MHB Finding the Product of Distinct Roots: A Complex Challenge

[lfdahl]

Let $r_1,r_2, …,r_7$ be the distinct roots (one real and six complex) of the equation $x^7-7= 0$.

Let \[p = (r_1+r_2)(r_1+r_3)…(r_1+r_7)(r_2+r_3)(r_2+r_4)…(r_2+r_7)…(r_6+r_7) = \prod_{1\leq i<j\leq 7}(r_i+r_j).\]

Evaluate $p^2$.

 

[lfdahl]

Here´s the suggested solution:

Spoiler

There are $21$ factors of $p$.

Let $a = 7^{1/7}$. Let $t = \frac{360^{\circ}}{14}$ and let $z(\theta) = \cos \theta + i\sin \theta$.

The first three factors are:

$r_1+r_2 = a + a z(2t) = a(\cos^2t + \sin^2 t)+a(\cos^2t-\sin^2t+i2\cos t \sin t) = 2a\cos (t) z(t)$.

$r_1+r_3 = a + az(4t) = 2a\cos(2t)z(2t)$.

$r_1 + r_4 = a + a z(6t) = 2a\cos(3t)z(3t)$.

In fact any sum of roots, $r_i+r_j$, $1 \leq i < j \leq 7$, has one of the three moduli:

$2a\cos(t)$, $2a\cos(2t)$ or $2a\cos(3t)$. This is due to the identity: $1+z(2\theta) = 2\cos \theta z(\theta)$.

Thus the product of all moduli is:

$p = (2a\cos(t)\cdot 2a \cos(2t) \cdot 2a \cos(3t))^7 = a^{21} = 7^3$, and we get the result: $p^2 = 7^6$.

Here we have used the identity $\cos(t)\cos(2t)\cos(3t) = \frac{1}{8}$.