Nth roots subtending right angle at origin Let $\ z_1$ and $\ z_2$ be nth roots of unity which subtend a right angle at the origin. Then n must be of the form(A) 4k + 1(B) 4k + 2(C) 4k + 3(D) 4k My approach Let the roots be represented as $Cos \frac{2kπ}{n}$+i$Sin \frac{2kπ}{n}$ At $\ z_1$ the coordinates are ($Cos \frac{2\ k_1 π}{n}$,$Sin \frac{2\ k_1 π}{n}$) At $\ z_2$ the coordinates are ($Cos \frac{2\ k_2 π}{n}$,$Sin \frac{2\ k_2 π}{n}$) I tried to use the product to two slope as -1 but not getting the answer

Let nth root be $z=\exp({2\pi i\frac{k}{n}})$. Now you can say other root of unity is $z'=\exp(2\pi i\frac{k+l}{n})$.

Now a number subtending angle $90^\circ$ anticlockwise from $z$ is $\exp(\frac{\pi i}{2})\cdot z$. So $z' = \exp(\frac{\pi i}{2})\cdot z$

$$\exp\left(2\pi i\frac{k+l}{n}\right) =\exp\left(\frac{\pi i}{2}\right) \exp\left(2\pi i\frac{k}{n}\right)$$

On solving

$$\frac{ 2\pi i(k+l)}{n} = \frac{\pi i}{2} + \frac{2\pi ik}{n}$$

you get $n = 4l$