State and prove the triangle inequality of complex numbers.

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Solution

Consider ${| z_{1} + z_{2} |}^{2} = (z_{1} + z_{2}) (¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2})$ (since $z ¯ ¯ ¯ z = {| z |}^{2}$
$= (z_{1} + z_{2}) (_{1} +_{2})$
$= (z_{1}_{1} + z_{1}_{2} + z_{2}_{1} + z_{2}_{2})$
$= {| z_{1} |}^{2} + z_{1}_{2} + ¯ ¯¯¯¯¯¯¯ ¯ z_{1}_{2} {| z_{2} |}^{2}$
$= {| z_{1} |}^{2} + {| z_{2} |}^{2} + 2. R e (z_{1}_{2})$ since $z_{1} + z_{2} = 2 R e (z)$
$\leq {| z_{1} |}^{2} + {| z_{2} |}^{2} + 2 ∣ ∣ z_{1}_{2} ∣ ∣$
$\leq {| z_{1} |}^{2} + {| z_{2} |}^{2} + 2 | z_{1} | | z_{2} |$
$\leq {(| z_{1} + z_{2} |)}^{2}$
Taking positive square root both sides
$| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} |$

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