The number of complex numbers z such that |z| = 1 and $\left|\frac{\mathrm{z}}{\overline{\mathrm{z}}}+\frac{\overline{\mathrm{z}}}{\mathrm{z}}\right|=1\text{ is }(\arg (\mathrm{z})\in [0,2\mathrm{\pi }\left]\right)$

Let $\mathrm{z}=\cos \mathrm{x}+\mathrm{isin}\mathrm{x},\mathrm{x}\in [0,2\mathrm{\pi })$. Then,

$\begin{array}{c}1=\left|\frac{\mathrm{z}}{\overline{\mathrm{z}}}+\frac{\overline{\mathrm{z}}}{\mathrm{z}}\right|\\ =\frac{\left|{\mathrm{z}}^2+{\overline{\mathrm{z}}}^2\right|}{|\mathrm{z}{|}^2}\\ =|\cos 2\mathrm{x}+\mathrm{isin}2\mathrm{x}+\cos 2\mathrm{x}-\mathrm{isin}2\mathrm{x}|\\ =2|\cos 2\mathrm{x}|\end{array}$

$\Rightarrow \cos 2\mathrm{x}=\pm 1/2$

Now, cos 2x = 1/2

$\begin{array}{l}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{\pi }}{6},{\mathrm{x}}_2=\frac{5\mathrm{\pi }}{6},{\mathrm{x}}_3=\frac{7\mathrm{\pi }}{6},{\mathrm{x}}_4=\frac{11\mathrm{\pi }}{6}\\ \cos 2\mathrm{x}=-\frac{1}{2}\\ \Rightarrow {\mathrm{x}}_5=\frac{\mathrm{\pi }}{3},{\mathrm{x}}_6=\frac{2\mathrm{\pi }}{3},{\mathrm{x}}_7=\frac{4\mathrm{\pi }}{3},{\mathrm{x}}_8=\frac{5\mathrm{\pi }}{3}\end{array}$