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Hometriangle inequality of complex numbers

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# triangle inequality of complex numbers

Theorem. All complex numbers $z_{1}$ and $z_{2}$ satisfy the triangle inequality

$\displaystyle|z_{1}\!+\!z_{z}|\;\leqq\;|z_{1}|+|z_{2}|.$ | (1) |

*Proof.*

$\displaystyle|z_{1}\!+\!z_{2}|^{2}$ | $\displaystyle\;=\;(z_{1}+z_{2})\overline{(z_{1}+z_{2})}$ | ||

$\displaystyle\;=\;(z_{1}+z_{2})(\overline{z_{1}}+\overline{z_{2}})$ | |||

$\displaystyle\;=\;z_{1}\overline{z_{1}}+z_{2}\overline{z_{2}}+z_{1}\overline{z% _{2}}+\overline{z_{1}}z_{2}$ | |||

$\displaystyle\;=\;|z_{1}|^{2}+|z_{2}|^{2}+z_{1}\overline{z_{2}}+\overline{z_{1% }\overline{z_{2}}}$ | |||

$\displaystyle\;=\;|z_{1}|^{2}+|z_{2}|^{2}+2\mbox{Re}(z_{1}\overline{z_{2}})$ | |||

$\displaystyle\;\leqq\;|z_{1}|^{2}+|z_{2}|^{2}+2|z_{1}\overline{z_{2}}|$ | |||

$\displaystyle\;=\;|z_{1}|^{2}+|z_{2}|^{2}+2|z_{1}|\cdot|\overline{z_{2}}|$ | |||

$\displaystyle\;=\;(|z_{1}|+|z_{2}|)^{2}$ |

Taking then the nonnegative square root, one obtains the asserted inequality.

Remark. Since the real numbers are complex numbers, the inequality (1) and its proof are valid also for all real numbers; however the inequality chain may be simplified to

$|x+y|^{2}\leqq(x+y)^{2}=x^{2}+2xy+y^{2}\leqq x^{2}+2|x||y|+y^{2}=|x|^{2}+2|x||% y|+|y|^{2}=(|x|+|y|)^{2}.$ |

Related:

Modulus, ComplexConjugate, SquareOfSum, TriangleInequality

Synonym:

triangle inequality

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

30-00*no label found*12D99

*no label found*

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Sep 28

new question: how to contest an entry? by zorba

new question: simple question by parag