## Saturday, January 12, 2008

### Vandermonde: Observations on the Eleventh Root of Unity

Alexandre-Theophile Vandermonde was able to solve the eleventh roots of unity in radicals by first observing relations between the roots of unity using trigonometric methods.

Lemma 1:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

then there exists y such that:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0

Proof:

(1) Let y = -(x + x-1) = -x + -x-1

(2) Then:

y2 = x2 + 2(-x)*(-x)-1 + x-2 = x2 + 2 + x-2

y3 = y*y2 = -x3 + -2x + -x-1 + -x + -2x-1 + -x-3 = -x3 + -3x + -3x-1 + -x-3

y4 = y*y3 = x4 + 3x2 + 3 + x-2 + x2 + 3 + 3x-2 + x-4 = x4 + 4x2 + 6 + 4x-2 + x-4

y5 = y*y4 = -x5 + -4x3 -6x - 4x-1 + -x-3 + -x3 + -4x + -6x-1 + -4x-3 + -x-5 =

= -x5 - 5x3 - 10x -10x-1 -5x-3 - x-5

(3) So:

y5 - y4 -4y3 + 3y2 + 3y - 1 =

= [-x5 - 5x3 - 10x -10x-1 -5x-3 - x-5] - [x4 + 4x2 + 6 + 4x-2 + x-4] - 4[-x3 + -3x + -3x-1 + -x-3] + 3[x2 + 2 + x-2] + 3[-x + -x-1] - 1 =

= -x5 - x4 + [-5x3 +4x3] + [-4x2 + 3x2] + [-10x +12x -3x] + [-6 + 6 - 1] + [-10x-1 + 12x-1 -3x-1] + [-4x-2 + 3x-2] + [-5x-3 + 4x-3] -x-4 -x-5 =

= -x5 -x4 -x3 -x2 -x - 1 -x-1 - x-2 -x-3 -x-4 -x-5

(4) We assume that:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

(5) We divide each side of the equation by -x5 to get:

-x5 -x4 -x3 -x2 -x - 1 -x-1 - x-2 -x-3 -x-4 -x-5 = 0

(6) Now, we apply step #3 to get:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0

QED

Lemma 2: The eleventh roots of unity using trigonometric methods

Using trigonometric methods, the eleventh roots of unity are:

cos 0 + isin 0
cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

Proof:

This follows directly from a previous result [See Corollary 1.1, here].

QED

Corollary 2.1:

The 10 solutions for:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

are:

cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

Proof:

(1) From Lemma 2 above, we can see that the solutions for:

x11 - 1 = 0

are:

cos 0 + isin 0
cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

(2) Using the Fundamental Theorem of Algebra (see Theorem, here), we can see that:

x11 - 1 = (x - cos0 - isin 0)(x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(3) Since, cos 0 - isin 0 = 1 - i*0 = 1, dividing both sides by (x - 1) gives us:

(x11-1)/(x - 1) = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(4) From a previous result (see Lemma 1, here):

(x11-1)/(x-1) = x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1

(5) So, combining step #3 and step #4 gives us:

(x11-1)/(x-1) = x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

QED

Lemma 3:

-2cos 12π/11 = -2cos(10π/11)

-2 cos 14π/11 = -2cos(8π/11)

-2 cos 16π/11 = -2cos(6π/11)

-2 cos 18π/11 = -2cos(4π/11)

-2cos 20π/11 = -2cos(2π/11)

Proof:

Using Property 5 here and the fact that 2π = 360 degrees [see Definition, here], we have:

-2cos 12π/11 = -2 cos (12π/11 - 2π) = -2cos([12π - 22π]/11) = -2cos(-10π/11) = -2cos(10π/11)

-2 cos 14π/11 = -2 cos (14π/11 - 2π) = -2cos([14π - 22π]/11) = -2cos(-8π/11) = -2cos(8π/11)

-2 cos 16π/11 = -2cos (16π/11 - 2π) = -2cos([16π - 22π]/11) = -2cos(-6π/11) = -2cos(6π/11)

-2 cos 18π/11 = -2cos (18π/11 - 2π) = -2cos([18π - 22π]/11) = -2cos(-4π/11) = -2cos(4π/11)

-2cos 20π/11 = -2cos(20π/11 - 2π) = -2cos([20π - 22π]/11) = -2cos(-2π/11) = -2cos(2π/11)

QED

Lemma 4

if:

x = cos u + isin u

then:

1/x = cos u - isin u

Proof:

(1) Let x = cos u + isin u

(2) 1/x = 1/[cos u + isin u] = [cos u - isinu]/[cos2 u - i2sin2 u] = [cos u - isin u]/[cos2 u + sin2 u] = cos u - isin u

QED

Corollary 4.1

if:

x = cos u + isin u

then:

-x + -x-1 = -2cos u

Proof:

(1) Let x = cos u + isin u

(2) Then x-1 = cosu - isin u

(3) x + x-1 = 2cos u

(4) -x + x-1 = -2cos u

QED

Corollary 4.2:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

and

y = -x + -x-1

then:

the possible values of y are:

-2cos 2π/11

-2cos 4π/11

-2cos 6π/11

-2cos 8π/11

-2cos 10π/11

Proof:

(1) Case 1: x = cos 2π/11 + isin 2π/11

-x + -x-1 = -2cos 2π/11

(2) Case 2: x = cos 4π/11 + isin 4π/11

-x + -x-1 = -2 cos 4π/11

(3) Case 3: x = cos 6π/11 + isin 6π/11

-x + -x-1 = -2 cos 6π/11

(4) Case 4: x = cos 8π/11 + isin 8π/11

-x + -x-1 = -2 cos 8π/11

(5) Case 5: x = cos 10π/11 + isin 10π/11

-x + -x-1 = -2cos 10π/11

(6) Case 6: x = cos 12π/11 + isin 12π/11

-x + -x-1 = -2cos 12π/11 = -2cos(10π/11) [See Lemma 3 above]

(7) Case 7: x = cos 14π/11 + isin 14π/11

-x + -x-1 = -2 cos 14π/11 = -2cos(8π/11) [See Lemma 3 above]

(8) Case 8: x = cos 16π/11 + isin 16π/11

-x + -x-1 = -2 cos 16π/11 = -2cos(6π/11) [See Lemma 3 above]

(9) Case 9: x = cos 18π/11 + isin 18π/11

-x + -x-1 = -2 cos 18π/11 = -2cos(4π/11) [See Lemma 3 above]

(10) Case 10: x = cos 20π/11 + isin 20π/11

-x + -x-1 = -2cos 20π/11 = -2cos(2π/11) [See Lemma 3 above]

QED

Lemma 5: 2cosαcosβ = cos(α + β) + cos(α - β)

Proof:

(1) cos(A+B) = cosA*cosB - sinA*sinB [See Theorem 2, here]

(2) Likewise:

cos (A - B) = cos (A + [-B]) = cosA*cos(-B) - sin(A)sin(-B) = cosAcosB + sinAsinB

(3) cos(A + B) + cos(A-B) = cosA*cosB - sinA*sinB + cosA*cosB + sinA*sinB = 2cosAcosB

QED

Corollary 5.1: 2cos2(α) = cos(2*α) + 1

Proof:

cos(2a) + 1 = cos(2a) + cos(0) = cos(a + a) + cos(a - a) = 2cos(α)cos(α) = 2cos2(α)

QED

Corollary 5.2:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

a2 = -b + 2
b2 = -d + 2
c2 = -e + 2
d2 = -c + 2
e2 = -a + 2

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) Using Corollary 5.1 above, we have:

2cos2(2π/11) = cos(4π/11) + 1

2cos2(4π/11) = cos(8π/11) + 1

2cos2(6π/11) = cos(12π/11) + 1

2cos2(8π/11) = cos(16π/11) + 1

2cos2(10π/11) = cos(20π/11) + 1

(3) Multiplying both sides by 2 gives us:

4cos2(2π/11) = 2cos(4π/11) + 2

4cos2(4π/11) = 2cos(8π/11) + 2

4cos2(6π/11) = 2cos(12π/11) + 2

4cos2(8π/11) = 2cos(16π/11) + 2

4cos2(10π/11) = 2cos(20π/11) + 2

(4) So that we have:

a2 = 4cos2(2π/11) = 2cos(4π/11) + 2 = -b + 2

b2 = 4cos2(4π/11) = 2cos(8π/11) + 2 = -d + 2

c2 = 4cos2(6π/11) = 2cos(12π/11) + 2 = 2cos(10π/11) + 2 = -e + 2

d2 = 4cos2(8π/11) = 2cos(16π/11) + 2 = 2cos(6π/11) + 2 = -c + 2

e2 = 4cos2(10π/11) = 2cos(20π/11) + 2 = 2cos(2π/11) + 2 = -a + 2

QED

Corollary 5.3:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

a2 = -b + 2

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: a2 = -b + 2 becomes b2 = -d + 2 { a --> b and b --> d } which becomes d2 = -c + 2 { b --> d and d -->c } which becomes c2 = -e + 2 { d --> c and c --> e } which becomes e2 = -a + 2 { c --> e and e --> a }]

Proof:

This follows directly from Corollary 6.2 above.

QED

Lemma 6:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

ab = -a -c
bc = -a -e
cd = -a -d
de = -a-b
ac = -b -d
bd = -b -e
ce = -b -c
be = -c -d
ae = -d -e

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) So, we have:

-a -b = (2cos 2π/11) + (2cos 4π/11)

-a -c = (2cos 2π/11) + (2cos 6π/11)

-a -d = (2cos 2π/11) + (2cos 8π/11)

-a -e = (2cos 2π/11) + (2cos 10π/11)

-b -c = (2cos 4π/11) + (2cos 6π/11)

-b -d = (2cos 4π/11) + (2cos 8π/11)

-b -e = (2cos 4π/11) + (2cos 10π/11)

-c -d = (2cos 6π/11) + (2cos 8π/11)

-c -e = (2cos 6π/11) + (2cos 10π/11)

-d -e = (2cos 8π/11) + (2cos 10π/11)

(3) Which gives us:

ab = (-2cos 2π/11)(-2cos 4π/11) = 4cos(2π/11)*cos(4π/11) = 2[2cos(2π/11)*cos(4π/11)] = 2[cos(4π/11 + 2π/11) + cos(4π/11 - 2π/11) = 2[cos(6π/11) + cos(2π/11)] = 2cos(6π/11) + 2cos(2π/11) = -a -c

ac = (-2cos 2π/11)(-2 cos 6π/11) = 4cos(2π/11)*cos(6π/11) = 2[2cos(2π/11)*cos(6π/11)] = 2[cos(6π/11 + 2π/11) + cos(6π/11 - 2π/11) = 2[cos(8π/11) + cos(4π/11)] = 2cos(8π/11) + 2cos(4π/11) = -b -d

ad = (-2cos 2π/11)(-2 cos 8π/11) = 4cos(2π/11)*cos(8π/11) = 2[2cos(2π/11)*cos(8π/11)] = 2[cos(8π/11 + 2π/11) + cos(8π/11 - 2π/11) = 2[cos(10π/11) + cos(6π/11)] = 2cos(10π/11) + 2cos(6π/11) = -c -e

ae = (-2cos 2π/11)(-2 cos 10π/11) = 4cos(2π/11)*cos(10π/11) = 2[2cos(2π/11)*cos(10π/11)] = 2[cos(10π/11 + 2π/11) + cos(10π/11 - 2π/11) = 2[cos(12π/11) + cos(8π/11)] = 2cos(12π/11) + 2cos(8π/11) = 2cos(10π/11) + 2cos(8π/11) = -d -e

bc = (-2cos 4π/11)( -2 cos 6π/11) = 4cos(4π/11)*cos(6π/11) = 2[2cos(4π/11)*cos(6π/11)] = 2[cos(6π/11 + 4π/11) + cos(6π/11 - 4π/11) = 2[cos(10π/11) + cos(2π/11)] = 2cos(10π/11) + 2cos(2π/11) = -a -e

bd = (-2cos 4π/11)(-2 cos 8π/11) = 4cos(4π/11)*cos(8π/11) = 2[2cos(4π/11)*cos(8π/11)] = 2[cos(8π/11 + 4π/11) + cos(8π/11 - 4π/11) = 2[cos(12π/11) + cos(4π/11)] = 2cos(12π/11) + 2cos(4π/11) = 2cos(10π/11) + 2cos(4π/11) = -b -e

be = (-2cos 4π/11)(-2 cos 10π/11) = 4cos(4π/11)*cos(10π/11) = 2[2cos(4π/11)*cos(10π/11)] = 2[cos(10π/11 + 4π/11) + cos(10π/11 - 4π/11) = 2[cos(14π/11) + cos(6π/11)] = 2cos(14π/11) + 2cos(6π/11) = 2cos(8π/11) + 2cos(6π/11) = -c -d

cd = ( -2 cos 6π/11)(-2 cos 8π/11) = 4cos(6π/11)*cos(8π/11) = 2[2cos(6π/11)*cos(8π/11)] = 2[cos(8π/11 + 6π/11) + cos(8π/11 - 6π/11) = 2[cos(14π/11) + cos(2π/11)] = 2cos(14π/11) + 2cos(2π/11) = 2cos(8π/11) + 2cos(2π/11) = -a -d

ce = (-2 cos 6π/11)(-2 cos 10π/11) = 4cos(6π/11)*cos(10π/11) = 2[2cos(6π/11)*cos(10π/11)] = 2[cos(10π/11 + 6π/11) + cos(10π/11 - 6π/11) = 2[cos(16π/11) + cos(4π/11)] = 2cos(16π/11) + 2cos(4π/11) = 2cos(6π/11) + 2cos(4π/11) = -b -c

de = (-2 cos 8π/11)(-2 cos 10π/11) = 4cos(8π/11)*cos(10π/11) = 2[2cos(8π/11)*cos(10π/11)] = 2[cos(10π/11 + 8π/11) + cos(10π/11 - 8π/11) = 2[cos(18π/11) + cos(2π/11)] = 2cos(18π/11) + 2cos(2π/11) = 2cos(4π/11) + 2cos(2π/11) = -a -b

QED

Corollary 6.1:

if:

x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = 0

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

ab = -a -c
ac = -b -d

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: ab = -a - c becomes bd = -b - e (a --> b and b--> d and c --> e) which becomes cd = -d -a ( b --> d and d --> c and e --> a) which becomes ce = -c -b ( c --> e and d --> c and a --> b) which becomes ae = -e -d etc.]

Proof:

This follows directly from Lemma 7 above.

QED

References
• Jean-Pierre Tignol, , World Scientific, 2001