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