a roots of unity geometry question (1 Viewer)

[spice girl]

How about this one:

Let A, B, C, ..., G represent 1, w, w^2, ..., w^6 where w = cis(2pi/7). Let H represent -1

Prove HA*HB*...*HG = 2

there's several answers to this one. I found the good solution only after I did the HSC. it involves polynomials i think...

 

[OLDMAN]

Shall let students try this one out. Already challenging, could be made more interesting if the question assumes a more general 2n+1 regular polygon rather than just a 7-sided one. Result is quite interesting. trigonometrically speaking.

 

[spice girl]

What's the trigonometric result from this question?

 

[OLDMAN]

2^n(1+cos(n2pi/(2n+1))(1+cos((n-1)2pi/(2n+1)...(1+cos2pi(n+1))=1 for any positive integer n. Hope its right.
Would be interested in a pure trigonometric proof, as this was borne out by the question using complex/polynomial approach. Maybe impossible.

 

[ND]

Let w^7 -1 = 0
(w - 1)(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) = 0
when w =/ (is tha what is used for 'not equal to'?) 1
(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) = 0 ...[1]

Now HA = 2
HB = (w + 1)
.
.
HG = (w^6 + 1)

so HA*HB*...*HG = 2(w +1)(w^2 + 1)...(w^6 + 1)
= 2[3w^10 + 5w^9 + 3w^8 + 2w^7 + 9w^6 + 9w^5 + 9w^4 + 6w^3 + 4w^2 + 6w + 8]
Now
w^7 = 1
w^8 = w
.
.
w^11 = w^4
so
HA*HB*...*HG = 2[9w^6 + 9w^5 + 4w^4 + 9w^3 + 9w^2 + 9w + 10)
= 2[9(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) + 1]
= 2[9*0 + 1] from [1]
= 2

I know there should be AV signs in there but i couldn't be bothered...

edit: i just realized that i converted some of the higher power roots to their low range equivalent during some of the lines which i didn't write, but there are too many lines to type.

 

[ND]

Another solution (i think):

Hx = x + 1

1, w, w^2, ... w^6 are the roots of equation:

x^7 - 1 = 0

This means that 1 + 1, w + 1, w^2 + 1, ... w^6 + 1 are roots of equation:

(x - 1)^7 - 1 = 0
(x^7 - 7x^6 + 21x^5 - 35x^4 + 35x^3 - 21x^2 + 7x - 1) - 1 = 0
x^7 - 7x^6 + 21x^5 - 35x^4 + 35x^3 - 21x^2 + 7x - 2 = 0

Now HA*HB*...*HG = product of roots of this equation
= 2

 

[spice girl]

Ahh...nice

that's heaps similar to the solution i was thinking of.

But here's wot I was thinking (it sorta is the same).

1, w, w^2, ..., w^6 are the roots of the polynomial P(z) = z^7 - 1

(-1-1)(-1-w)(-1-w^2)...(-1-w^6) = P(-1) = -2

thus (1+1)(1+w)(1+w^2)...(1+w^6) = 2

 

[OLDMAN]

Ahh...nicer! But must also congratulate ND for the way he muscles in towards a solution.

Just a (generalized) variation on the same problem:

Let A_1,A_2,...A_2n+1 represent 1,w,w^2,...w^2n where
w=cis(2pi/(2n+1)). Let H represent -1.

(i) Prove HA_1.HA_2...HA_2n+1=2.
(ii) Hence prove
2^n[(1+cos(n2pi/(2n+1))(1+cos((n-1)2pi/(2n+1))...(1+cos(1)2pi/(2n+1))]=1

 

[ND]

Originally posted by spice girl
Ahh...nice

that's heaps similar to the solution i was thinking of.

But here's wot I was thinking (it sorta is the same).

1, w, w^2, ..., w^6 are the roots of the polynomial P(z) = z^7 - 1

(-1-1)(-1-w)(-1-w^2)...(-1-w^6) = P(-1) = -2

thus (1+1)(1+w)(1+w^2)...(1+w^6) = 2

ooo, nice.

 

[ND]

Originally posted by OLDMAN
Ahh...nicer! But must also congratulate ND for the way he muscles in towards a solution.

Just a (generalized) variation on the same problem:

Let A_1,A_2,...A_2n+1 represent 1,w,w^2,...w^2n where
w=cis(2pi/(2n+1)). Let H represent -1.

(i) Prove HA_1.HA_2...HA_2n+1=2.
(ii) Hence prove
2^n[(1+cos(n2pi/(2n+1))(1+cos((n-1)2pi/(2n+1))...(1+cos(1)2pi/(2n+1))]=1

Hmmm... part (i) could be done by spice girl's method, but that wouldn't really do anything for part (ii)... I don't know...

PS, where do you guys get these types of questions from? I've looked in textbooks but they don't seem to have ones like these. Of course there is q7/q8 of past papers, but from where else can they be obtained?

 

[spice girl]

I think (ii) has something to do with z + z(bar) = 2Re(z)

Remembering also that sum of conjugates = conjugate of sum, and that if z = cisx then z(bar) = cis(-x)

 

[OLDMAN]

ND: "where do you get these types of questions?"

I'll answer this later and it might help you in your quest for the holy grail of elegance.

Meanwhile following is a laundry list of complex number "truths" which SpiceGirl started off with z+zbar=re(z) :
1) z^n-1=0 : if n is odd, roots are 1 and exact number of pairs of conjugates, eg. z^3-1=0, roots are 1 and cis(2pi/3),cis(2pi/3)bar.
2) if n is even, roots are 1, -1 and exact number of pairs of conjugates eg. z^4-1=0, roots are 1,-1,cis(2pi/4),cis(2pi/4)bar.
3) conjugate of rcis(@) could be written as rcis(@)bar, rcis(-@) or r(cos@-isin(@)
4) the product of a pair of conjugates is real, z.zbar=r^2
5) if you add a real constant to each of a pair of conjugates, the resulting pair will also be a pair of conjugates eg. (-1-cis(@)) and (-1-cis(@)bar) will also be conjugates of each other -hence their products will also be real.

 

[ND]

Well this q is out of my league. It doesn't look like anyone is having a go, so how about posting the soln? i am interested in seeing it.

 

[OLDMAN]

Using the result for Part (i) HA_1.HA_2...HA_2n+1 =2

(cis((1)2pi/(2n+1)+1)(cis((2)2pi/(2n+1)...(cis((2n+1)2pi/(2n+1)+1)=2

Equation z^2n+1-1=0 has 2n+1 roots : 1 and n pairs of conjugate roots; note all complex roots polygon are symmetrical on the "x axis".

above expression could be paired up :

(cis((1)2pi/(2n+1)+1)(cis((-1)2pi/(2n+1)+1)(cis((2)2pi/(2n+1)+1)(cis((-2)2pi/(2n+1)+1).....(cis((n)2pi/(2n+1)+1)(cis((-n)2pi/(2n+1)*2=2

(since the last one (cis((2n+1)2pi/(2n+1)+1)=2)

multiplying out the each pair gets the desired result.

 

[ND]

Ah i see.