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Prove ||x|-|y||≤|x-y|, where x and y are complex

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    If x, y are complex, prove that

    | |x| - |y| | ≤ |x - y|

    2. Relevant equations

    If x = a + ib, |x| = √(a2+b2)

    |x + y| ≤ |x| + |y| (works for both complex and real numbers)

    3. The attempt at a solution

    Maybe start with the left side

    | |x| - |y| | = | |x| + (-|y|) ||x| + |-|y|| = |x| + |y|

    ....... Maybe almost there is I can show |x| + |y| ≤ |x - y| ........


    How can I not get this problem?

    feels-bad-man-.jpg
     
  2. jcsd
  3. Oct 11, 2011 #2

    micromass

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    Use that

    [tex]|x|=|(x-y)+y|[/tex]
     
  4. Oct 12, 2011 #3
    screen-capture-2-23.png


    feelsgoodman.png









    Stay around here. I'm gonna hit you up with another question later.
     
  5. Oct 12, 2011 #4

    micromass

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    What you wrote down is incorrect. Specifically, the first inequality is wrong.

    You need to prove two things:

    [tex]|x|-|y|\leq |x-y|~\text{and}~|x|-|y|\geq -|x-y|[/tex]

    These two together would imply your inequality.
     
  6. Oct 12, 2011 #5
    screen-capture-3-29.png
     
  7. Oct 12, 2011 #6

    micromass

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    This is better :smile:
     
  8. Oct 12, 2011 #7
    Think you could help me with the Schwartz equality problem? My homework is due in 1 hour and that's the only one I have left.
     
  9. Oct 12, 2011 #8

    micromass

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    Well, what is the problem and what did you try?
     
  10. Oct 12, 2011 #9
    Figuring out under what condition equality holds in the Schwartz inequality. (I know the answer is when a and b are linearly independent)

    I let aj = xj + iyj, bj = uj + ivj

    and after some simplification came up with

    ∑(xj2+bj2)(uj2+vj2) = ∑(xj2+bj2)∑(uj2+vj2)

    which somehow shows that a is a scalar multiple of b. Not sure how, though.
     
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