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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.

Community
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3 Answers3

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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.

Eric Wofsey
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  • Thanks. A follow up question: Do you think the statement is correct even if some of the $z_i$ are $0$ (in which case their argument is not defined)? I am not sure how to formulate the statement in total generality. –  Apr 14 '16 at 10:41
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    What you should really say is that there exists a ray starting from the origin which contains all the $z_i$. – Eric Wofsey Apr 14 '16 at 17:55
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Let $z_m= r_m e^{i t_m}$ with $r_m\geq 0$ and $t\in R.$ $$\text {We have }\quad \sum_{m\ne m'}|z_m|\cdot |z_{m'}|-\sum_{m\ne m'}z_m \bar z_{m'}=\sum_{1\leq m<m'\leq n} f(m,m')$$ $$\text {where }\quad f(m,m')=2 |z_m|\cdot |z_{m'}|-z_m\bar z_{m'}-z_{m'}\bar z_m.$$ $$ \text {We have }\quad f(m,m')=2 r_m r_{m'}-r_m r_{m'} (\;e^{i(t_m-t_m')}+e^{i(t_m'-t_m)}\;)=$$ $$= r_m r_{m'} (\;2-2\cos (t_m-t_m')\;)\geq 0$$ with equality only when $(t_m-t_m')/2\pi \in Z$ or $r_m r_{m'}= 0.$

Note that $\sum_{m\ne m'}$ denotes the sum over all ordered pairs $(m,m')$ with $m\ne m'$. So for example, with $n\geq 2,$ the terms with $(m,m')=(1,2)$ and with $(m,m')=(2,1)$ both occur in the sum.

DanielWainfleet
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If you look at complex numbers as being vectors and use the polygon addition of vectors (ie, you put them one after the other, head of next to tail of the previous, tail being the arrow part, and the final result, aka the sum, being the vector from the first head to the last tail), you can immediately see that the statements is true: the vectors must be aligned and one of them should have a modulo $\pi$ phase. Hope this doesn’t sound too hand-waving, as I believe is valid proof.

Chip
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