No, for example pick $\zeta = \exp \frac{i\pi}{15}$, a primitive $30$th root of unity.
Then $0 = (1 + \zeta^{10} + \zeta^{20}) + \zeta^{15}(1 + \zeta^6 + \zeta^{12} + \zeta^{18} + \zeta^{24}) - (1 + \zeta^{15}) \\
= \zeta^3 + \zeta^9 + \zeta^{10} + \zeta^{20} + \zeta^{21} + \zeta^{27}$
But you can't partition those $6$ roots into vertices of regular $k$-gons.
However, it is true that you can always add vertices of regular $k$-gons to your set of roots (possibly adding the same root multiple times) to obtain a new multiset that you can then arrange into a sum of vertices of regular $k$-gons (possibly using the same $k$-gon multiple times).
Let $\zeta_n = \exp{\frac{2i\pi}n}$ and $f$ be the map $\Bbb Z^n \to \Bbb Z[\zeta_n]$ given by $(a_i) \mapsto \sum a_i \zeta_n^i$.
The minimal polynomial of $\zeta_n$ over $\Bbb Q$ is the cyclotomic polynomial $\Phi_n$, of degree $\varphi(n)$. Hence $\Bbb Z[\zeta_n]$ is free of rank $\varphi(n)$.
If $n = \prod p_i^{d_i}$, then $\zeta$ is a primitive $n$th root of unity iff it is of order exactly $n$, and not $n/p$ for any $p$, iff $\zeta^{n/p}$ is a primitive $p$th root, for all $p$. Hence $\Phi_n$ is the gcd of the $\Phi_p(X^{n/p}) = 1 + X^{n/p} + X^{2n/p} + \ldots + X^{(p-1)n/p}$, and $\ker f$ is generated by those relations, which correspond to vertices of regular $p$-gons.
Part 1 is still kind of nice; depending on what you need this for this may or may not be useful:
For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate roots of unity
$$e^{\frac{2i\pi k}{n}},\ e^{-\frac{2i\pi k}{n}},$$
then the quadratic polynomial with them as roots is
$$P(x)=x^2-2\cos\left(\frac{2\pi k}{n}\right)x+1.$$
So, any real-coefficiented polynomial (whose roots are all roots of unity) has the form
$$(x+1)^m(x-1)^n\prod_{k=1}^N \left(x^2-2\cos\left(\frac{2\pi a_k}{b_k}\right)x+1\right),$$
where $m,n$ are nonnegative integers and all $a_k,b_k$ are positive integers (with $a_k<b_k$, if you wish).
For part 2, I mean, you can just pick the roots to be
$$x_k=e^{\frac{2i\pi a_k}{b_k}},$$
and your polynomial will be
$$\prod_{k=1}^N \left(x-e^{\frac{2i\pi a_k}{b_k}}\right),$$
but this probably isn't that useful. Another thing that may or may not be useful for either of these is that all polynomials with only roots of unity as roots have roots that all satisfy $x^M=1$ for some (possibly large with respect to the degree of the polynomial) integer $M$. In the case above this $M$ can be taken to be $b_1b_2\cdots b_k$. So, all polynomials with only roots of unity as roots are factors of $x^M-1$ for some $M$.
Best Answer
For a generic polynomial over the complex numbers, you can factor it into linear factors: $$p(z) = a(z-r_1)(z-r_2)...(z-r_n)$$ If you multiply out that polynomial, the coefficient of $z^n$ is $1$ and the coefficient of $z^{n-1}$ is $-r_1-r_2-...-r_n$, the negative of the sum of the roots (times $a$), just like the constant term is $(-1)^n\cdot$(product of roots times $a$), and so forth for all the other coefficients.
The geometric explanation for why the sum of the $n$-th roots of unity is zero is that they form a regular $n$-gon around the origin when plotted in the complex plane.