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. $4p+2,p\in Z^{+},p\le n-1$
• B. $4p+1,p\in Z^{+},p\le n-1$
• C. $4p,p\in Z^{+},p\le n-1$
• D. $4p+3,p\in Z^{+},p\le n-1$

Answer 1

The correct option is C $4p,p\in Z^{+},p\le n-1$

The $n^{th}$ roots of unity are given by

$e^{i\left(\frac{2k\pi }{n}\right)},k=0,1,2,......,n-1$
$\text{Let,}{z}_1=e^{i\left(\frac{2k_1\pi }{n}\right)}\text{and}{z}_2=e^{i\left(\frac{2k_2\pi }{n}\right)}$
where, $k_1$ and $k_2$ are distinct integers from the set $\{0,1,2,3,....,n-1\}$
As, $z_1$ and $z_2$ subtend a right angle at the origin, therefore, difference $\text{between there arguments should be}\frac{\pi }{2}$
$\Rightarrow |\frac{2k_1\pi }{n}-\frac{2k_2\pi }{n}|=\frac{\pi }{2},k_1\ne k_2$
$\Rightarrow \frac{2\pi }{n}|k_1-k_2|=\frac{\pi }{2}$
$\Rightarrow 4|k_1-k_2|=n$
So, $n=4p\text{let,}|k_1-k_2|=p$
$\therefore n=4p,p\in Z^{+},p\le n-1(\because k_1\ne k_2)$