$\sum_j e^{i\phi_j}$ vs $\sum_j e^{ip\phi_j}$

Question

Let $\phi_j$ be a collection of angles. If $p$ is a positive integer, how is the sum $\sum_je^{i\phi_j}$ related to $\sum_je^{ip\phi_j}$?

Answer 1

Let us introduce $x_j=e^{i\phi_j}$, then the sums $$S_p=\sum_{j=1}^n x_j^p$$ generate the ring of symmetric polynomials in $n$ variables $x_1,\ldots,x_n$. These generators are called power sum symmetric polynomials. They are algebraically independent, i.e. do not satisfy any relation (for $p=1,\ldots,n$). However, $S_{p>n}$ can be written in terms of $S_1,\ldots,S_n$.

Answer 2

Take the case of two angles.

$s_1=e^{i\phi_0}+e^{i\phi_1}$ and $s_2=e^{i2\phi_0}+e^{i2\phi_1}$ are such that $$s_2-s_1^2=2e^{i(\phi_0+\phi_1)}.$$ As the RHS represents a complex number of modulus $2$ and arbitrary phase, we can conclude $$|s_2-s_1^2|=2.$$ In the case of three angles, $$s_2-s_1^2=2e^{i(\phi_0+\phi_1)}+2e^{i(\phi_1+\phi_2)}+2e^{i(\phi_2+\phi_0)}.$$ The RHS can be any complex number with modulus $\le6$, and $$|s_2-s_1^2|\le6.$$ Indeed, taking $\phi_0=\frac\phi2, \phi_1=\frac\phi2-\theta, \phi_2=\frac\phi2+\theta$, any point on a circle of radius $r=2(1+2\cos\theta)$ and angle $\phi$ can be reached.

More generally, $$|s_2-s_1^2|\le n(n-1).$$