Proposition VI.3.4 in Silverman's Arithmetic of Elliptic Curves proves that for elliptic curves over $\mathbb{C}$ (given as a lattice $\Lambda$), any divisor $D$ of degree $0$ such that the sum of the divisor summed as points in $\mathbb{C}$ lies in $\Lambda$ is principal.
I am confused on the first step in the proof: Let $n_1, \dots, n_r \in \mathbb{Z}$ and $z_1, \dots, z_r \in \mathbb{C}$ such that $\sum n_i = 0$ and $\sum n_i z_i \in \Lambda$. Silverman has already constructed a function $\sigma$ whose zeros are exactly the points $\Lambda$, all with order 1. He then says:
Let $\lambda = \sum n_i z_i \in \Lambda$. Replacing $n_1 (z_1) + \cdots + n_r (z_r)$ by $n_1(z_1) + \cdots + n_r (z_r) + (0) - (\lambda)$, we may assume that $\sum n_i z_i = 0$,
and goes on to define $f(z) = \prod \sigma(z - z_i)^{n_i}$, which has the correct zeros and poles. I don't see why this first step of reducing to the case that $\sum n_i z_i = 0$ is necessary. Since we already know the zeros of $\sigma(z)$, we know that the zeros of $\sigma(z - z_i)$ are at exactly the points $z_i + \Lambda$, and all have order $1$, so the function $f(z)$ has the correct zeros and poles. Why do we need to reduce to the case that $\sum n_i z_i = 0$?