# Prove equality in triangle inequality for complex numbers

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?

• Well, thanks to my brother (a fellow mathematician), I figured it out. – Misunderstood Observer May 20 '13 at 21:56

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?

• 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. – Patrick Da Silva May 20 '13 at 22:02

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.

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 …