Is My Proof Correct? Understanding the Complex Modulus Inequality

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The discussion centers on the proof of the complex modulus inequality |z + z̄| ≤ 2|z|. The original proof attempted by the user involved manipulating the expressions but ultimately led to an incorrect conclusion due to a misunderstanding of the properties of complex numbers. The correct proof utilizes the triangle inequality, demonstrating that the modulus of the sum of a complex number and its conjugate is at most twice the modulus of the number itself. The user acknowledges confusion between the rules of real and complex numbers, particularly regarding squaring negative values. Overall, the conversation highlights the importance of understanding complex number properties in mathematical proofs.
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I would like to prove \mid z + \overline{z} \mid \leq 2 \mid z \mid

The first way I could think of:
<br /> \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}<br />

I now know the correct proof is as follow:
<br /> \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}\\<br />

But what is wrong with my original proof?
 
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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...

you just got the signs wrong to start with.
 
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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&gt;0
your last line had the wrong signs... or i just missed something.
 
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When ever you do something like this, always step back and think: what happens in a simple example. For instance why not put z=i in and see what happens?

But you are confusing some rules of real numbers with complex ones. Sure, if you square a real number it is positive, but the whole raison d'etre of complex number is that you have things that square to negative numbers.
 
I see... thx~
Too much used to real no... have never thought that there would be a problem on that line
(never thought of posting a question in a forum could get immediate response too~ ^^)
 
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