# 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 may be simplified to

 $|x+y|^{2}\leqq(x+y)^{2}=x^{2}+2xy+y^{2}\leqq x^{2}+2|x||y|+y^{2}=|x|^{2}+2|x||% y|+|y|^{2}=(|x|+|y|)^{2}.$
 Title triangle inequality of complex numbers Canonical name TriangleInequalityOfComplexNumbers Date of creation 2013-03-22 18:51:47 Last modified on 2013-03-22 18:51:47 Owner pahio (2872) Last modified by pahio (2872) Numerical id 11 Author pahio (2872) Entry type Theorem Classification msc 30-00 Classification msc 12D99 Synonym triangle inequality Related topic Modulus Related topic ComplexConjugate Related topic SquareOfSum Related topic TriangleInequality