Triangle Inequality/Complex Numbers/Proof 4

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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
\(\ds \leadsto \ \ \) \(\ds \cmod {z + w}\) \(\le\) \(\ds \cmod z + \cmod w\)

$\blacksquare$


Proof

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