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, $\arg \frac{z_{1}}{z_{2}}=\frac{\pi}{2}$
$\Rightarrow \frac{z_{1}}{z_{2}}=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}=i$
$\therefore \frac{z_{1}^{n}}{z_{2}^{n}}=(i)^{n} \Rightarrow i^{n}=1$
$\left[\because\left|z_{2}\right|=\left|z_{1}\right|=1\right]$
$\Rightarrow n=4 k$
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.
$z_{1}^{n}=z_{2}^{n}=1 \Rightarrow\left(\frac{z_{2}}{z_{1}}\right)^{n}=1 \Rightarrow i^{n}=1$
$\Rightarrow n=4 k$, where $k$ is an integer.