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Let $\alpha_1, \ldots, \alpha_n > 0$ be positive real numbers and $\omega_1, \ldots, \omega_n \in \mathbb{S}^1$ some complex numbers of modulus one. My question is when exactly do there exist some other $\omega_1', \ldots, \omega_n' \in \mathbb{S}^1$ such that $$\alpha_1\omega_1 + \cdots + \alpha_n\omega_n = \alpha_1\omega_1' + \cdots + \alpha_n\omega_n'$$ and $(\omega_1, \ldots, \omega_n) \ne (\omega_1', \ldots, \omega_n')$?

For example, if $\omega_1 = \cdots = \omega_n$, then clearly such $\omega'_i$ do not exist since we would have $$|\alpha_1\omega_1' + \cdots + \alpha_n\omega_n'| = |\alpha_1\omega_1 + \cdots + \alpha_n\omega_n| = \alpha_1 + \cdots + \alpha_n = |\alpha_1\omega_1'| + \cdots + |\alpha_n\omega_n'|$$ and hence $\omega_1' = \cdots = \omega_n'$ so it follows that $(\omega_1, \ldots, \omega_n) = (\omega_1', \ldots, \omega_n')$.

On the other hand, if $$\alpha_1\omega_1 + \cdots + \alpha_n\omega_n = 0$$ then for any $\omega \in \mathbb{S}^1, \omega \ne 1$ we have $$\alpha_1\omega_1\omega + \cdots + \alpha_n\omega_n\omega = 0 = \alpha_1\omega_1 + \cdots + \alpha_n\omega_n$$ and clearly $(\omega_1,\ldots, \omega_n) \ne (\omega_1\omega,\ldots, \omega_n\omega).$

My conjecture is that such $\omega_1', \ldots, \omega_n' \in \mathbb{S}^1$ will exist if and only if $\omega_1 = \cdots = \omega_n$ is false, but I'm not able to prove such a result. I'm assuming the continuity of the function $$(\omega_1,\ldots, \omega_n) \mapsto \alpha_1\omega_1 + \cdots + \alpha_n\omega_n$$ will somehow come into play.

mechanodroid
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    You may need more restrictions. For example, if some/all $a_j$ are equal then some/all permutations of $(\omega_1, \ldots, \omega_n)$ would work. – Martin R Oct 16 '21 at 08:51
  • Another example: If $\alpha_1\omega_1 + \cdots + \alpha_n\omega_n = 0$ then all $\omega_j$ can be multiplied with a factor of modulus one. – Martin R Oct 16 '21 at 09:43
  • @MartinR Yes, I mentioned that last example in the question already. True about the permutations. What if we assume, say, $0 < \alpha_1 < \alpha_2 < \cdots < \alpha_n$? – mechanodroid Oct 16 '21 at 10:07
  • @MartinR What I actually want is to construct a continuous function $$F : [0,+\infty\rangle^n \times (\mathbb{S}^1)^n \to (\mathbb{S}^1)^n$$ such that for all $(\alpha_1,\ldots,\alpha_n) \in [0,+\infty\rangle^n$ and $(\omega_1, \ldots, \omega_n) \in (\mathbb{S}^1)^n$, if we denote $$F(\alpha_1,\ldots,\alpha_n,\omega_1, \ldots, \omega_n) = (\omega_1', \ldots, \omega_n')$$ we have $$\alpha_1\omega_1 + \cdots + \alpha_n\omega_n = \alpha_1\omega_1' + \cdots + \alpha_n\omega_n'$$ – mechanodroid Oct 16 '21 at 10:15
  • ...and that there exists at least one choice $(\alpha_1,\ldots,\alpha_n) \in [0,+\infty\rangle^n$ and $(\omega_1, \ldots, \omega_n) \in (\mathbb{S}^1)^n$ such that $$F(\alpha_1,\ldots,\alpha_n,\omega_1, \ldots, \omega_n) \ne (\omega_1, \ldots, \omega_n).$$ Hence if $F$ acts as a permutation for some choices of $\alpha_1,\ldots, \alpha_n$ and identity for other choices, there is no way it could be continuous. – mechanodroid Oct 16 '21 at 10:15

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