Solution
$\mathrm{z}=1+\mathrm{i} \Rightarrow \mathrm{z}_{1}=\frac{1+\mathrm{i} \overline{\mathrm{z}}}{\overline{\mathrm{z}}(1-\mathrm{z})+1 / \mathrm{z}} \Rightarrow \mathrm{z}_{1}=\frac{1+\mathrm{i}(1-\mathrm{i})}{(1-\mathrm{i})(1-1-\mathrm{i})+\frac{1}{1+\mathrm{i}}}$
\[
\begin{aligned}
& =\frac{1+\mathrm{i}-\mathrm{i}^{2}}{(1-\mathrm{i})(-\mathrm{i})+\frac{1-\mathrm{i}}{2}}=\frac{2+\mathrm{i}}{-\frac{3 \mathrm{i}-1}{2}}=\frac{4+2 \mathrm{i}}{-3 \mathrm{i}-1}=-\frac{(4+2 \mathrm{i})(3 \mathrm{i}-1)}{(3 \mathrm{i})^{2}-(1)^{2}}=-1+\mathrm{i} \\
& \operatorname{Arg}\left(\mathrm{z}_{1}\right)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \therefore \frac{12}{\pi} \arg \left(\mathrm{z}_{1}\right)=\frac{12}{\pi} \times \frac{3 \pi}{4}=9
\end{aligned}
\]
.
