I have an unknown function $\psi(\xi_1,\dots,\xi_n)$, such that $\psi$ satisfy an (unknown) polynomial equation with coefficients polynomials in the $\xi_i$.

The function is homogeneous, that is, $\psi(t \xi_1,\dots, t \xi_n) = t^c \psi(\xi_1,\dots,\xi_n)$. I also know that $|\psi(e^{i \theta_1},\dots,e^{i \theta_n})|=1$ when $\sum_i \theta_i =0$, which implies that $$\psi(e^{i \theta_1},\dots,e^{i \theta_n}) = e^{i A(\theta_1,\dots,\theta_n)}$$ for some real-valued function $A$.

Clearly, one option is that $\psi(\xi_1,\dots,\xi_n) = \xi_1^{p_1} \cdots \xi_n^{p_n}$ such that $p_1+\cdots+p_n = c$, but can one exclude any other form? How do one take advantage of the fact that $\psi$ is algebraic in the $\xi_i$?

2 Replaced g' withc'.

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# Algebraig function with extra condition, what can it be?

I have an unknown function $\psi(\xi_1,\dots,\xi_n)$, such that $\psi$ satisfy an (unknown) polynomial equation with coefficients polynomials in the $\xi_i$.

The function is homogeneous, that is, $\psi(t \xi_1,\dots, t \xi_n) = t^c \psi(\xi_1,\dots,\xi_n)$. I also know that $|\psi(e^{i \theta_1},\dots,e^{i \theta_n})|=1$ which implies that $$\psi(e^{i \theta_1},\dots,e^{i \theta_n}) = e^{i A(\theta_1,\dots,\theta_n)}$$ for some real-valued function $A$.

Clearly, one option is that $\psi(\xi_1,\dots,\xi_n) = \xi_1^{p_1} \cdots \xi_n^{p_n}$ such that $p_1+\cdots+p_n = c$, but can one exclude any other form? How do one take advantage of the fact that $\psi$ is algebraic in the $\xi_i$?