# Product of differences of roots of unity

$$\DeclareMathOperator{\lcm}{lcm}$$ Denote by $$\mu_n$$ a primitive n-th root of unity. Examples suggest that $$\prod_{0\le k\lt a \\ 0 \le l \lt b \\ k/a \neq l/b} (\mu_a^k - \mu_b^l)=\pm \lcm(a,b)^{\gcd(a,b)} \text{.}$$ Is this formula known? How to prove it?

• – dxiv
Apr 1 at 4:18

I will use $$\Omega_n$$ to denote the set of $$n$$-th roots of unity, so $$\Omega_n = \{ \mu_n^k : 0 \le k < n \}$$.

Divide the LHS to two parts: \begin{aligned} \mathrm{LHS} &= \prod_{z \in \Omega_a} \prod_{\substack{z' \in \Omega_b \\ z' \ne z}} (z - z') \\ &= \prod_{\substack{z \in \Omega_a \\ z \notin \Omega_b}} \prod_{z' \in \Omega_b} (z - z') \cdot \prod_{z \in \Omega_{\gcd(a, b)}} \prod_{\substack{z' \in \Omega_b \\ z' \ne z}} (z - z') \text{.} \end{aligned}

Define $$\displaystyle p_n(z) = \prod_{z' \in \Omega_n} (z - z')$$, we've known that $$p_n(z) = z^n - 1$$.

For the first part, \begin{aligned} \prod_{\substack{z \in \Omega_a \\ z \notin \Omega_b}} \prod_{z' \in \Omega_b} (z - z') &= \prod_{\substack{z \in \Omega_a \\ z \notin \Omega_b}} p_b(z) = \prod_{\substack{z \in \Omega_a \\ z \notin \Omega_b}} (z^b - 1) \\ &= \Biggl( \prod_{\substack{z \in \Omega_{a / \gcd(a, b)} \\ z \ne 1}} (z - 1) \Biggr)^{\gcd(a, b)} \\ &= \biggl( (-1)^{a / \gcd(a, b) - 1} \frac{p_{a / \gcd(a, b)}(z)}{z - 1} \Biggr\rvert_{z = 1} \biggr)^{\gcd(a, b)} \\ &= \biggl( (-1)^{a / \gcd(a, b) - 1} \frac{z^{a / \gcd(a, b)} - 1}{z - 1} \Biggr\rvert_{z = 1} \biggr)^{\gcd(a, b)} \\ &= \biggl( (-1)^{a / \gcd(a, b) - 1} (a / \gcd(a, b)) \biggr)^{\gcd(a, b)} \\ &= (-1)^{a - \gcd(a, b)} \biggl( \frac{\operatorname{lcm}(a, b)}{b} \biggr)^{\gcd(a, b)} \text{.} \end{aligned}

For the second part, \begin{aligned} \prod_{z \in \Omega_{\gcd(a, b)}} \prod_{\substack{z' \in \Omega_b \\ z' \ne z}} (z - z') &= \prod_{z \in \Omega_{\gcd(a, b)}} z^{b - 1} \prod_{\substack{z' \in \Omega_b \\ z' \ne z}} (1 - z' / z) \\ &= \prod_{z \in \Omega_{\gcd(a, b)}} z^{b - 1} \prod_{\substack{z' \in \Omega_b \\ z' \ne 1}} (1 - z') \\ &= \prod_{z \in \Omega_{\gcd(a, b)}} z^{b - 1} \cdot b \\ &= \prod_{z \in \Omega_{\gcd(a, b)}} b / z \\ &= b^{\gcd(a, b)} \prod_{z \in \Omega_{\gcd(a, b)}} \bar{z} \\ &= b^{\gcd(a, b)} \prod_{z \in \Omega_{\gcd(a, b)}} z \\ &= b^{\gcd(a, b)} \cdot (-1)^{\gcd(a, b)} p_{\gcd(a, b)}(0) \\ &= (-1)^{\gcd(a, b) + 1} b^{\gcd(a, b)} \text{.} \end{aligned}

Therefore, $$\mathrm{LHS} = (-1)^{a + 1} \operatorname{lcm}(a, b)^{\gcd(a, b)}$$.

• Thank you. It's a nice computation :-)
– vuur
Apr 2 at 4:30