Question

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

• A. $4k+1$
• B. $4k+2$
• C. $4k+3$
• D. $4k$

Answer 1

Answer: (D)

$n$th roots of unity are given by $\cos \left(\frac{2m\pi }{n}\right)+i\sin \left(\frac{2m\pi }{n}\right)=e^{2m\pi i/n}$ for $m=0,1,2,\cdots ,n-1$.
Let $z_1=e^{2m_1\pi i/n}$ and $z_2=e^{2m_2\pi i/n}$ where $0\le m_1,m_2<n,m_1\ne m_2$.
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
$\frac{e^{2m_1\pi i/n}}{e^{2m_2\pi i/n}}=il\text{ for some real }l\Rightarrow e^{2\left(m_1-m_2\right)\pi i/n}=il$
$\Rightarrow \cos \left[\frac{2\left(m_1-m_2\right)\pi }{n}\right]+i\sin \left[\frac{2\left(m_1-m_2\right)\pi }{n}\right]=il$
$\Rightarrow \cos \left[\frac{2\left(m_1-m_2\right)\pi }{n}\right]=0\Rightarrow \frac{2\left(m_1-m_2\right)\pi }{n}=\frac{\pi }{2}$
$\Rightarrow n=4\left(m_1-m_2\right)$
Thus, $n$ must be of the form $4k$.