Question

Let $z_{1}$ and $z_{2}$ be the $n^{t h}$ roots of unity which are ends of a line segment that 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)

We know that $1^{\frac{1}{n}}=cos \frac{2 r \pi }{n}+isin ⁡ \frac{2 r \pi }{n},r=0,1,\ldots \ldots ,n-1$
Let, $z_{1}=cos \frac{2 r_{1} \pi }{n}+ \, isin ⁡ \frac{2 r_{1} \pi }{n}$
And, $z_{2}=cos \frac{2 r_{2} \pi }{n}+isin ⁡ \frac{2 r_{2} \pi }{n}$ .
Then, $\angle z_{1}oz_{2}=amp\left(\frac{z_{1}}{z_{2}}\right)=amp\left(z_{1}\right)-amp\left(z_{2}\right)$
$\Rightarrow amp\left(z_{1}\right)-amp\left(z_{2}\right)=\frac{2 \left(r_{1} - r_{2}\right) \pi }{n}=\frac{\pi }{2}$
$\therefore $ $n=4 \, \left(r_{1} - r_{2}\right)=4\times $ integer.
So, $n$ is of the form $4k$ .