# Triangle Inequality/Complex Numbers/Proof 4

## 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

 $\ds \cmod {z + w}^2$ $=$ $\ds \paren {z + w} \paren {\overline z + \overline w}$ Product of Complex Number with Conjugate $\ds$ $=$ $\ds z \overline z + w \overline w + w \overline z + z \overline w$ $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + w \overline z + z \overline w$ Product of Complex Number with Conjugate $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + w \overline z + \overline {\paren {\overline z} } \overline w$ Complex Conjugation is Involution $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + \overline z w + \overline {\paren {\overline z w} }$ Product of Complex Conjugates and Complex Multiplication is Commutative $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + 2 \map \Re {z \overline w}$ Sum of Complex Number with Conjugate $\ds$ $\le$ $\ds \cmod z^2 + \cmod w^2 + 2 \cmod {z \overline w}$ Modulus Larger than Real Part $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + 2 \cmod z \cmod {\overline w}$ Complex Modulus of Product of Complex Numbers $\ds$ $=$ $\ds \cmod z^2 + \cmod w^2 + 2 \cmod z \cmod w$ Complex Modulus equals Complex Modulus of Conjugate $\ds$ $=$ $\ds \paren {\cmod z + \cmod w}^2$ Square of Sum $\ds \leadsto \ \$ $\ds \cmod {z + w}$ $\le$ $\ds \cmod z + \cmod w$

$\blacksquare$