Let $z_1$ and $z_2$ be two non-zero complex numbers such that $\frac{z_1}{z_2}+\frac{z_2}{z_1}=1,$ then the origin and points represented by $z_1$ and $z_2$

Let $z=\frac{z_1}{z_2},$ and $z+\frac{1}{z}=1\Rightarrow z^2-z+1=0$

$\Rightarrow z=\frac{1\pm \sqrt{3}i}{2}\Rightarrow \frac{z_1}{z_2}=\frac{1\pm \sqrt{3}i}{2}$

If $z_1$ and $z_2$ are represented by $A$ and $B$ respectively and $O$ be the origin, then

$\begin{array}{l}\frac{OA}{OB}=\frac{\left|z_1\right|}{\left|z_2\right|}=\left|\frac{1\pm \sqrt{3i}}{2}\right|=\sqrt{\frac{1}{4}+\frac{3}{4}=1}\\ \Rightarrow OA=OB\end{array}$

Also, $\frac{AB}{OB}=\frac{\left|z_2-z_1\right|}{\left|z_2\right|}=\left|1-\frac{z_1}{z_2}\right|$

   $=\left|1-\left(\frac{1}{2}\pm \frac{\sqrt{3}}{2}i\right)\right|=\left|\frac{1}{2}\mp \frac{\sqrt{3}}{2}i\right|=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
$\Rightarrow AB=OB$
Thus, $OA=OB=AB.$

$\therefore ∆OAB$ is an equilateral triangle.