If z is a complex number of unit modulus and argument $\mathrm{\theta }, \mathrm{then} \arg \left(\frac{1+\mathrm{z}}{1+\overline{z}}\right)$ equals:

$$\begin{gathered} \left|\mathrm{z}\right|=1, \mathrm{Arg} \mathrm{z}=\mathrm{\theta } \Rightarrow \mathrm{z}=\mathrm{cos\theta }+\mathrm{i} \mathrm{sin\theta } \\ \therefore \frac{1+\mathrm{z}}{1+\overline{\mathrm{z}}}=\frac{1+\mathrm{cos\theta }+\mathrm{i} \mathrm{sin\theta }}{1+\mathrm{cos\theta }-\mathrm{i} \mathrm{sin\theta }}=\frac{2{\cos }^2{\displaystyle \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\mathrm{i}\left(2\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\right(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}\left)\right)}{2{\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-\mathrm{i}\left(2\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\cos {\displaystyle \frac{\mathrm{\theta }}{2}}\right)} \\ =\frac{2\cos {\displaystyle \frac{\mathrm{\theta }}{2}}(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}+\mathrm{isin}{\displaystyle \frac{\mathrm{\theta }}{2}})}{2\cos {\displaystyle \frac{\mathrm{\theta }}{2}}(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}-\mathrm{i} \sin {\displaystyle \frac{\mathrm{\theta }}{2}})}=\frac{{\mathrm{e}}^{-\mathrm{i} {\displaystyle \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\mathrm{e}}^{-\mathrm{i} {\displaystyle \raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}={\mathrm{e}}^{\mathrm{i}\left(2\right)\frac{\mathrm{\theta }}{2}}={\mathrm{e}}^{\mathrm{i\theta }} \\ =\cos \mathrm{\theta }+i\sin \mathrm{\theta } \\ \boxed{Arg \left(\frac{1+\mathrm{z}}{1+\overline{\mathrm{z}}}\right) =\mathrm{\theta } } \end{gathered}$$