Question

Let $z_1$ and $z_2$ be $n^{th}$ roots of unity which subtend a right angled at the origin, then $n$ must be of the form (where, $k$ is an integer)

• A. 4k + 1
• B. 4k + 2
• C. 4k + 3
• D. 4k

Answer 1

Answer: (D)

Since, $\arg \frac{z_1}{z_2}=\frac{\pi }{2}$
$\Rightarrow \frac{z_1}{z_2}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i$
$\therefore \frac{z_1^n}{z_2^n}=(i{)}^n\Rightarrow i^n=1$
$\left[\because \left|z_2\right|=\left|z_1\right|=1\right]$
$\Rightarrow n=4k$
Alternate Solution

Since, $\arg \frac{z_2}{z_1}=\frac{\pi }{2}$
$\therefore \frac{z_2}{z_1}=\left|\frac{z_2}{z_1}\right|e^{i\frac{\pi }{2}}$
$\Rightarrow \frac{z_2}{z_1}=i$
$\left[\because \left|z_1\right|=\left|z_2\right|=1\right]$
$\Rightarrow {\left(\frac{z_2}{z_1}\right)}^n=i^n$
$\therefore z_1$ and $z_2$ are $n$th roots of unity.
$z_1^n=z_2^n=1\Rightarrow {\left(\frac{z_2}{z_1}\right)}^n=1\Rightarrow i^n=1$
$\Rightarrow n=4k$, where $k$ is an integer.