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.
Thus from Sum of Two Sides of Triangle Greater than Third Side:
- $\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$
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.
Thus from Sum of Two Sides of Triangle Greater than Third Side:
- $\cmod {z_1 + z_2 + z_3} \le \cmod {z_1 + z_2} + \cmod {z_3}$
The result follows.
$\blacksquare$