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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