# Complex no & conjugate

I would like to prove $\mid z + \overline{z} \mid \leq 2 \mid z \mid$

The first way I could think of:
$$\begin{multline} RHS^2 - LHS^2\\ =4\mid z \mid^2 - \mid z + \overline{z} \mid ^ 2\\ =4z\overline{z} - (z + \overline{z})(\overline{z}+z)\\ =4z\overline{z} - z\overline{z}-z^2-\overline{z}^2-z\overline{z}\\ =2z\overline{z}-z^2-\overline{z}^2\\ =-(z^2-2z\overline{z}+\overline{z}^2)\\ =-(z-\overline{z})^2\\ \leq 0 ??? \end{multline}$$

I now know the correct proof is as follow:
$$\begin{multline} \mid z + \overline{z} \mid\\ \leq \mid z \mid + \mid \overline{z} \mid\\ = \mid z \mid + \mid z \mid\\ = 2 \mid z \mid\\ \end{multline}\\$$

But what is wrong with my original proof?

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matt grime
Homework Helper
Nothing is wrong with it. what is z-z*? It's not a real number, is it.....

okay got it..
lets say b is Im(z).
then b^2 is always positive, and (ib)^2 is always negative...

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according to my proof, instead of $\mid z + \overline{z} \mid \leq 2 \mid z \mid$, it is $\mid z + \overline{z} \mid \geq 2 \mid z \mid$...

z is a complex no. and $\overline{z}$ is its conjugate. Have I mixed up some basic rules in complex no. with those in real no.?

Sorry forgoth I haven't noticed your reply when i post mine... but I do not understand... do you mean that I cannot square a complex no?

$$-(z-z^*)^2=-(a+ib-a+ib)^2=4b^2>0$$
your last line had the wrong signs... or i just missed something.

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matt grime