triangle inequality of complex numbers

Theorem.  All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality

$|z_1+z_z|\leqq |z_1|+|z_2|.$ (1)

Proof.

${|z_1+z_2|}^2$ $\mathrm{\hspace{0.33em}}=(z_1+z_2)\overline{(z_1+z_2)}$

$\mathrm{\hspace{0.33em}}=(z_1+z_2)(\overline{z_1}+\overline{z_2})$

$\mathrm{\hspace{0.33em}}=z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}{z}_2$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+2\text{Re}(z_1\overline{z_2})$

$\mathrm{\hspace{0.33em}}\leqq {|z_1|}^2+{|z_2|}^2+2|z_1\overline{z_2}|$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+2|z_1|\cdot |\overline{z_2}|$

$\mathrm{\hspace{0.33em}}={(|z_1|+|z_2|)}^2$

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