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Jun 23, 2017 at 20:09 comment added Federico Poloni @FedorPetrov Good point. Another quick way to find the absolute value of the determinant is via the relation $\Phi^*\Phi=nI$.
Jun 23, 2017 at 20:04 comment added Fedor Petrov note that for fixed $i$ the product $\prod_{j\ne i} (\xi^i-\xi^j)$ is the derivative of the polynomial $z^n-1$ at $z=\xi^i$, which equals $n\xi^{-i}$. Multiplying by all $i$ gives you the value of the square of $\det \Phi$. And it is routine to find the argument, since we know the arguments of all brackets.
Jun 23, 2017 at 19:54 history answered Federico Poloni CC BY-SA 3.0