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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


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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


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    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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