triangle inequality of complex numbersTriangleInequalityOfComplexNumbers2009-03-21 16:12:572009-04-03 15:18:23Theoremmodulus of complex number% this is the default PlanetMath preamble. as your knowledge
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\textbf{Theorem.}\, All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality
\begin{align}
|z_1\!+\!z_z| \;\leqq\;|z_1|+|z_2|.
\end{align}
\emph{Proof.}
\begin{align*}
|z_1\!+\!z_2|^2 &\;=\; (z_1+z_2)\overline{(z_1+z_2)}\\
&\;=\; (z_1+z_2)(\overline{z_1}+\overline{z_2})\\
&\;=\; z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2\\
&\;=\; |z_1|^2+|z_2|^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}\\
&\;=\; |z_1|^2+|z_2|^2+2\mbox{Re}(z_1\overline{z_2})\\
&\;\leqq\; |z_1|^2+|z_2|^2+2|z_1\overline{z_2}|\\
&\;=\; |z_1|^2+|z_2|^2+2|z_1|\cdot|\overline{z_2}|\\
&\;=\; (|z_1|+|z_2|)^2
\end{align*}
Taking then the nonnegative square root, one obtains the asserted inequality.\\
\textbf{Remark.}\, Since the real numbers are complex numbers, the inequality (1) and its proof are valid also for all real numbers; however the inequality \PMlinkescapetext{chain} may be simplified to
$$|x+y|^2 \leqq (x+y)^2 = x^2+2xy+y^2 \leqq x^2+2|x||y|+y^2 = |x|^2+2|x||y|+|y|^2 = (|x|+|y|)^2.$$