| Version 4 |
Version 3 |
| \textbf{Theorem.}\, All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality |
\textbf{Theorem.}\, All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality |
| $$|z_1\!+\!z_z| \;\leqq\;|z_1|+|z_2|.$$\\ |
$$|z_1\!+\!z_z| \;\leqq\;|z_1|+|z_2|.$$\\ |
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| \emph{Proof.} |
\emph{Proof.} |
| \begin{align*} |
\begin{align*} |
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|z_1\!+\!z_2|^2 &\;=\; (z_1+z_2)\overline{(z_1+z_2)}\\
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|z_1\!+\!z_2|^2 &\;=\; (z_1+z_2)\bar{(z_1+z_2)}\\
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&\;=\; (z_1+z_2)(\overline{z_1}+\overline{z_2})\\
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&\;=\; (z_1+z_2)(\bar{z_1}+\bar{z_2})\\
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&\;=\; z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2\\
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&\;=\; z_1\bar{z_1}+z_2\bar{z_2}+z_1\bar{z_2}+\bar{z_1}z_2\\
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&\;=\; |z_1|^2+|z_2|^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}\\
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&\;=\; |z_1|^2+|z_2|^2+z_1\bar{z_2}+\bar{z_1\bar{z_2}}\\
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&\;=\; |z_1|^2+|z_2|^2+2\mbox{Re}(z_1\overline{z_2})\\
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&\;=\; |z_1|^2+|z_2|^2+2\mbox{Re}(z_1\bar{z_2})\\
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&\;\leqq\; |z_1|^2+|z_2|^2+2|z_1\overline{z_2}|\\
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&\;\leqq\; |z_1|^2+|z_2|^2+2|z_1\bar{z_2}|\\
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&\;=\; |z_1|^2+|z_2|^2+2|z_1|\cdot|\overline{z_2}|\\
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&\;=\; |z_1|^2+|z_2|^2+2|z_1|\cdot|\bar{z_2}|\\
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| &\;=\; (|z_1|+|z_2|)^2 |
&\;=\; (|z_1|+|z_2|)^2 |
| \end{align*} |
\end{align*} |
| Taking then the nonnegative square root, one obtains the asserted inequality. |
Taking then the nonnegative square root, one obtains the asserted inequality. |
