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Complex numbers

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Hey guys.
    I got this question and I don't know where to start.
    There is a little hint, which is to show that the numerator is smaller then 5 and the enumerator is bigger then 3(1-abs(z)).
    Any idea guys?

    2. Relevant equations

    3. The attempt at a solution

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  3. Nov 23, 2008 #2


    Staff: Mentor

    What exactly is the question? Are you supposed to prove that
    [tex]|\frac{3 + z^{3n} - |z|}{3 - z^{5n} + x^{4n} -z}| \leq \frac{5}{3(1 - |z|)}[/tex]?
    If so, that's not much of a hint, because it's essentially what you need to do.
    If that's not what you need to do, then I have no clue.
  4. Nov 24, 2008 #3
    Yeah, that's the question.

    Well, how do you do that?
  5. Nov 24, 2008 #4


    Staff: Mentor

    I don't know how to prove this. I've spent a little time, but nothing seems to pop out at me as an approach. Before I spend any more time on it, is there any more to this problem? All I know so far is that z is complex, |z| <= 1, and that you are to prove the inequality above, which you neglected to mention in your first post.

    Is there anything else you have neglected to mention?
  6. Nov 24, 2008 #5


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    A major problem is that is it NOT true! If z= 1, the denominator on the right side is 0 while the denominator on the left is not. If z is close to 1, then the right side will be extremely large while the left side is close to 3/2.
  7. Nov 24, 2008 #6

    I didn't posted the entire problem because it's in Hebrew :smile:.
    I'm sorry if I didn't make my self clear. Yeah, I need to prove this inequality.
    I didn't neglect any thing else.

  8. Nov 24, 2008 #7


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    I don't think that's really a problem because of the direction of the inequality.... surely [tex]\frac{3}{2} \leq \infty[/tex] ?
  9. Nov 24, 2008 #8


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    You did neglect something....are there any restrictions on [itex]n[/itex]? Surely it has to be a positive integer?
  10. Nov 24, 2008 #9
    This is the entire problem (in Hebrew), I don't see any restrictions on n.
    But lets say it's positive, any ideas?


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  11. Nov 24, 2008 #10
    The things you need to know:

    1: Since |z|≤1, we have for any positive integer k that |z|^(kn) ≤ |z| (assuming that n is positive)
    2: Use the triangle inequality -- |a+b| ≤ |a| + |b| -- and its counterpart, |a-b| ≥ ||a| - |b||
    3: As the problem hint suggests, prove the absolute value of the numerator is at most 5 and that the absolute value of the denominator is at least 3 - 3|z|.
  12. Nov 25, 2008 #11
    Ok, but in the problem, it's not |z|^(3n), it's z^(3n).
    Lets talk about the numerator, I know that 3 > 3-|z| >= 2 so what I need to prove is that z^(3n) <= 2, right?
    Well, how can I prove just a thing?

  13. Nov 25, 2008 #12
    |z^(3n)| = |z|^(3n)

    No, you're trying to provide an upper bound on the magnitude of the numerator, so proving 3 ≥ 3-|z| ≥ 2 is not helpful. Also, you do not need to prove that z^(3n) ≤ 2, and indeed, since z is a complex number, it may not be comparable to 2 (a≤b only makes sense if a and b are both real).

    What you need to prove is: |3 + z^(3n) - |z|| ≤ 5

    The first step here is to apply the triangle inequality, so we have:

    |3 + z^(3n) - |z|| ≤ |3| + |z^(3n)| + |-|z|| = 3 + |z|^(3n) + |z|.

    so it suffices to show that 3 + |z|^(3n) + |z| ≤ 5. Can you see how to do that with what I've showed you?

    Obvious hint:
    Note that |z|^(3n) ≤ |z| ≤ 1
  14. Nov 25, 2008 #13
    Yeah, I can see that, 10x a lot.
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