De Moivre: Fifth Root of Unity and the Seventh Root of Unity

Abraham de Moivre was able to solve Φ₅ and Φ₇ [see here for Definition of Cyclotomic Polynomial if needed] in order to express the primitive 5th roots of unity and the primitive 7th roots of unity as radicals.

In today's blog, I will show de Moivre's solutions.

Theorem 1: fifth root of unity can be solved by radicals.

Proof:

(1) Since 5 is prime, from a previous result (see Lemma 1, here), we have:

Φ₅ = x⁴ + x³ + x² + x + 1 = 0

(2) Dividing both sides by x², the gives us:

x² + x + 1 + x^-1 + x^-2 = 0

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

(3) Then:

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

(4) So,

y² + y - 1 = x² + x^-2 + x + x^-1 + 1

(5) Using the quadratic equation to solve for y² + y - 1 = 0 gives us (see Theorem, here if needed):

y = (-1 ± √1 + 4)/2 = (-1 ± √5)/2

(6) Since y = x + x^-1, we have:

x² - [(-1 ± √5)/2]x + 1 = 0

(7) Again, using the quadratic equation, we have:

x = { [(-1 ± √5)/2] ± √[(-1 ± √5)/2]² - 4}/2

(8) We can simplify this by noting that:

[(-1 ± √5)/2]² = [1 ± 2√5 + 5]/4 = [6 ± 2√5]/4

(9) So,

x = { [(-1 ± √5)/2] ± √[(-1 ± √5)/2]² - 4}/2 =

= [(-1 ± √5)/2 ± √(6 ± 2√5)/4 - 16/4]/2 =

= [(-1 ± √5)/2 ± √(-10 ± 2√5)/4]/2 =

= [(-1 ± √5) ± √(-10 ± 2√5)]/4

QED

Theorem 2: Seventh root of unity can be solved by radicals.

Proof:

(1) Since 7 is prime, from a previous result (see Lemma 1, here), we have:

Φ₇ = x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0

(2) Dividing both sides by x³, the gives us:

x³ + x² + x + 1 + x^-1 + x^-2 + x^-3 = 0

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

(3) Then:

y² = (x + x^-1)² = x² + 2xx^-1 + x^-2 = x² + 2 + x^-2

y³ = (x + x^-1)(x² + 2 + x^-2) = x³ + 2x + x^-1 + x + 2x^-1 + x^-3 = x³ + 3x + 3x^-1 + x^-3

(4) So,

y³ + y² - 2y - 1 = x³ + 3x + 3x^-1 + x^-3 + x² + 2 + x^-2 -2(x + x^-1) - 1 =

= x³ + x² + (3x - 2x) + (2 -1) + (3x^-1 - 2x^-1) + x^-2 + x^-3 =

= x³ + x² + x + 1 + x^-1 + x^-2 + x^-3

(5) So using the cubic equation (see Theorem, here), we can solve for y in:

y³ + y² - 2y - 1 = 0

to get:



where

p = (-2)/(1) - (1)²/3(1)² = -2 - (1/3) = -7/3

q = (-2(1)³)/(27(1)³) + (1)(-2)/(3(1)²) + 1/1 = -2/27 -2/3 + 1 = (-2 -18 + 27)/27 = 7/27

r = (1)/(3) = 1/3

(6) Once a solution for y is found, then it is possible to solve for x using the quadratic equation:

x² - [(-1 ± √5)/2]y + 1 = 0

QED

It would have to wait until Alexandre-Theophile Vandermonde to solve for the eleventh root of unity in terms of radicals. Later, Carl Friedrich Gauss would demonstrate that all roots of unity are expressible as radicals.

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001