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Complex no & conjugate |
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| Apr21-06, 09:49 AM | #1 |
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Complex no & conjugate
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? |
| Apr21-06, 09:59 AM | #2 |
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Nothing is wrong with it. what is z-z*? It's not a real number, is it.....
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| Apr21-06, 10:03 AM | #3 |
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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. |
| Apr21-06, 10:07 AM | #4 |
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Complex no & conjugate
according to my proof, instead of [itex]\mid z + \overline{z} \mid \leq 2 \mid z \mid[/itex], it is [itex]\mid z + \overline{z} \mid \geq 2 \mid z \mid[/itex]...
z is a complex no. and [itex]\overline{z}[/itex] is its conjugate. Have I mixed up some basic rules in complex no. with those in real no.? |
| Apr21-06, 10:10 AM | #5 |
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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?
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| Apr21-06, 10:11 AM | #6 |
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[tex]-(z-z^*)^2=-(a+ib-a+ib)^2=4b^2>0[/tex]
your last line had the wrong signs... or i just missed something. |
| Apr21-06, 10:14 AM | #7 |
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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. |
| Apr21-06, 10:16 AM | #8 |
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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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