Let  $Z=1+i$  and  $z_1=\frac{1+i\overline{z}}{\overline{z}(1-z)+\frac{1}{z}}$ then the value of  $\frac{12}{\pi }Arg\left(z_1\right)$  is

$$\begin{gathered} z_1=\frac{1+i\overline{z}}{\overline{z}(1-z)+\frac{1}{z}} \\ {z}_1=\frac{z\left(1+i\overline{z}\right)}{z\overline{z}(1-z)+1} \\ {z}_1=\frac{z+i{\left|z\right|}^2}{{\left|z\right|}^2(-i)+1} \\ {z}_1=\frac{1+i+2i}{1-2i}= \\ {z}_1=\frac{\left(1+3i\right)\left(1+2i\right)}{5} \\ {z}_1=\frac{-5+5i}{5}=-1+i \\ Arg\left(z_1\right)=\frac{3\mathrm{\pi }}{4} \end{gathered}$$

$\frac{12}{\pi }Arg\left(z_1\right)=\frac{12}{\pi }\left(\frac{3\mathrm{\pi }}{4}\right)=9$