# Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 3

## 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

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:

\(\displaystyle OA\) | \(=\) | \(\displaystyle \cmod {z_1}\) | |||||||||||

\(\displaystyle AD\) | \(=\) | \(\displaystyle \cmod {z_2}\) | |||||||||||

\(\displaystyle OD\) | \(=\) | \(\displaystyle \cmod {z_1 + z_2}\) |

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:

\(\displaystyle OC\) | \(=\) | \(\displaystyle \cmod {z_3}\) | |||||||||||

\(\displaystyle CE\) | \(=\) | \(\displaystyle \cmod {z_2 + z_2}\) | |||||||||||

\(\displaystyle OE\) | \(=\) | \(\displaystyle \cmod {z_1 + z_2 + z_3}\) |

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.

$\blacksquare$

## Sources

- 1981: Murray R. Spiegel:
*Theory and Problems of Complex Variables*(SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(b)}$