The 5th Roots of Unity¶

Let $\zeta=\exp(2\pi i/5)$. The primitive fifth roots satisfy

$$ \Phi_5(x)=x^4+x^3+x^2+x+1=0. $$

In particular,

$$ \cos\frac{2\pi}{5}=\frac{\sqrt5-1}{4}. $$

We recover this radical expression by organizing powers of $\zeta$ into orbits rather than asking a black-box polynomial solver for four roots at once.

Pair conjugate powers¶

Complex conjugation pairs $\zeta$ with $\zeta^4$ and $\zeta^2$ with $\zeta^3$. Their symmetric sums are real. Adding the two pairs gives $-1$, the sum of all primitive fifth roots.

def z : MathValue := rtu 5

def a11 : MathValue := z ^ 1 + z ^ 4
def a12 : MathValue := z ^ 2 + z ^ 3

def b10 : MathValue := a11 + a12
def b11 : MathValue := a11 - a12
def b12 : MathValue := a12 - a11

(b10, b11, b12)

$(-1, -2 rtu(5)^{3} - 2 rtu(5)^{2} - 1, 2 rtu(5)^{3} + 2 rtu(5)^{2} + 1)$

Invert the first two-point transform¶

The sum is already $b_{10}=-1$. Squaring the difference removes its sign ambiguity; choosing a square-root branch and applying the inverse transform reconstructs the two real periods.

def b10' : MathValue := b10
def b11' : MathValue := sqrt (b11 ^ 2)

def a11' : MathValue := (b10' + b11') / 2
def a12' : MathValue := (b10' - b11') / 2

(a11', a12')

$(\frac{1}{2} \sqrt{5} + \frac{-1}{2}, \frac{-1}{2} \sqrt{5} + \frac{-1}{2})$

Recover the imaginary parts¶

The antisymmetric pairs $\zeta-\zeta^{-1}$ and $\zeta^2-\zeta^{-2}$ carry the imaginary parts. A second two-point transform reduces them to square roots whose radicands depend only on the real periods above.

def a21 : MathValue := z ^ 1 - z ^ 4
def a22 : MathValue := z ^ 2 - z ^ 3

def b20 : MathValue := a21 + a22
def b21 : MathValue := a21 - a22
def b22 : MathValue := a22 - a21

def b20' : MathValue := sqrt ((-3) + 4 * a12')
def b21' : MathValue := sqrt ((-3) + 4 * a11')

def a21' : MathValue := (b20' + b21') / 2
def a22' : MathValue := (b20' - b21') / 2

def z1' : MathValue := (a11' + a21') / 2

z1'

$\frac{1}{4} \sqrt{2 \sqrt{5} + 5} i + \frac{1}{4} \sqrt{-2 \sqrt{5} + 5} i + \frac{1}{4} \sqrt{5} + \frac{-1}{4}$

Verify the radical relation¶

Nested square roots introduce branch atoms that are algebraically related. The following ideal records their defining relations. Reducing $(z_1')^5-1$ modulo that ideal gives an exact symbolic check, rather than a floating-point approximation.

def radicalRels : [MathValue] :=
  [ '((sqrt (5 + 2 * sqrt 5)) ^ 2 - 5 - 2 * sqrt 5)
  , '((sqrt (5 - 2 * sqrt 5)) ^ 2 - 5 + 2 * sqrt 5)
  , '((sqrt (5 + 2 * sqrt 5))
        * (sqrt (5 - 2 * sqrt 5))
        - sqrt 5)
  , '((sqrt 5) ^ 2 - 5)
  , '(i ^ 2 + 1) ]

idealNF radicalRels (z1' ^ 5 - 1)

$0$

Takeaway¶

The familiar golden-ratio square root appears because the Galois group of $\Phi_5$ can be resolved by two successive two-point transforms. Egison keeps the orbit sums and the exact radical verification in the same symbolic calculation.