The problem is to show that if $z_1,\dots,z_n$ are complex numbers then $$|z_1+\cdots+z_n|=\sum |z_i|$$ if and only if $$\mbox{arg}(z_i)\equiv \mbox{arg}(z_j)\mod 2\pi$$ for all $i,j$.
I can establish the case $n=2$. I thought of using induction to prove the general case but didn't get anywhere. Following the hint in this question I tried proceeding as follows:
Assume $|z_1+\cdots+z_n|=\sum |z_i|$. Squaring both sides yields $\sum_{i\ne j}z_i\overline{z_j}=\sum_{i\ne j}|z_i||z_j|$. Since all $z_i\ne 0$ for otherwise we may use induction so we have
$$\frac{\sum_{i\ne j}z_i\overline{z_j}}{\sum_{i\ne j}{|z_i||z_j|}}=1.$$
At this point I cannot think of anything else to do.
Answer
I don't see an easy way to finish your attempted proof. But assuming you can do the case $n=2$, you can get the general case by induction as follows. Let $w=z_1+\dots+z_{n-1}$. Then note that $|w|\leq |z_1|+\dots+|z_{n-1}|$, and so $$|w+z_n|=|z_1|+\dots+|z_{n-1}|+|z_n|$$ implies $$|w+z_n|\geq |w|+|z_n|.$$ The only way this can hold is if in fact $|w+z_n|=|w|+|z_n|$, and in this case we must also have $|w|=|z_1|+\dots+|z_{n-1}|$. The induction hypothesis now gives that the $z_i$ for $i=1,\dots,n-1$ all have the same argument, and the case $n=2$ applied to $w$ and $z_n$ gives that $z_n$ also has the same argument.
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