z+1)(\overline{z}+1)}\\ =\frac{2\cos 2\theta -2}{z\overline{z}+z+\overline{z}+1}\end{array}$

$=\frac{2(\cos 2\theta -1)}{2(\cos \theta +1)}$

$=\frac{-2{\sin }^2\left(\theta \right)}{2{\cos }^2(\theta /2)}=-4{\sin }^2\left(\frac{\theta }{2}\right)$

$\Rightarrow \mathrm{Re}\left(w\right)=-2{\sin }^2\left(\frac{\theta }{2}\right)$