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Q.
Let $z_1$ and $z_2$ be $n$th roots of unity which subtend a right angle at the origin. Then $n$ must be of the form
Complex Numbers and Quadratic Equations
Solution:
$n$th roots of unity are given by $\cos \left(\frac{2 m \pi}{n}\right)+i \sin \left(\frac{2 m \pi}{n}\right)=e^{2 m \pi i / n}$ for $m=0,1,2, \cdots, n-1$.
Let $z_1=e^{2 m_1 \pi i / n}$ and $z_2=e^{2 m_2 \pi i / n}$ where $0 \leq m_1, m_2< n, m_1 \neq m_2$.
As the join of $z_1$ and $z_2$ subtend a right angle at the origin $z_1 / z_2$ is purely imaginary we get
$\frac{e^{2 m_1 \pi i / n}}{e^{2 m_2 \pi i / n}}=i l \text { for some real } l \Rightarrow e^{2\left(m_1-m_2\right) \pi i / n}=i l $
$\Rightarrow \cos \left[\frac{2\left(m_1-m_2\right) \pi}{n}\right]+i \sin \left[\frac{2\left(m_1-m_2\right) \pi}{n}\right]=i l$
$\Rightarrow \cos \left[\frac{2\left(m_1-m_2\right) \pi}{n}\right]=0 \Rightarrow \frac{2\left(m_1-m_2\right) \pi}{n}=\frac{\pi}{2}$
$\Rightarrow n=4\left(m_1-m_2\right)$
Thus, $n$ must be of the form $4 k$.