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.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: $(2.1)$
• 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: $7 \ \text{(a)}$