Question

Let $z_{1}and{z}_2$ be $n^{th}$ roots of unity which subtend a right angle at origin. Then 𝑛 must be of the form

• A. 4k+1
• B. 4k
• C. 4k+2
• D. 4k+3

Answer 1

The correct option is B 4k

$Letz=(1{)}^{1/n}=(\cos 2k\pi +i\sin 2k\pi {)}^{1/n}$
$=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n},k=0,1,2,\cdots ,n-1$

Let $z_1=\cos \frac{2k_1\pi }{n}+i\sin \frac{2k_1\pi }{n}$
$z_2=\cos \frac{2k_2\pi }{n}+i\sin \frac{2k_2\pi }{n}$
be the two values of $z$ such that they subtend angle of ${90}^{\circ }$ at origin.

Then, $\frac{2k_1\pi }{n}-\frac{2k_2\pi }{n}=\pm \frac{\pi }{2}$
⇒ $4(k_1-k_2)=\pm n$
As, $k_1$ and $k_2$ are integers and $k_1\ne k_2$
Therefore, $n=4k,k\in Z$