Đặt $z=a+bi,\left(a,b\in ℝ\right).$  Ta có

$\left|z\right|=\sqrt{a^2+b^2}=1\Rightarrow a^2+b^2=1.$

$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right|=\left|\frac{z^2+{\overline{z}}^2}{z.\overline{z}}\right|=\frac{\left|{\left(a+bi\right)}^2+{\left(a-bi\right)}^2\right|}{{\left|z\right|}^2}=\left|2a^2-2b^2\right|=1.$

Ta có hệ: $\left\{\begin{array}{l}{a}^2+b^2=1\\ \left|2a^2-2b^2\right|=1\end{array}\right.\iff \left\{\begin{array}{l}{a}^2+b^2=1\\ a^2-b^2=\frac{1}{2}\end{array}\right.$ hoặc $\left\{\begin{array}{l}{a}^2+b^2=1\\ a^2-b^2=-\frac{1}{2}\end{array}\right.$

$\iff \left\{\begin{array}{l}{a}^2=\frac{3}{4}\\ b^2=\frac{1}{4}\end{array}\right.$ hoặc $\left\{\begin{array}{l}{a}^2=\frac{1}{4}\\ b^2=\frac{3}{4}\end{array}\right..$

Suy ra $\left(a;b\right)\in \left\{\left(\frac{1}{2};\pm \frac{\sqrt{3}}{2}\right);\left(-\frac{1}{2};\pm \frac{\sqrt{3}}{2}\right);\left(\frac{\sqrt{3}}{2};\pm \frac{1}{2}\right);\left(-\frac{\sqrt{3}}{2};\pm \frac{1}{2}\right)\right\}.$

Vậy có 8 cặp số (a;b) do đó có 8 số phức thỏa mãn.