Complex Roots of Unity and Regular Polygons

In the HSC Mathematics Extension 2 course, complex numbers provide a powerful bridge between algebra and geometry. The $n$-th roots of unity lie equally spaced on the unit circle, forming the vertices of a regular $n$-gon. By exploring their algebraic properties—such as their sum and product—we uncover elegant connections between polynomial equations and geometric distances. This problem demonstrates two classic results: the sum of the non‑trivial roots is zero, and the product of the chord lengths from one vertex to all others equals $n$.

Problem Statement

Let $\omega$ be a complex number such that $\omega^n = 1$ and $\omega \neq 1$, meaning $\omega$ is an $n$-th root of unity. The roots of the equation $z^n - 1 = 0$ are given by $1, \omega, \omega^2, \dots, \omega^{n-1}$.

1. Show that $1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0$.

2. Suppose these $n$ roots are represented by the points $P_0, P_1, P_2, \dots, P_{n-1}$ on the complex plane, which form a regular $n$-gon inscribed in the unit circle. Let $P_0$ correspond to the root $1$. Prove that the product of the lengths of all chords from $P_0$ to the other vertices is equal to $n$. That is, prove that:

   $$|P_0 P_1| \cdot |P_0 P_2| \cdots |P_0 P_{n-1}| = n$$

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Hints

• For the first part, consider the geometric progression or the sum of roots in the polynomial $z^n - 1$.
• For the second part, factorize $z^n - 1$ over the complex numbers into linear factors.
• The length of the chord $|P_0 P_k|$ corresponds to the modulus $|1 - \omega^k|$.
• Substitute an appropriate value for $z$ after dividing out the factor $(z-1)$ from the polynomial.

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Solutions

1. We are given that $\omega^n = 1$ and $\omega \neq 1$. Consider the sum of a geometric progression with $n$ terms, first term $a=1$, and common ratio $r=\omega$:

   $$S_n = 1 + \omega + \omega^2 + \dots + \omega^{n-1}$$

   Using the sum formula $S_n = \frac{a(1 - r^n)}{1 - r}$:

   $$S_n = \frac{1 - \omega^n}{1 - \omega}$$

   Since $\omega^n = 1$ and $\omega \neq 1$, we have:

   $$S_n = \frac{1 - 1}{1 - \omega} = 0$$

   Therefore, $1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0$. (Alternatively, this is the sum of roots taken one at a time for the equation $z^n - 1 = 0$, where the coefficient of $z^{n-1}$ is 0).

2. Geometrically, the distance from $P_0$ (the number $1$) to $P_k$ (the number $\omega^k$) is $|1 - \omega^k|$, and we need the product of these distances for $k=1,\dots,n-1$. We’ll connect this product to the polynomial $z^n-1$. The roots of $z^n - 1 = 0$ are $1, \omega, \omega^2, \dots, \omega^{n-1}$. Therefore, we can factorize $z^n - 1$ as:

   $$z^n - 1 = (z - 1)(z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1})$$

   For $z \neq 1$, we can divide both sides by $(z - 1)$ to get:

   $$\frac{z^n - 1}{z - 1} = (z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1})$$

   Using the factorization $z^n - 1 = (z-1)(z^{n-1} + z^{n-2} + \dots + z + 1)$, we rewrite the left side:

   $$z^{n-1} + z^{n-2} + \dots + z + 1 = (z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1})$$

   This is an identity for all complex numbers $z$. Let $z = 1$:

   $$1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \omega)(1 - \omega^2) \cdots (1 - \omega^{n-1})$$

   $$n = (1 - \omega)(1 - \omega^2) \cdots (1 - \omega^{n-1})$$

   Taking the modulus of both sides, and noting that the modulus of a product is the product of the moduli:

   $$|n| = |1 - \omega| \cdot |1 - \omega^2| \cdots |1 - \omega^{n-1}|$$

   Since $n$ is positive, $|n| = n$. The point $P_0$ represents 1, and $P_k$ represents $\omega^k$. The distance $|P_0 P_k|$ is exactly $|1 - \omega^k|$. Thus, substituting these distances yields:

   $$n = |P_0 P_1| \cdot |P_0 P_2| \cdots |P_0 P_{n-1}|$$

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Takeaways

• The $n$-th roots of unity sum to $0$, which reflects the physical intuition of $n$ equal forces pulling outwards symmetrically.
• Polynomial factorization is a powerful tool to translate geometric lengths into algebraic products.
• Dividing a polynomial by $(z-1)$ and then evaluating at $z=1$ is a standard trick to find products of distances from the vertex at $(1, 0)$ to all other vertices of a regular polygon.

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Further Readings

HSC Complex Numbers, HSC Polynomials, HSC Sequences