Suppose you have a regular pentagon inscribed in a unit circle, and connect one vertex to each of the other four vertices. Then the product of the lengths of these four lines is 5.

More generally, suppose you have a regular *n*-gon inscribed in a unit circle. Connect one vertex to each of the others and multiply the lengths of each of these line segments. Then the product is *n*.

I ran across this theorem recently thumbing through Mathematical Diamonds and had a flashback. This was a homework problem in a complex variables class I took in college. The fact that it was a complex variables class gives a big hint at the solution: Put one of the vertices at 1 and then the rest are *n*th roots of 1. Connect all the roots to 1 and use algebra to show that the product of the lengths is *n*. This will be much easier than a geometric proof.

Let ω = exp(2π*i*/*n*). Then the roots of 1 are powers of ω. The products of the diagonals equals

|1 – ω| |1 – ω^{2}| |1 – ω^{3}| … |1 – ω^{n-1}|

You can change the absolute value signs to parentheses because the terms come in conjugate pairs. That is,

|1 – ω^{k}| |1 – ω^{n-k}| = (1 – ω^{k}) (1 – ω^{n-k}).

So the task is to prove

(1 – ω)(1 – ω^{2})(1 – ω^{3}) … (1 – ω^{n-1}) = *n*.

Since

(*z*^{n} – 1) = (*z* – 1)(*z*^{n-1} + *z*^{n-2} + … 1)

it follows that

(*z*^{n-1} + *z*^{n-2} + … 1) = (z – ω)(z – ω^{2}) … (z – ω^{n-1}).

The result follows from evaluating the expression above at *z* = 1.

**Related**: Applied complex analysis

Reminds me very much of the IB HL portfolio task for 2012-2013 (page 6). Cool stuff!

Amusing result. Maybe my morning coffee hadn’t kicked in… but I found the second the last step was a bit murky until I realized that what you were saying was that (z^n-1) must vanish at all the powers of omega, and thus given (z^n-1)=(z-1)(z^n-1 + … + 1), it follows that the sum must factor as stated. (If one worries about it, a few moments thought will also show that the constant factor must be one.)

GlennF: You’re right. That part could use more explanation.

Yes, I stumbled for a moment over that second to last step too. What a beautiful little gem of a proof!

Just a slight clarification : for n even, -1 is a root and does not come in a conjugate pair, but you can still remove the absolute value.

No need for any consideration of conjugate pairs. The product of the absolute values = the absolute value of the product, so just work out the product. It happens that it’s positive and real and hence is its own absolute value, but you don’t need to know ahead of time that it’s going to turn out that way.

Can you tell me why the length of the diagonal is given by |1-ω|?

Darvish: The distance between two complex numbers

aandbis |a–b|.Thanks for the quick reply :)

Just to clarify, |a-b| means the modulus of the complex number formed by subtracting complex number b from another complex number a?

Also, could you please elaborate on the second step, specifically why we can remove the modulus/absolute value signs and also the last step?

Thanks :)

Nevermind figured it out :)

John, surely if you say

“Since

(z^n – 1) = (z – 1)(z^n-1 + z^n-2 + … 1)

it follows that

(z^n-1 + z^n-2 + … 1) = (z – ω)(z – ω2) … (z – ωn-1).

The result follows from evaluating the expression above at z = 1.”

if you put z=1 into the expression you just prove that 0 = 0

since (z^n – 1) = 1 – 1 = 0

and (z – 1)(z^n-1 + z^n-2 + … 1) = (0)(z^n-1 + z^n-2 + … 1) = 0

Maybe my maths is a bit rusty after 50 years!

You and I are substituting z=1 into different equations. Stick z=1 into

(z^n-1 + z^n-2 + … 1) = (z – ω)(z – ω^2) … (z – ωn-1) and you get 1 + 1 + … + 1 = n on the left side and the product of the diagonal lenghts on the right.

The proof by complex numbers is elegant, but it does not help me to understand the result. Does anyone know a pure mathematical explanation of the beautyfull theses?

If by pure mathematical you mean pure geometrical, yes, you can prove the result geometrically, though the proof is much longer. I believe the book I cite gives a geometrical proof in a special case.

Hey people :) i just wanted to ask that if i would put z=1 and prove that 0=0 , is it a valid proof? Cause usually if you prove something you should not put values, but rather do it in general … so is there any other way of proving this conjecture? Would it be helpfull to further simplify the left hand side and then the right hand side or wouldn’t i get a solution??

Thanks for the answer in advance! :)

I have a question: how do you get from

z^n-1 /(z–1)

to

(z-w)(z-w^2 )(z-w^3 )…(z-w^(n-1))

@ Ralph:

Because

(z-1)(z-w)(z-w^2)…(z-w^(n-1))is an alternate factorization ofz^(n-1), since all the different values ofw^kare roots ofz^(n-1) – 1 = 0. Dividing by(z – 1)on both sides results in the last equation in the blog post.Essentially, this proof is setting two different (but equally valid) factorizations of

z^(n – 1)against each other, and plugging inz = 1.I have discussed your proof with a Maths professor and unfortunately him and me both agree that it is not valid, because of the impossibility of the substitution of z=1 in the last part of your proof.

The part:

(z^(n-1) + z^(n-2) + … 1) = (z – ω)(z – ω^2) … (z – ω^(n-1)).

Is only valid if z is not equal to 1, making the subsitution of z=1 into the equation impossible.

I wrote a paper recently with formulas connected to this identity and others. See

http://www.math.grin.edu/~chamberl/papers/trig_products.pdf

The substitution of z = 1 is correct because both sides of the equality are polynomials that are equal in all points except z = 1, therefore their coefficients are equal and the equality is satisfied in z = 1 as well.

This is a question in an old A level book that I have.