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Triangle inequality w/ Complex Numbers

  1. Sep 9, 2010 #1
    given z, w[tex]\in[/tex]C, and |z|=([conjugate of z]z)1/2 , prove ||z|-|w|| [tex]\leq[/tex] |z-w| [tex]\leq[/tex] |z|+|w|

    I squared all three terms and ended up with :
    -2|z||w| [tex]\leq[/tex] |-2zw| [tex]\leq[/tex] 2|z||w|
    I know this leaves the right 2 equal to each other but i figured if i show that since there exists a z[tex]\geq[/tex]w[tex]\geq[/tex]0, then |z-w| > |z|+|w| would be impossible.

    Can someone tell me if they think I screwed up or I am not done?
     
  2. jcsd
  3. Sep 9, 2010 #2

    jgens

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    You assumed what you're trying to prove, so that doesn't really work.

    Note that for any two complex numbers z and w, we have that zw* + z*w ≤ 2|z||w|. Now,
    |z+w|2 = (z+w)*(z+w) = (z*+w*)(z+w) = zz* + zw* + z*w + ww* ≤ |z|2 + 2|z||w| + |w|2 = (|z|+|w|)2. This proves that |z+w| ≤ |z|+|w|. Can you prove the rest from here?

    Edit: Fixed less than or equal to signs.
     
  4. Sep 9, 2010 #3
    Thank You for the help. I have redone the proof. here is what I've got:

    It is obvious that |z|-|w|[tex]\leq[/tex]|z|, which yields |z|-|w|[tex]\leq[/tex]|z-w|. From this, we have that ||z|-|w||[tex]\leq[/tex]|z-w|. This shows the left inequality. The right-side I will show by contradiction. Assume |z-w|>|z|+|w|, this implies that w<0[tex]\Rightarrow[/tex]|z-w|=|z|+|w|>|z|+|w|, which is a contradiction showing that |z-w| is at most equal to |z|+|w|. Done.

    Could anyone let me know if they think this is ok?
     
  5. Sep 9, 2010 #4

    jgens

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    Gold Member

    How do you use the first inequality to arrive at the second one? I'm not following your reasoning here.

    You can do better than that. Use the fact that |z-w| ≥ |z|-|w| and |z-w| ≥ |w|-|z| to arrive at your conclusion.

    First, you need to remember that you're working with complex numbers here, so you can't say anything concerning their ordering. Second, you've assumed a lot of things which aren't necessarily true so I would scrap this approach. Everything that you want to derive for this part follows from the inequality |z+w| ≤ |z|+|w|.
     
  6. Oct 8, 2010 #5
    [tex]||z|-|w|| \leq |z-w| \leq |z|+ |w|[/tex]

    pf: Left Side:
    [tex]|z| \leq |z-w+w|[/tex]
    [tex]|z|-|w| \leq |z-w|[/tex]
    [tex]||z|-|w|| \leq |z-w|[/tex]

    Right Side:
    [tex]|z| \leq |z| + 2|w|[/tex]
    [tex]|z-w| \leq |z| + |w|[/tex]

    Hows that?
     
  7. Oct 9, 2010 #6
    I'm not sure if you are hiding the use of the actuall formula for |z| in your proof but to me, it looks like you are just assuming the result is true to prove it is true.
     
  8. Oct 9, 2010 #7
    [tex]|z| = (z\bar{z})^{\frac{1}{2}}[/tex]

    Is that what you were asking for?

    Also, I now see that the right side is true by the triangle inequality.
     
    Last edited: Oct 9, 2010
  9. Oct 10, 2010 #8
    You can't say it's true by the triangle inequality because the triangle inequality is what you are proving. You need to prove that your definition of |z| implies [tex] ||z|-|w||\leq|z-w|\leq|z|+|w|[/tex].
     
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