Product of Diagonals in Regular N-gon

Let ε₀ = 1, ..., ε_N-1 be the vertices of a regular N-gon inscribed on the unit circle. Show that the product of the lengths of all diagonals of the N-gon emanating from ε₀ is N.

Solution

Let ε₀ = 1, ..., ε_N-1 be the vertices of a regular N-gon inscribed on the unit circle. Show that the product of the lengths of all diagonals of the N-gon emanating from ε₀ is N.

Solution

Let's shift the N-gon one unit to the left so that the vertices will become ω = ε - 1. In particular, ω₀ = ε₀ - 1 = 0 falls onto the origin. We are thus interested in the product Πω_k, where k ranges from 1 to N-1.

Vertices ε are the N^th roots of unity: ε^N = 1. Except for ε₀, all other ε's are the roots of the equation

ε^N-1 + ε^N-2 + ... + ε + 1 = 0.

This means that ω's satisfy the equation

(ω + 1)^N-1 + (ω + 1)^N-2 + ... + (ω + 1) + 1 = 0.

This is a polynomial equation of degree N-1 with the free coefficient equal to N and the roots ω₁, ω₂, ..., ω_N-1. As is well known, the product of all roots of a polynomial equation is given by its free coefficient (up to a sign), in particular

Πω_k = (-1)^N-1N,

which shows that

Π|ω_k| = |Πω_k| = N.

Exactly what we set out to prove.

Complex Numbers

71851904