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We need to show that $$ |z_{1}+z_{2}+\cdots+z_{n}|=|z_{1}|+|z_{2}|+\cdots+|z_{n}|$$ if and only if $z_{1},z_{2},\dots,z_{n}$ have the same argument (i.e. $z_{j}=r_{j}e^{i\theta}$ for $j=1,\dots,n$).

This way [$\Longleftarrow$] is easy, but for the other way around, I get stuck proving it for the sum of two complex numbers, even though the induction itself isn't that big a problem.

Does anyone happen to know how to prove this for two complex numbers?

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  • $\begingroup$ Well, thanks to my brother (a fellow mathematician), I figured it out. $\endgroup$ – Misunderstood Observer May 20 '13 at 21:56
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Let $z_1=a_1+ib_1$ and $z_2=a_2+ib_2$ for $a_1,a_2,b_1,b_2\in\mathbb{R}$. By simplification the equation $$\sqrt{(a_1+a_2)^2+(b_1+b_2)^2}=\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2},$$ you will get $$\frac{a_1}{a_2}=\frac{b_1}{b_2}.$$ What does it tell about arguments?

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  • $\begingroup$ You should keep it at $a_1 b_2 = a_2 b_1$ because you cannot assume $a_2$ or $b_2$ are non-zero in general. Or otherwise you should treat that case separately. $\endgroup$ – Patrick Da Silva May 20 '13 at 22:02
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Suppose $z_1,z_2\ne0$ (otherwise, it's trivial). Then $$\begin{align}|z_1+z_2|^2 &= (z_1+z_2)\overline{(z_1+z_2)}\\ &= (z_1+z_2)\left(\overline{z_1}+\overline{z_2}\right)\\ &= z_1\overline{z_1}+z_1\overline{z_2}+\overline{z_1}z_2+z_2\overline{z_2}\\ &= |z_1|^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}+|z_2|^2\\ &=|z_1|^2+2\Re(z_1\overline{z_2})+|z_2|^2\\ &= \bigl(|z_1|+|z_2|\bigr)^2+2\Re(z_1\overline{z_2})-2|z_1||z_2|.\end{align}$$ Thus, you must show that $\Re(z_1\overline{z_2})=|z_1||z_2|$ if and only if $z_1,z_2$ have the same argument.

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Let's try to solve the equation

$$ |a+b|=|a|+|b| $$

for $a,b\in\mathbb{C}$. It's not restrictive to assume that both $a$ and $b$ are non-zero. We can square it, because both sides are non-negative:

$$ |a+b|^2=(|a|+|b|)^2 $$

so, taking into account that $|z|^2=z\bar{z}$, we get

$$ (a+b)(\bar{a}+\bar{b})=a\bar{a}+2|a|\,|b|+b\bar{b} $$

that simplifies in

$$ \bar{a}b+a\bar{b}=2|a|\,|b| $$

Now, write $a=|a|u$ and $b=|b|v$; the equation becomes

$$ \bar{u}v+u\bar{v}=2 $$

Recall now that $\bar{u}=u^{-1}$, since by hypothesis $|u|=1$. Then …

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