Question

Let $z_1$ and $z_2$ be nth roots of unity which subtend a right angle 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

The correct option is D 4k

Since, $arg\frac{z_1}{z_2}=\frac{\pi }{2}\Rightarrow \frac{z_1}{z_2}=cos\frac{\pi }{2}+isin\frac{\pi }{2}=i\therefore \frac{z_1^n}{z_2^n}=(i{)}^n\Rightarrow i^n=1\Rightarrow n=4k$
Alternate Solution

arg$arg\frac{z_2}{z_1}=\frac{\pi }{2}\therefore \frac{z_2}{z_1}=\left|\frac{z_2}{z_1}\right|e^{i\frac{\pi }{2}}\Rightarrow \frac{z_2}{z_1}=i\Rightarrow {\left(\frac{z_2}{z_1}\right)}^n=i^n$
$\therefore z_1$ and $z_2$ are nth 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.