Roots of Unity

Theorem. For each positive integer $n$, the equation $z^n = 1$ has exactly $n$ complex solutions, namely $$z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n} = e^{2\pi i k/n}, \quad k = 0, 1, 2, \dots, n-1.$$

Derivation. Write $z = r(\cos\theta + i\sin\theta)$ and $1 = 1(\cos 0 + i\sin 0)$. By De Moivre, $z^n = r^n(\cos n\theta + i\sin n\theta)$, so $z^n = 1$ requires

• $r^n = 1$, and since $r \geq 0$, $r = 1$.
• $n\theta \equiv 0 \pmod{2\pi}$, i.e. $n\theta = 2\pi k$ for some integer $k$.

So $\theta = 2\pi k / n$. As $k$ ranges over $0, 1, \dots, n-1$ we get $n$ distinct values of $\theta$ in $[0, 2\pi)$; further integers $k$ recycle the same roots (e.g., $k = n$ gives $\theta = 2\pi$, the same as $k = 0$).

Geometric picture. All $n$ roots have modulus $1$, so they lie on the unit circle. Consecutive roots differ in argument by $2\pi/n$, so the chord between any two consecutive roots subtends the same angle at the centre. The roots therefore form the vertices of a regular $n$-gon, with one vertex at $z_0 = 1$.

Hook revisited — cube roots of unity: $z^3 = 1$ gives $z_k = e^{2\pi i k/3}$ for $k = 0, 1, 2$:

• $z_0 = 1$
• $z_1 = \cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3} = -\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}$
• $z_2 = \cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3} = -\tfrac{1}{2} - i\tfrac{\sqrt{3}}{2}$

Their sum: $1 + (-\tfrac{1}{2} + i\tfrac{\sqrt 3}{2}) + (-\tfrac{1}{2} - i\tfrac{\sqrt 3}{2}) = 0$. ✓

Formula: $z_k = e^{2\pi i k / n}$ for $k = 0, 1, \dots, n-1$ (the $n$th roots of unity) · Geometry: $n$ vertices of a regular $n$-gon on the unit circle, one at $z = 1$ · Cube roots: $1,\; -\tfrac{1}{2} \pm i\tfrac{\sqrt{3}}{2}$ · Identity: $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \dots + z + 1)$

Pause — copy the $n$th roots of unity formula $z_k = e^{2\pi ik/n}$, the regular $n$-gon geometry, the cube roots, and the factorisation $z^n-1 = (z-1)(z^{n-1}+\cdots+1)$ into your book.

We just saw that the $n$th roots of unity are $z_k = e^{2\pi ik/n}$ for $k = 0, \ldots, n-1$ — equally spaced on the unit circle with one root always at $z = 1$. That raises a question: what do Vieta's formulas say about the sum and product of these $n$ roots? This card answers it → the sum is 0 (for $n \geq 2$) and the product is $(-1)^{n-1}$, read from coefficients of $z^n - 1$.

Two identities about the $n$th roots of unity are essential for HSC questions:

• Sum. $1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$ for $n \geq 2$, where $\omega = e^{2\pi i / n}$.
• Product. $1 \cdot \omega \cdot \omega^2 \cdots \omega^{n-1} = (-1)^{n-1}$.

Proof of the sum. The roots of $z^n - 1 = 0$ are exactly $1, \omega, \omega^2, \dots, \omega^{n-1}$. By Vieta's formulas, the sum of the roots equals minus the coefficient of $z^{n-1}$ divided by the leading coefficient. In $z^n - 1$ the coefficient of $z^{n-1}$ is $0$, so the sum is $0$.

Alternatively: $\sum_{k=0}^{n-1} \omega^k = \dfrac{\omega^n - 1}{\omega - 1} = \dfrac{1 - 1}{\omega - 1} = 0$ (geometric series, valid since $\omega \neq 1$).

Proof of the product. Vieta gives the product equal to $(-1)^n \cdot (\text{constant})/\text{leading} = (-1)^n \cdot (-1)/1 = (-1)^{n+1} = (-1)^{n-1}$.

Sum of $n$th roots of unity is $0$ (for $n \geq 2$) · Product of $n$th roots of unity is $(-1)^{n-1}$ · Proof via Vieta's: read off coefficients of $z^n - 1$ · Real and imaginary parts of the sum identity give trig identities

Pause — copy the sum-of-roots identity ($= 0$ for $n \geq 2$), the product identity ($= (-1)^{n-1}$), the Vieta derivation from $z^n-1$, and the trig identity consequence (Re and Im parts of sum) into your book.