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One Question .Complex Numbers

  1. Jul 7, 2008 #1
    One Question.....Complex Numbers

    1. The problem statement, all variables and given/known data

    verify that: Square root (2) * lzl >= l Re (z) l + l Im ( z ) l

    suggestion : reduce this inequality to ( lxl - lyl )^2 >=0

    note : lxl <<<< modulus x
  2. jcsd
  3. Jul 7, 2008 #2


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    Re: One Question.....Complex Numbers

    Well, what have you TRIED? If z= x+ iy, what is |z|? Whar are Re(z) and Im(z)?

    I assume your note simply means that |x| and "modulus x" are the same thing. I had first interpreted it as "much less than" which makes no sense!
  4. Jul 7, 2008 #3
    Re: One Question.....Complex Numbers

    lxl <<<< modulus x

    <<<< i dosent mean less thab but i mean modulus x ---->lxl
    i mean modulus x = lxl

    ((2x2+2y2)1/2 - lxl - lyl )2 >= 0

    Substitute lxl - lyl = k
    2x2 + 2y2 - 2k * (2x2 + 2y2)1/2 + k2 >= 0

    Substitute 2x2 + 2y2 = m

    m - 2k * m1/2 + k2 >= 0

    this is my try ...
    i know its wrong ...
    and i cant continue ...
    Last edited: Jul 7, 2008
  5. Jul 7, 2008 #4


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    Re: One Question.....Complex Numbers

    You're going the wrong way. From your attempt it looks like you know |z|=sqrt(x^2+y^2) for z=x+iy. Start with what you want to prove. sqrt(2)*sqrt(x^2+y^2)>=|x|+|y|. Square both sides. Now move everything to one side and look at your suggestion.
  6. Jul 7, 2008 #5


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    Welcome to PF!

    Hi selena! Welcome to PF! :smile:

    (have a square root: √ and a square: ² :wink:)

    Do you mean 2(x² + y²) ≥ (|x| + |y|)²?

    hmm … let's keep it simple … :rolleyes:

    Hint: 2(x² + y²) = |x|² + |x|² + |y|² + |y|² :smile:
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