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I would like to prove [itex]\mid z + \overline{z} \mid \leq 2 \mid z \mid[/itex]
The first way I could think of:
[tex] \begin{multline}<br /> RHS^2 - LHS^2\\<br /> =4\mid z \mid^2 - \mid z + \overline{z} \mid ^ 2\\<br /> =4z\overline{z} - (z + \overline{z})(\overline{z}+z)\\<br /> =4z\overline{z} - z\overline{z}-z^2-\overline{z}^2-z\overline{z}\\<br /> =2z\overline{z}-z^2-\overline{z}^2\\<br /> =-(z^2-2z\overline{z}+\overline{z}^2)\\<br /> =-(z-\overline{z})^2\\<br /> \leq 0 ?<br /> \end{multline}[/tex]
I now know the correct proof is as follow:
[tex] \begin{multline}<br /> \mid z + \overline{z} \mid\\<br /> \leq \mid z \mid + \mid \overline{z} \mid\\<br /> = \mid z \mid + \mid z \mid\\<br /> = 2 \mid z \mid\\<br /> \end{multline}\\[/tex]
But what is wrong with my original proof?
The first way I could think of:
[tex] \begin{multline}<br /> RHS^2 - LHS^2\\<br /> =4\mid z \mid^2 - \mid z + \overline{z} \mid ^ 2\\<br /> =4z\overline{z} - (z + \overline{z})(\overline{z}+z)\\<br /> =4z\overline{z} - z\overline{z}-z^2-\overline{z}^2-z\overline{z}\\<br /> =2z\overline{z}-z^2-\overline{z}^2\\<br /> =-(z^2-2z\overline{z}+\overline{z}^2)\\<br /> =-(z-\overline{z})^2\\<br /> \leq 0 ?<br /> \end{multline}[/tex]
I now know the correct proof is as follow:
[tex] \begin{multline}<br /> \mid z + \overline{z} \mid\\<br /> \leq \mid z \mid + \mid \overline{z} \mid\\<br /> = \mid z \mid + \mid z \mid\\<br /> = 2 \mid z \mid\\<br /> \end{multline}\\[/tex]
But what is wrong with my original proof?
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