# Triangle Inequality/Complex Numbers/Proof 3

## Theorem

Let $z_1, z_2 \in \C$ be complex numbers.

Let $\cmod z$ denote the modulus of $z$.

Then:

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

## Proof

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

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

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

As $OACB$ is a parallelogram, we have that $OB = AC$.

The lengths of $OA$, $AC$ and $OC$ are:

 $\ds OA$ $=$ $\ds \cmod {z_1}$ $\ds AC$ $=$ $\ds \cmod {z_2}$ $\ds OC$ $=$ $\ds \cmod {z_1 + z_2}$

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

The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.

$\blacksquare$