Find nth roots of unity and prove they form vertices of a regular polygon

Question

Find all the $n$ th roots of unity. Prove that these roots, when plotted on the Argand plane, form the vertices of a regular $n$ -sided polygon inscribed in the unit circle.

(JEE Advanced 2020, similar pattern)

Solution — Step by Step

Write $1$ in polar form: $1 = e^{i \cdot 2k\pi}$ for any integer $k$ .

So $z^n = e^{i \cdot 2k\pi}$ , which gives:

z = e^{i \cdot 2k\pi/n}, \quad k = 0, 1, 2, \ldots, n-1

Let $\omega = e^{i \cdot 2\pi/n}$ . Then the $n$ roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$ .

For each root $\omega^k = e^{i \cdot 2k\pi/n}$ :

|\omega^k| = |e^{i \cdot 2k\pi/n}| = 1

Every root has modulus 1, so all $n$ points lie on the unit circle centred at the origin.

The argument of $\omega^k$ is $\frac{2k\pi}{n}$ .

The angular gap between consecutive roots $\omega^k$ and $\omega^{k+1}$ is:

\frac{2(k+1)\pi}{n} - \frac{2k\pi}{n} = \frac{2\pi}{n}

This gap is constant for all consecutive pairs. Equal angular spacing on a circle means equal arc lengths, which means equal chord lengths.

Since all $n$ points lie on the same circle (radius 1) and are equally spaced at angles of $\frac{2\pi}{n}$ , they form the vertices of a regular $n$ -sided polygon.

The first root is always at $(1, 0)$ — the point where the positive real axis meets the unit circle.

Why This Works

The $n$ th roots of unity are solutions to $z^n - 1 = 0$ , a polynomial of degree $n$ that must have exactly $n$ roots (counting multiplicity). De Moivre’s theorem gives us all $n$ roots explicitly in exponential form, and the geometry follows from the fact that multiplication by $\omega$ is a rotation by $2\pi/n$ .

Think of it this way: starting from the point $1$ on the unit circle, each multiplication by $\omega$ rotates you by the same angle $2\pi/n$ . After $n$ rotations, you’re back to $1$ (since $\omega^n = 1$ ). This uniform rotation traces out a regular polygon.

Alternative Method — Distance calculation

To prove regularity without angles, compute the distance between consecutive roots:

|\omega^{k+1} - \omega^k| = |\omega^k||\omega - 1| = 1 \cdot |\omega - 1| = |\omega - 1|

This distance is the same for every consecutive pair. For a convex polygon inscribed in a circle, equal consecutive side lengths implies regularity.

For $n = 3$ : equilateral triangle with vertices at $1, \omega, \omega^2$ where $\omega = e^{i2\pi/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ . For $n = 4$ : square with vertices $1, i, -1, -i$ . Visualising these specific cases helps in MCQs where you need to quickly check properties of roots of unity.

Common Mistake

A subtle error: students write $z = e^{i k\pi/n}$ instead of $z = e^{i \cdot 2k\pi/n}$ — missing the factor of 2. This gives $2n$ roots instead of $n$ , half of which are roots of $z^n = -1$ , not $z^n = 1$ . The full angle for one revolution is $2\pi$ , not $\pi$ . Always start from $1 = e^{i \cdot 2k\pi}$ with the $2\pi$ clearly visible.

Question

Solution — Step by Step

Why This Works

Alternative Method — Distance calculation

Common Mistake

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