## Thursday, January 24, 2008

In the previous blog, I showed an algorithm for constructing the heptadecagon (a regular 17-sided figure). In today's blog, I will show a proof that based on the Gauss's solution for the seventeenth roots of unity in terms of radicals, the algorithm in the previous blog really works.

Lemma 1:

if α = 2π/17

and

ζk = cos kα + isin k α

then

ζk + ζ17 - k = 2 cos kα

Proof:

(1) Using De Moivre's Formula for roots of unity (see Corollary 1.1, here), we note that:

if α = 2π/17, then:

ζ = cos α + isinα is a seventeenth root of unity.

(2) Using De Moivre's Formula (see Theorem 1, here), we note that:

ζk = (cos α + isin α)k = cos kα + isin k α

ζ17-k = (cos α + isin α)17-k = cos ([17 - k]α) + isin([17 -k]α) = cos ([17 - k]2π/17) + isin([17 - k]2π/17) = cos(2π - kα) + isin(2π - kα) = cos(-kα) + isin(-kα) = cos(kα) - isin(kα)

(3) So that we have:

ζk + ζ17-k = cos kα + isin kα + cos kα - isink α = 2cos kα

QED

Lemma 2:

If C is the least positive acute angle such that tan 4C = 4

with

α = 2π/17

ζk = cos kα + isin k α

and:

x1 = ζ + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2 x2 = ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6

then:

x1 = 2 tan 2C x2 = -2cot 2C

Proof:

(1) Using step #1 thru step #7 in Theorem 1, here, we have:

x1 and x2 are the solutions to:

x2 + x - 4 = 0

(2) Since tan 4C = 4, it follows that cot 4C = 1/(tan 4C) = 1/4

(3) So 4xcot 4C = 4x(1/4) = x
(4) And we have:

x2 + 4xcot 4C - 4 = 0
(5) Using the quadratic equation (see Theorem, here), we have:

x = (1/2)[-4cot4c ± √(4cot 4c)2 + 16]

(6) Using Lemma 1, here, we have:

x = (1/2)[-4(1/2)(cot 2c - tan2c) ± √16(1/4)(cot 2c - tan 2c)2 + 16) ]= = (tan 2c - cot2c) ± (1/2)√4[cot2(2c) - 2cot(2c)tan(2c) + tan2(2c)] + 16 = = (tan 2c - cot 2c) ± √cot2(2c) - 2 + tan2(2c) + 4 = = (tan 2c - cot 2c) ± √cot2(2c) + tan2(2c) + 2 = = (tan 2c - cot 2c) ± √[cot(2c) + tan(2c)]2 = = (tan 2c - cot 2c) ± [cot(2c) + tan(2c)]

(7) Since x1 is greater than x2, we have:

x1 = 2 tan 2c x2 = -2 cot 2c

QED

Lemma 4:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17

ζk = cos kα + isin k α

and:

y1 = ζ1 + ζ13 + ζ16 + ζ4

y2 = ζ9 + ζ15 + ζ8 + ζ2

y3 = ζ3 + ζ5 + ζ14 + ζ12

y4 = ζ10 + ζ11 + ζ7 + ζ6

then:

y1 = tan(C + π/4) y2 = tan(C - π/4) y3 = tan C y4 = -cot C

Proof:

(1) From steps #9 thru #13 in Theorem 1, here, we have:

y1, y2 are solutions to:

y2 - x1y - 1 = 0

and

y3, y4 are solutions to:

y2 - x2y - 1 = 0

(2) Let's solve first for y1, y2

(3) Using the quadratic equation (see Theorem 1, here), we get:
y = (1/2)(x1 ± √x12 + 4)

(4) From Lemma 2 above, we know that x1 = 2 tan 2c

(5) So that we have:

y = (1/2)(2 tan 2c ± √4(tan 2c)2 + 4 = = tan 2c ± √(tan 2c)2 + 1

(6) Using Lemma 4, here, we have:

tan 2c ± √(tan 2c)2 + 1 = (2 tan c)/(1 - tan2c) ± √(4 tan2 c)/(1 - 2tan2c + tan4 c) + 1 = = (2 tan c)/(1 - tan2c) ± √(4 tan2 c + 1 - 2tan2c + tan4 c )/(1 - 2tan2c + tan4 c) = = (2 tan c)/(1 - tan2c) ± √(2tan2 c + 1 + tan4 c )/(1 - 2tan2c + tan4 c) = = (2 tan c)/(1 - tan2c) ± √(1 + tan2 c )2/(1 - tan2c)2 = = (2 tan c)/(1 - tan2c) ± (1 + tan2c)/(1 - tan2c) = = (2 tan c ± [1 + tan2c])/(1 - tan2c)
(7) This then gives us:

(2 tan c + 1 + tan2c)/(1 - tan2c) = = (tan c + 1)(tan c + 1)/(1 + tan c)(1 - tan c) = (tan c + 1)/(1 - tan c)

and

(2 tan c - 1 - tan2c)/(1 - tan2c) = (tan2c - 2tan c + 1)/(tan2c - 1) = = (tan c - 1)(tan c - 1)/(tan c - 1)(tan c + 1) = (tan c - 1)/(tan c + 1)
(8) Since tan(π/4) = 1 (see Lemma 1, here), we can restate this as:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) (tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4))

(9) Using Lemma 3, here, we have:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) = tan(c + π/4)

(10) Using Corollary 3.1, here, we have:

(tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4)) = tan(c - π/4)

(11) Since y1 is greater than y2,

y1 = tan(c + π/4) y2 = tan(c - π/4)

(11) Now, let's solve for y3, y4

(12) Using the quadratic equation (see Theorem 1, here), we get:

y = (1/2)(x2 ± √x22 + 4)

(13) From Lemma 2 above, we know that x2 = -2 cot 2c

(14) So that we have:

y = (1/2)(-2 cot 2c ± √4(cot 2c)2 + 4 = = -cot 2c ± √(cot 2c)2 + 1

(15) Using Lemma 1, here, we have:

-cot 2c ± √(cot 2c)2 + 1 = -(1/2)(cot c - tan c) ± √(1/4)(cot c - tan c)2 + 1 = = (1/2)(tan c - cot c) ± √(1/4)(cot2c + tan2c - 2(cot c)(tan c)) + (4/4) = = (1/2)(tan c - cot c) ± √(1/4)(cot2c + tan2c - 2 + 4) = = (1/2)(tan c - cot c) ± √(1/4)(cotc + tanc)2 = = (1/2)(tan c - cot c) ± (1/2)(tan c + cot c)
(16) Since y3 is greater than y4, this gives us:

y3 = tan c y4 = -cot c

QED
Corollary 4.1:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17 tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α

Proof:

(1) Let us define the following:

y3 = ζ3 + ζ5 + ζ14 + ζ12

and

y2 = ζ9 + ζ15 + ζ8 + ζ2

where

ζk = cos kα + isin k α

(2) Using Lemma 1 above:

y3 =(ζ3 + ζ14) + (ζ5 + ζ12) = 2cos 3α + 2cos 5α

y2 = 8 + ζ9) + 2 + ζ15) = 2 cos 8α + 2 cos 2 α

(3) Using Lemma 5, here, we have:

2 cos(3α) * 2cos(5α) = 2*[2 cos(3α)*cos(5α)] = 2*[cos(5 + 3)α + cos(5 - 3)α] = 2[cos(8α) + cos(2α)] = 2cos(8α) + 2cos(2α)

(4) From Lemma 4 above, we then have:

tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α = 2cos(3α)*2cos(5α)
QED

Theorem 5: Constructibility of Heptadecagon using compass and ruler

Proof:

(1) Let CD, CP0 two radii of a circle C1 that are perpendicular to each other.
(2) Let CG be 1/4 the length of CD (bisector of a bisector).
(3) Let ∠ CGL be 1/4 ∠ CGP0 with L on CP0 (bisector of a bisector).
(4) Let ∠ LGP = π/4 (bisector a right angle since right angle = π/2, see here for review of radians) where P is on CP0.
(5) Let Q be the midpoint of PP0 and draw a circle C2 with center Q and radius QP0.
(6) Let R be the point where the circle C1 intersects with CD.
(7) Draw a circle C3 with center L and radius equal to RL.
(8) Let S,Q be the points where the circle C3 intersects with CP0.
(9) Let SP5, QP3 be lines parallel to CP0 that intersect with circle C1.
(10) Let C = measurement ∠ CGL
(11) Then 4C = measurement ∠ CGP0
(12) cos P0CP3 = CQ/CP3 = CQ/CP0 [See here for definition of cosine]
(13) cos P0CP5 = cos (π/2 + π/2 - SCP5) = cos (π - SCP5)
(14) Using cos(a + b) = (cos a)(cos b) - (sin a)(sin b) (see Theorem 2, here) and cos(-x) = cos(x) (see Property 9, here), we have:

cos(π - SCP5) = (cos π)(cos -SCP5) - (sin π)(sin -SCP5) = (-1)cos(SCP5) - (0)(-SCP5) = -cos(SCP5)

(15) So, cos P0CP5 = -cos(SCP5) = -SC/CP5 = -SC/CP0

(16) This gives us that:

2cos P0CP3 + 2cos P0CP5 = 2(CQ - SC)/CP0

(17) Since CQ = CL + LQ and SC = SL - CL and SL=LQ:

SC = SL - CL = LQ - CL

CQ - SC = (CL + LQ) - (LQ - CL) = CL + LQ - LQ + CL = 2CL

(18) This gives us that:

2(CQ - SC)/CP0 = 2(2CL)/CP0 = 4CL/CP0

(19) Since CG is 1/4 of CD and CD=CP0, we have:

4CL/CP0 = 4CL/CD = 4CL/(4CG) = CL/CG

(20) Since GCL is a right angle and C = measurement ∠ CGL, we have:

tan CGL = tan c = sin CGL/cos CGL = (LC/GL)/(GC/GL) = LC/GC = 2cos P0CP3 + 2cos P0CP5

(21) Since cos P0CP3 =CQ/CP0 and cos P0CP5 = -SC/CP0, we have:

2 cos(P0CP3)*2cos(P0CP5) = -4*(SC)(CQ)/(CP0)(CP0)

(22) Using the Pythagorean Theorem (see Theorem, here):

CR2 = CL2 + RL2 = (CL - RL)(CL + RL)

(23) Since LQ=LR and LS=RL:

CL + RL = CL + LQ = CQ

CL - RL = CL - LS = SC

(24) So we have:

CR2 = (CL - LR)(CL + LR) = (CQ)(SC)

(25) Using the Pythagorean Theorem:

(RC)2 = RQ2 - CQ2 = (RQ - CQ)(RQ + CQ)

(26) Now RQ = QP and RQ=QP0 so that:

(RQ - CQ) = (QP - CQ) = PC

(RQ + CQ) = (QP1 + CQ) = CP1

(27) This gives us that (RC)2 = (PC)(CP0) so that:

-4(RC)2/(CP0)2 = -4(PC)(CP0)/(CP0)(CP0) = -4(PC)/(CP0)

(28) Since CG = (1/4)CD = (1/4)CP0, we have:

-4(PC)/(CP0) = -4(PC)/(4CG) = -PC/CG

(29) Also:

tan ∠ CGP = sin ∠ CGP/cos ∠ CGP = (PC/GP)/(CG/GP) = PC/GC

(30) Since ∠ CGP = ∠ LGP - ∠ CGL, we have:

tan ∠ CGP = tan (∠ LGP - ∠ CGL) = tan(π/4 - C)

and this means:

-PC/GC = -tan(π/4 - C) = tan(C - π/4)

(31) Using Corollary 4.1 above, we have:

2cos ∠ P0CP3 + 2cos ∠ P0CP5 = 2 cos 3α + 2 cos 5α = tan c where α = 2π/17

2 cos(∠P0CP3)*2cos(∠P0CP5) = 2cos(3α)*2cos(5α) = tan (c - π/4)

(32) Now, it is clear that ∠ P0CP3 and ∠ P0CP5 map onto and .

(33) Since ∠ P0CP5 is greater than ∠P0CP3, it is clear that:

∠ P0CP3 = 3α

∠ P0CP5 = 5α

QED

References

## Wednesday, January 23, 2008

### Gauss: Construction of the Heptadecagon

Carl Friedrich Gauss not only solved the seventeenth root of unity in terms of radicals but also realized that his solution indicated that a heptadecagon (a regular, seventeen-sided polygon) could be constructed by compass and ruler.

In my previous blog, I showed Gauss's solution of the seventeenth root of unity. In today's blog, I will show how this solution provides a recipe for constructing a heptadecagon.

For those ready to cut to the chase, a YouTube video of construction of the heptadecagon can be found here.

The algorithm below is a bit more complicated than the above YouTube video. It is taken from Hardy and Wright's Introduction to the Theory of Numbers (see the reference below).

Algorithm: Construction of a heptadecagon using compass and ruler

(1) Draw a line connecting point A and P1
(2) Let C be the midpoint
(3) Draw a circle with C as the center and the radius equal to CP0
(4) Let DE be a line perpendicular to CP0 which passes through C.
(5) Let F be the midpoint on CD and G be be the midpoint on CF.
(6) Draw a line connecting G and P0.
(7) Find a point H on GE where GH = GP0
(8) Draw an arc with center G and radius GP0 from P0 to H.
(9) Draw a line connecting P0 to H
(10) Let I be the midpoint on HP0.
(11) Draw a line passing from G through point I onto the arc and label the point on the arc J.
(12) Draw a line connecting J and H.
(13) Let K be the midpoint on JH.
(14) Draw a line from G to K and let L be the point where GK intersects with CP0.
(15) Draw a line from point G that is perpendicular to GL and label the point N where this line intersects with AC.
(16) Bisect the angle ∠ NGL and label P the point where this angle bisector intersects with AC.
(17) Let Q be the midpoint between P and P0.
(18) Draw a circle with center Q and radius QP0.
(19) Label R the point where this circle intersects with CD
(20) Draw a circle with center L and radius LR
(21) Label S the point where this circle intersects with AC.
(22) Draw a line from S that is perpendicular to AC and label P5 the point where this line intersects with the top of the circle.
(23) Draw a line from Q that is perpendicular to CP1 and label P3 the point where this line intersects with the top of the circle.
(24) Draw a line connecting P3 and P5 and label the midpoint W.
(25) Draw a line from C to W and label the point P4 where CW intersects with the top of the circle.
(26) Now, all the rest of the points can be completed in the following way:
(a) Draw a circle with center P3 and radius P3P4.
(b) Label P2 where this circle intersects with the larger circle.
(27) Repeat this same step for all missing remaining points.

In my next blog, I will show the proof that the above construction really works. This proof is based on the solution for the seventeenth root of unity.

References

## Sunday, January 20, 2008

### Gauss: Seventeenth Root of Unity Expressed As Radicals

Carl Friedrich Gauss was 18 when he was able to solve the seventeenth root of unity in terms of radicals. He not only made this discovery but also realized that this implied that it would be possible to construct a seventeen-sided polygon using only compass and ruler. Gauss later generalized these results on two levels:

(1) He proved that all roots of unity are expressible as radicals.
(2) A regular polygon is constructible by ruler and compass if and only if its number of sides is a product of distinct Fermat primes Fn (where Fn = 22n + 1) and a power of 2.

In today's blog, I will show Gauss's solution for the seventeenth root of unity. In the next blog, I will prove that it is possible to construct a seventeen sided polygon known as the heptadecagon using only compass and ruler.

Gauss was so fascinated by the construction of the heptadecagon that he decided that he would dedicate his life to mathematics instead of philology.

Theorem 1: The seventeen root of unity is expressible as radicals

Proof:

(1) Let ζ be a primitive seventeen-root of unity.

(2) Then, the sixteen roots of unity not equal to 1 are:

ζ1, ζ2, ζ3, ζ4, ζ5, ζ6, ζ7, ζ8, ζ9, ζ10, ζ11, ζ12, ζ13, ζ14, ζ15, ζ16

(3) We can divide the above roots into the following sums:

x1 = ζ1 + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2

x2 = ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6

(4) Now, since (see Lemma 1, here):

1 + ζ + ζ2 + ... + ζ16 = 0

We have:

x1 + x2 + 1 = 0

So that:

x1 + x2 = -1

(5) x1 * x2 = ( ζ1 + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2)(ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) =

= ζ13 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ93 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ133 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ153 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ163 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ83 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ43 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) + ζ23 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6) =

= (ζ4 + ζ11 + ζ6 + ζ12 + ζ15 + ζ8 + ζ13 + ζ7) + (ζ12 + ζ2 + ζ14 + ζ3 + ζ6 + ζ16 + ζ4 + ζ15) +
16 + ζ6 + ζ + ζ7 + ζ10 + ζ3 + ζ8 + ζ2) + (ζ + ζ8 + ζ3 + ζ9 + ζ12 + ζ5 + ζ10 + ζ4) +
2 + ζ9 + ζ4 + ζ10 + ζ13 + ζ6 + ζ11 + ζ5) + (ζ11 + ζ + ζ13 + ζ2 + ζ5 + ζ15 + ζ3 + ζ14) +
7 + ζ14 + ζ9 + ζ15 + ζ + ζ11 + ζ16 + ζ10) + (ζ5 + ζ12 + ζ7 + ζ13 + ζ16 + ζ9 + ζ14 + ζ8) =

= 4[ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16] = 4[x1 + x2] = -4

(6) We can now use the two equations to form a quadratic equation:

x2 = -1 -x1

x1*x2 = -4

x1(-1 -x1) = -4

which gives us:

-x12 - x1 + 4 = 0

or if we multiply each side by -1

x12 + x1 - 4 = 0

(7) Now, it should also be clear that since each equation is symmetric, we could easily have said that:

x1 = -1 - x2

so it is clear that we also have:

x22 + x2 - 4 = 0

(8) This means that x1, x2 are found using the quadratic equation (see Theorem, here) to get:

x = (-1 ±√1 - (-16))/2 =

= (-1 ±√17)/2

So, we that we have:

x1 = (-1 +√17)/2

x2 = (-1 -√17)/2

(9) Now, let's define the following values:

y1 = ζ1 + ζ13 + ζ16 + ζ4

y2 = ζ9 + ζ15 + ζ8 + ζ2

y3 = ζ3 + ζ5 + ζ14 + ζ12

y4 = ζ10 + ζ11 + ζ7 + ζ6

(10) Now, it is clear that:

y1 + y2 = x1

and

y3 + y4 = x2

(11) Further, we have:

y1y2 = (ζ1 + ζ13 + ζ16 + ζ4)(ζ9 + ζ15 + ζ8 + ζ2) =

= ζ19 + ζ15 + ζ8 + ζ2) + ζ139 + ζ15 + ζ8 + ζ2) +
ζ16 9 + ζ15 + ζ8 + ζ2) + ζ49 + ζ15 + ζ8 + ζ2) =

= (ζ10 + ζ16 + ζ9 + ζ3) + (ζ5 + ζ11 + ζ4 + ζ15) +
8 + ζ14 + ζ7 + ζ) + (ζ13 + ζ2 + ζ12 + ζ6) =

= ζ + ζ2 + ζ3 + ζ4 + ζ56 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = x1 + x2 = -1

y3y4 = (ζ3 + ζ5 + ζ14 + ζ12)(ζ10 + ζ11 + ζ7 + ζ6) =

= ζ310 + ζ11 + ζ7 + ζ6) + ζ510 + ζ11 + ζ7 + ζ6) +
ζ1410 + ζ11 + ζ7 + ζ6) + ζ1210 + ζ11 + ζ7 + ζ6) =

= (ζ13 + ζ14 + ζ10 + ζ9) + (ζ15 + ζ16 + ζ12 + ζ11) +
7 + ζ8 + ζ4 + ζ3) + (ζ5 + ζ6 + ζ2 + ζ) =

= ζ +ζ23 + ζ4 + ζ56 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = -1

(12) So, this gives us:

y2 = x1 - y1
y4 = x2 - y3

y1(x1 - y1) = -1
y3(x2 - y3) =-1

So that:

x1y1 - y12 + 1 = 0
x2y3 - y32 + 1 = 0

(13) So that we can find the value for y1, y2 by solving:

y2 - (x1)y - 1 = 0

So that y = [(x1) ± √(x12 + 4)]/2

So we assign:

y1 = [(x1) + √(x12 + 4)]/2

y2 = [(x1) - √(x12 + 4)]/2

(14) We can find the value for y3, y4 by solving:

Y2 - (x2)Y - 1 = 0

So that Y = [(x2) ± √(x22 + 4)]/2

So we assign:

y3 = [(x2) + √(x22 + 4)]/2

y4 = [(x2) - √(x22 + 4)]/2

(15) Now, let's define the following values:

z1 = ζ + ζ16

z2 = ζ13 + ζ4

z3 = ζ8 + ζ9

z4 = ζ2 + ζ15

z5 = ζ3 + ζ14

z6 = ζ5 + ζ12

z7 = ζ7 + ζ10

z8 = ζ6 + ζ11

(16) We note that:

z1 + z2 = y1

z3 + z4 = y2

z5 + z6 = y3

z7 + z8 = y4

(17) If we multiply combinations together, we get:

z1*z2 = (ζ + ζ16)(ζ13 + ζ4) = ζ14 + ζ5 + ζ12 + ζ3 = y3

z3*z4 = (ζ8 + ζ9)( ζ2 + ζ15) = ζ10 + ζ6 + ζ11 + ζ7 = y4

z5*z6 = (ζ3 + ζ14)(ζ5 + ζ12) = ζ8 + ζ15 + ζ2 + ζ9 = y2

z7*z8 = (ζ7 + ζ10)(ζ6 + ζ11) = ζ13 + ζ + ζ16 + ζ4 = y1

(18) So, this gives us:

z2 = y1 - z1

z4 = y2 - z3

z6 = y3 - z5

z8 = y4 - z7

which means:

z1(y1 - z1) = y3

z3(y2 - z3) = y4

z5(y3 - z5) = y2

z7( y4 - z7 ) = y1

(19) So, we find z1, z2 by solving for:

z12 - z1y1 + y3 = 0

so that:

z1 = [y1 + √(y1)2 - 4y3)]/2

z2 = [y1 - √(y1)2 - 4y3)]/2

(20) Finally, we can now solve for the seventeen root of unity since:

ζ16 = z1 - ζ [From step #15 above]

ζ(z1 - ζ) = 1 [This follows since ζ17 = 1]

z1ζ - ζ2 - 1 = 0

ζ2 - z1ζ + 1 = 0 [Multiplying both sides by -1]

ζ = (z1 + √(z1)2 - 4)/2 [Using the solution to the quadratic equation, see here]

Working all this out gives us:

ζ =

References