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Problem :

Let $z_1$ and $z_2$ be the $nth$ roots of unity which subtend a right angle at origin, then prove that n must be of the form $4k$

Solution : Here $arg \frac{z_1}{z_2}=\frac{\pi}{2}$

$\Rightarrow \frac{z_1}{z_2} =cos\frac{\pi}{2}+isin\frac{\pi}{2}$=i I didn't understand this step can anybody please help me on this , will be greatful to him/her.

Thanks.

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since the $n^{\text{th}}$ roots of unity are $e^{\frac{2\pi ik}{n}}$ we have integers $p,q$ which satisfy $$ (p-q)\frac{2\pi}{n} = \frac{\pi}2 $$ so $$ n = 4(p-q) $$

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