Product of Diagonals in Regular N-gon
Let ε_{0} = 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 ε_{0} is N. |
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Let ε_{0} = 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 ε_{0} is N. |
Solution
Let's shift the N-gon one unit to the left so that the vertices will become
Vertices ε are the N^{th} roots of unity:
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 ω_{1}, ω_{2}, ..., 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-1}N, |
which shows that
Π|ω_{k}| = |Πω_{k}| = N. |
Exactly what we set out to prove.
Complex Numbers
- Algebraic Structure of Complex Numbers
- Division of Complex Numbers
- Useful Identities Among Complex Numbers
- Useful Inequalities Among Complex Numbers
- Trigonometric Form of Complex Numbers
- Real and Complex Products of Complex Numbers
- Complex Numbers and Geometry
- Plane Isometries As Complex Functions
- Remarks on the History of Complex Numbers
- Complex Numbers: an Interactive Gizmo
- Cartesian Coordinate System
- Fundamental Theorem of Algebra
- Complex Number To a Complex Power May Be Real
- One can't compare two complex numbers
- Riemann Sphere and Möbius Transformation
- Problems
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