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
 A: 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 $n = 4l$
A: Here is a hint for a different approach, if you know that the result of dividing one complex number by another gives you a result whose argument is the difference in arguments of the original numbers (counterpart to the fact that multiplying adds arguments). You also need to know that the complex numbers on the unit circle with arguments equal to a right-angle are $\pm i$.
Hint: show that $\cfrac {z_1}{z_2}$ is an $n^{th}$ root of unity (not necessarily primitive).

 Then show that $\cfrac {z_1}{z_2}=\pm i$ and note that both $(\pm i)^4=1$ and $(\pm i)^n=1$

