Complex Numbers
- Algebraic Structure of Complex Numbers
- Division of Complex Numbers
- Product of Diagonals in Regular N-gon
- Useful Inequalities Among Complex Numbers
- Trigonometric Form of Complex Numbers
- Real and Complex Products of Complex Numbers
- Complex Numbers and Geometry
- Remarks on the History of Complex Numbers
- Complex Numbers: A Dynamic Tool
- Cartesian Coordinate System
- Fundamental Theorem of Algebra
- Complex Number To a Complex Power May Be Real
- One Can't Compare Two Complex Numbers
- Problems
- Product of Diagonals in Regular N-gon
Product of Diagonals in Regular N-gon
| |
Let ε0 = 1, ..., εN-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
Copyright © 1996-2008 Alexander Bogomolny
| |
Let ε0 = 1, ..., εN-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 Πωk, where k ranges from 1 to N-1.
Vertices ε are the Nth roots of unity: εN = 1. Except for ε0, all other ε's are the roots of the equation
| |
εN-1 + εN-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, ..., ωN-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
which shows that
Exactly what we set out to prove.
Copyright © 1996-2008 Alexander Bogomolny
|
