# Triangle Inequality/Complex Numbers/Examples/3 Arguments

## Example of Use of Triangle Inequality for Complex Numbers

For all $z_1, z_2, z_3 \in \C$:

$\cmod {z_1 + z_2 + z_3} \le \cmod {z_1} + \cmod {z_2} + \cmod {z_3}$

## Proof 1

This is an instance of the General Triangle Inequality for Complex Numbers:

$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$

setting $n = 3$.

$\blacksquare$

## Proof 2

 $\ds \cmod {z_1 + z_2 + z_3}$ $=$ $\ds \cmod {z_1 + \paren {z_2 + z_3} }$ $\ds$ $\le$ $\ds \cmod {z_1} + \cmod {z_2 + z_3}$ Triangle Inequality for Complex Numbers $\ds$ $\le$ $\ds \cmod {z_1} + \cmod {z_2} + \cmod {z_3}$ Triangle Inequality for Complex Numbers

$\blacksquare$

## Proof 3

Let $z_1$, $z_2$ and $z_3$ be represented by the points $A$, $B$ and $C$ respectively in the complex plane.

From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OADB$ where:

$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
$OD$ represents $z_1 + z_2$.

Also from Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OCED$ where:

$OC$ and $OD$ represent $z_3$ and $z_1 + z_2$ respectively
$OD$ represents $z_1 + z_2 + z_3$.

As $OADB$ is a parallelogram, we have that $OB = AD$.

The lengths of $OA$, $AD$ and $OD$ are:

 $\ds OA$ $=$ $\ds \cmod {z_1}$ $\ds AD$ $=$ $\ds \cmod {z_2}$ $\ds OD$ $=$ $\ds \cmod {z_1 + z_2}$

But $OA$, $OB$ and $OD$ form the sides of a triangle.

$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$

Similarly, as $OCED$ is a parallelogram, we have that $OD = CE$.

The lengths of $OC$, $CE$ and $OE$ are:

 $\ds OC$ $=$ $\ds \cmod {z_3}$ $\ds CE$ $=$ $\ds \cmod {z_2 + z_2}$ $\ds OE$ $=$ $\ds \cmod {z_1 + z_2 + z_3}$

But $OC$, $CE$ and $OE$ form the sides of a triangle.

$\cmod {z_1 + z_2 + z_3} \le \cmod {z_1 + z_2} + \cmod {z_3}$

The result follows.

$\blacksquare$