If z₁ and z₂ be the nth root of unity which subtend right angled at the origin. Then, n must be of the form

The nth roots of  unity is given by

$\cos \left(\frac{2\mathrm{r\pi }}{\mathrm{n}}\right)+\mathrm{isin}\left(\frac{2\mathrm{r\pi }}{\mathrm{n}}\right)={\mathrm{e}}^{2\mathrm{riin}}$

Where r= 0,1,2,….(n-1)

So, let ${\mathrm{z}}_1={\mathrm{e}}^{2{\mathrm{\pi r}}_1\mathrm{i}/\mathrm{n}}$ and ${\mathrm{z}}_2={\mathrm{e}}^{2{\mathrm{\pi r}}_2\mathrm{i}/\mathrm{n}}$ where $0\le {\mathrm{r}}_1{\mathrm{r}}_2<{\mathrm{n}}_2{\mathrm{r}}_{\mathrm{i}}\ne {\mathrm{r}}_2$.

It is given that the line segment joining the points having affixes z₁ and z₂  subtends a right angled at the origin. Therefore,
$\begin{array}{l}\arg \left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=\pm \frac{\mathrm{\pi }}{2}\\ \Rightarrow \frac{2{\mathrm{\pi r}}_1}{\mathrm{n}}-\frac{2{\mathrm{\pi r}}_2}{\mathrm{n}}=\pm \frac{\mathrm{\pi }}{2}\\ \Rightarrow \mathrm{n}=\pm 4\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)\\ \therefore \mathrm{n}=4\mathrm{k},\text{ where }\mathrm{k}=\pm \left({\mathrm{r}}_1-{\mathrm{r}}_2\right)\end{array}$