# triangle inequality of complex numbers

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