triangle inequality of complex numbers


Theorem.  All complex numbersMathworldPlanetmathPlanetmath z1 and z2 satisfy the triangle inequalityMathworldMathworldPlanetmath

|z1+zz||z1|+|z2|. (1)

Proof.

|z1+z2|2 =(z1+z2)(z1+z2)¯
=(z1+z2)(z1¯+z2¯)
=z1z1¯+z2z2¯+z1z2¯+z1¯z2
=|z1|2+|z2|2+z1z2¯+z1z2¯¯
=|z1|2+|z2|2+2Re(z1z2¯)
|z1|2+|z2|2+2|z1z2¯|
=|z1|2+|z2|2+2|z1||z2¯|
=(|z1|+|z2|)2

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

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(x+y)2=x2+2xy+y2x2+2|x||y|+y2=|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