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Triangle inequality

  1. Apr 12, 2006 #1
    Edit: Nevermind i got it, thanks anyway

    for complex numbers z1 and z2

    prove that [tex] |z_{2}| - |z_{1}| \leq |z_{2} - z_{1}| [/tex]

    the left hand side becomes
    [tex] \sqrt{x_{2}^2 + y_{2}^2} - \sqrt{x_{1}^2 + y_{1}^2} [/tex]

    the right hand side becomes
    [tex] \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2} [/tex]

    now i tried squaring both sides and i get
    left hand side
    [tex] x_{2}^2 + y_{2}^2 + x_{1}^2 + y_{1}^2 - 2 \sqrt{(x_{2}x_{1})^2 + (x_{2}y_{1})^2 + (x_{1}y_{2})^2 + (y_{1}y_{2})^2} [/tex]

    right hand side
    [tex] (x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 [/tex]
    i put the two of them not equal to eah other and reduce and i ended up with
    [tex] x_{2}^2 y_{1}^2 + x_{1}^2 y_{2}^2 \neq 2x_{1} x_{2} y_{1} y_{2} [/tex]
    im stuck now...

    please help
    is there a simpler... less tedious way of doing this... by the way?
    Last edited: Apr 12, 2006
  2. jcsd
  3. Apr 12, 2006 #2


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    The less tedious way is to use the triangle inequality: [itex]|z-w|\leq |z|+|w|[/itex]
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