# Triangle inequality for n complex numbers

• Feb 10th 2011, 06:38 PM
Jame
Triangle inequality for n complex numbers
I am trying to prove that
$|z_1+z_2+...+z_n| = |z_1| + |z_2| + ... |z_n|$

iff $z_i/z_j$ is a positive real number $\forall$ integers i and j, s.t. $i,j \in$ $\left\{ 1,...,n \right\}$

I really don't see how these two ideas imply each other. After looking at this for several hours the only thing I managed to come up with (which I'm sure could be extended to n variables) is

$|z_1|^2 + m|z_2|^2 = |z_1+z_2|^2$ where $m \in \mathbb{R}$

however, that real number m is not always $z_i/z_j$

• Feb 10th 2011, 07:06 PM
chisigma
If $\forall i,j \in {1,2,...,n \}$ is $\displaystyle \frac{z_{i}}{z_{j}} = \alpha_{i,j}$ and any $\alpha_{i,j}$ is positive real, then $\forall i$ is $z_{i}= e^{\sigma\ \theta}\ \beta_{i}$, being $\sigma= \sqrt{-1}$ , $\theta$ real and any $\beta_{i}$ positive real. In this case is...

$\displaystyle |z_{1} + z_{2} + ...+ z_{n}|= |e^{\sigma \theta}|\ |\beta_{1} + \beta_{2} +...+ \beta_{n}|= |z_{1}|+|z_{2}|+...+|z_{n}|$ (1)

The inverse however is not true and a symple 'counterexample' is when one of the $z_{j}$ is zero and the terms $\frac{z_{i}}{z_{j}}$ for $i \ne j$ don't exist...

Kind regards

$\chi$ $\sigma$
• Feb 10th 2011, 07:35 PM
Jame
Sorry, I meant to specify that $z_j \neq 0$. Thank you very much for your help though. I will try to figure out the other direction on my own.

-Cheers
• Feb 10th 2011, 09:02 PM
xxp9
Use the real part and imaginary part of the complex numbers to re-write the inequality then apply the Cauchy–Schwarz inequality.