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'triangle inequality of complex numbers'
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| Title of object: |
triangle inequality of complex numbers |
| Canonical Name: |
TriangleInequalityOfComplexNumbers |
| Type: |
Theorem |
| Created on: |
2009-03-21 16:12:57 |
| Modified on: |
2009-03-21 17:18:12 |
| Classification: |
msc:12D99, msc:30-00 |
| Synonyms: |
triangle inequality of complex numbers=triangle inequality |
Revision comment (for changes between this and next version):
Preamble:
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Content:
\textbf{Theorem.}\, All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality
$$|z_1\!+\!z_z| \;\leqq\;|z_1|+|z_2|.$$\\
\emph{Proof.}
\begin{align*}
|z_1\!+\!z_2|^2 &\;=\; (z_1+z_2)\bar{(z_1+z_2)}\\
&\;=\; (z_1+z_2)(\bar{z_1}+\bar{z_2})\\
&\;=\; z_1\bar{z_1}+z_2\bar{z_2}+z_1\bar{z_2}+\bar{z_1}z_2\\
&\;=\; |z_1|^2+|z_2|^2+z_1\bar{z_2}+\bar{z_1\bar{z_2}}\\
&\;=\; |z_1|^2+|z_2|^2+2\mbox{Re}(z_1\bar{z_2})\\
&\;\leqq\; |z_1|^2+|z_2|^2+2|z_1\bar{z_2}|\\
&\;=\; |z_1|^2+|z_2|^2+2|z_1|\cdot|\bar{z_2}|\\
&\;=\; (|z_1|+|z_2|)^2
\end{align*}
Taking then the nonnegative square root, one obtains the asserted inequality.
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