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 (where $k$ is an integer)
Step-by-Step Solution
Key Concept: A right angle at the origin means the ratio $z_1/z_2=\pm i$, which requires $e^{2\pi i(a-b)/n}=\pm i$, forcing $4\mid n$.
**Step 1: Express z₁ and z₂ as nth roots**
$z_1 = e^{2\pi i a/n}$, $z_2 = e^{2\pi i b/n}$ for integers $a,b$.
**Step 2: Right angle condition**
Vectors $z_1,z_2$ from origin are perpendicular iff $\arg(z_1/z_2)=\pm\dfrac{\pi}{2}$, i.e. $\dfrac{2\pi(a-b)}{n}=\pm\dfrac{\pi}{2}$.
**Step 3: Conclude**
$\dfrac{a-b}{n}=\pm\dfrac{1}{4}$, so $n = 4(a-b)$, meaning $4\mid n$, i.e. $n=4k$.
Correct Answer: 4