## Complex attitude!

Let us a discuss yet one more complex number based IITJEE mains problem.

Problem:

Let $z_{1}$ and $z_{2}$ be nth roots of unity, which subtend a right angle at the origin. Then, integer n must be of the form __________________. (fill in the blank).

Solution:

nth roots of unity are given by $\cos{\frac{2m\pi}{n}}+ i \sin{\frac{2m\pi}{n}}= e^{\frac{2m\pi}{n}}$

where $m=0,1,2, \ldots, n-1$.

Let $z_{1}=\cos{\frac{2m_{1}\pi}{n}}+ i\sin{\frac{2m_{1}\pi}{n}}=e^{2m_{1}\pi i/n}$

Let $z_{2}=e^{2m_{2}\pi i/n}$ where $0 \leq m_{1}, m_{2}< n$, $m_{1} \neq m_{2}$

As the join of $z_{1}$ and $z_{2}$ subtends a right angle at the origin, we deduce that $\frac{z_{1}}{z_{2}}$ is purely imaginary. $\Longrightarrow \frac{e^{2m_{1}\pi i/n}}{e^{2m_{2}\pi i/n}}=ik$, for some real k $\Longrightarrow e^{2(m_{1}-m_{2})\pi i/n}=ik$ $\Longrightarrow n=4(m_{1}-m_{2})$. Thus, n must be of the form 4k.

More later,

Nalin Pithwa

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