Those $z_i$ are odd numbered $2n$th root of unity. In the figure regular $n$-gon.
Consider following function's decomposition. $f(x) = (1+x)(1+x^2)\ldots(1+x^n)$. Then $x^k$'s coefficient is a set whose sum of elements. We need to find $x^k$'s sum of coefficients, which is denoted by $A_n$. Here $n = 2009$. Now consider that sum $f(\varepsilon) + f(\varepsilon^2) + \ldots + f(\varepsilon^n)$, here $\varepsilon = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$. We know that $\varepsilon + \varepsilon^2 + \ldots + \varepsilon^n = 0$ then we can easily see that $\sum_{k=1}^{n} f(\varepsilon^k) = n \cdot A_n$. Now compute the $f(\varepsilon^k)$, Assume $d = (k, n)$. Then $f(\varepsilon^d) = f(\varepsilon^k)$. Otherwise, numbers of all $d$ such that $d = (k, n)$ is $\varphi\left(\frac{n}{d}\right)$. Also,
$$f(\varepsilon^d) = (1+\varepsilon^d)^d (1+\varepsilon^{2d})^d \dots (1+\varepsilon^{n/d})^d$$
and easy calculation, we get
$$f(\varepsilon^d) = \left[ (1 + \varepsilon^d)(1 + \varepsilon^{2d})\dots(1 + \varepsilon^{\frac{n}{d} \cdot d}) \right]^d$$
Because of $x^n - 1 = \prod_{k=1}^{n} (x - \varepsilon^k)$ then substituting $x = -1$, we get $(-1)^n - 1 = (-1)^n f(\varepsilon)$, In other word $f(\varepsilon) = 1 + (-1)^{n+1}$. Observe that $(\varepsilon^d)^{\frac{n}{d}} = 1$, we get $f(\varepsilon^d) = (1 + (-1)^{\frac{n}{d}+1})^d$. Finally
$$n \cdot A_n = \sum_{d|n} \varphi\left(\frac{n}{d}\right) \cdot \left(1 + (-1)^{\frac{n}{d}+1}\right)^d.$$