When $|z_1+\ldots + z_n| = |z_1| + \ldots + |z_n|$. Problem: Prove that the equality $|z_1+\ldots + z_n| = |z_1| + \ldots + |z_n|$ holds if and only if $z_k/z_l \ge 0$ for any integer $k$ and $l$, $1 \le k, l \le n$, for which $z_l\ne0$.
My attempt: First, when we think geometrically the statement above makes sense because $z_k/z_l \ge 0$ means that all the numbers are on a line. Therefore, the equality holds. However, I am trying to prove this analytically, what I've done is below:

Assume $z_k/z_l \ge 0$ for any integer $k$ and $l$, $1 \le k, l \le
> n$, for which $z_l\ne0$.Then, for particular $k$ and $l$ there is a
  positive real number $m$ such that $z_k=mz_l$. Thus, $z_k+z_l=(m+1)z_l
> \Rightarrow |z_k+z_l|=(m+1)|z_l|=|z_k|+|z_l|$.

However, the result I reached above is only for 2 complex numbers, and I cannot proceed to show that it holds for $n$ complex numbers. Could you give a hint about the solution?
 A: This is a hint as you asked (the full proof is in Ahlfor's Complex Analysis).
Suppose equality holds for any $n\ge 2$. Then we must have $|z_1+z_2|=|z_1|+|z_2|$ by cancelling terms. Now by the already established result we have $z_1/z_2\ge 0$ (assuming $z_2\ne 0$). But the numberings of $z_i$ are arbitrary and so the ratio of any two non zero terms is positive.
For the converse use the fact that $|1+\frac{z_2}{z_1}+\cdots+\frac{z_n}{z_1}|=1+\frac{z_2}{z_1}+\cdots+\frac{z_n}{z_1}$ and write $|z_1+\cdots+z_n|$ as $|z_1|(1+\frac{z_2}{z_1}+\cdots+\frac{z_n}{z_1})$ (assuming $z_1\ne 0$).
A: The comment about induction by Daniel Fischer looks fine. For a direct proof:
Find $\ell$ so that $z_\ell =|z_\ell| e^{i\theta} \neq 0$. Then (show and) use the inequality:
  $$ {\rm Re} \left( e^{-i \theta} (z_1 + \cdots + z_n ) \right) \leq \sum_k |z_k| $$
and study the case when there is equality. Alternative: Look at equality in:
$$ \left| \sum_k z_k \right|^2 = \sum_{k,\ell} z_k \bar{z}_\ell \leq \sum_{k,\ell} |z_k|\; |z_\ell| = \left( \sum_k |z_k|\right)^2 $$
