Let $z_1$ and $z_2$ be the nth roots of unity which subtend a right angle at the origin, then n = 

Let $\begin{array}{r}z=(1{)}^{1/n}=[\cos \left(2r\pi \right)+i\sin \left(2r\pi \right){]}^{1/n}\\ =\left(\cos \left(\frac{2r\pi }{n}\right)+i\sin \left(\frac{2r\pi }{\mathrm{n}}\right)\right)\end{array}$

where $r=0,1,2,3,\dots ,(n-1)$

Let 

$z_1=1\text{ and }{z}_2=e^{i\frac{2k\pi }{n}}$

It is given that

$\begin{array}{l}\left(z_2-0\right)=\left(z_1-0\right)e^{i\frac{\pi }{2}}\\ \Rightarrow e^{i\frac{2k\pi }{n}}=e^{i\frac{\pi }{2}}\\ \Rightarrow \frac{2k\pi }{n}=\frac{\pi }{2}\\ \Rightarrow n=4k\end{array}$