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[parent] triangle inequality of complex numbers (Theorem)

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

$\displaystyle \vert z_1\!+\!z_z\vert \;\leqq\;\vert z_1\vert+\vert z_2\vert.$ (1)

Proof.
$\displaystyle \vert z_1\!+\!z_2\vert^2$ $\displaystyle \;=\; (z_1+z_2)\overline{(z_1+z_2)}$    
  $\displaystyle \;=\; (z_1+z_2)(\overline{z_1}+\overline{z_2})$    
  $\displaystyle \;=\; z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2$    
  $\displaystyle \;=\; \vert z_1\vert^2+\vert z_2\vert^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}$    
  $\displaystyle \;=\; \vert z_1\vert^2+\vert z_2\vert^2+2$Re$\displaystyle (z_1\overline{z_2})$    
  $\displaystyle \;\leqq\; \vert z_1\vert^2+\vert z_2\vert^2+2\vert z_1\overline{z_2}\vert$    
  $\displaystyle \;=\; \vert z_1\vert^2+\vert z_2\vert^2+2\vert z_1\vert\cdot\vert\overline{z_2}\vert$    
  $\displaystyle \;=\; (\vert z_1\vert+\vert z_2\vert)^2$    

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



"triangle inequality of complex numbers" is owned by pahio.
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See Also: modulus, complex conjugate, square of sum, triangle inequality

Other names:  triangle inequality

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Cross-references: valid, real numbers, inequality, square root, proof, complex numbers, theorem
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This is version 8 of triangle inequality of complex numbers, born on 2009-03-21, modified 2009-04-03.
Object id is 11678, canonical name is TriangleInequalityOfComplexNumbers.
Accessed 1768 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 30-00 (Functions of a complex variable :: General reference works )
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