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Homework Help: Describing Set of Complex Numbers for which This Converges

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Describe the set of all [tex]z \in \mathbb{C}[/tex] such that the series [tex]\sum_{n=1}^{\infty} (1-z^2)^n[/tex] converges

    2. Relevant equations

    Basic analytic techniques.

    3. The attempt at a solution

    This is from a graduate complex analysis class, and I just have a feeling my answer is too obvious to be correct...

    Basically, I used the necessary condition that the the sequence given by [tex]a_n = |(1-z^2)^n| [/tex] must tend to 0 as n tends to infinity. In other words, we know the modulus of the terms must go to zero. Then, viewing the series as a geometric series, [tex]|(1-z^2)| < 1[/tex]. From there it was a few simple steps to conclude the set is [tex]\lbrace z\in \mathbb{C} \: | \: 0<|z|<2 \rbrace [/tex].

    Even as I read over this, it makes sense. But like I said, it's much too simple to be correct...what am I forgetting here?
     
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  3. Sep 16, 2010 #2

    Dick

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    Your 'simple steps' are wrong. Whatever they are. |(1-z^2)|<1 looks ok. 0<|z|<2 doesn't. z=3/2 doesn't converge. z=i doesn't converge. I'd write z in polar form and try that again.
     
  4. Sep 16, 2010 #3

    Office_Shredder

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    The reasoning described for determining the conditions necessary is missing a detail as well. Just because an goes to zero does not mean that the series will converge necessarily. It only works in this case specifically because we have a geometric series (and you can write out an explicit solution for the partial sums to see that it does converge)
     
  5. Sep 16, 2010 #4
    Ah of course! A silly mistake...thanks for pointing it out.
     
  6. Sep 19, 2010 #5
    Actually, it wasn't a silly mistake. Upon revisiting this problem, for some reason I'm not seeing the answer. Where should I go after [tex]\vert 1-z^2\vert<1 \leftrightarrow\vert 1- r^2e^{2i\theta} \vert < 1[/tex]? I've tried using the triangle inequality and I've also tried expanding into sine's and cosine's to no avail. I know this must have some relatively simple trick, but I don't know what...a nudge in the right direction would be greatly appreciated.
     
  7. Sep 19, 2010 #6
    [tex] -1 < 1- z^{2} < 1[/tex]
    [tex] -2 < -z^{2} < 0[/tex]
    [tex] 0 <z^{2} < 2[/tex]
    [tex] 0< \left|z^{2} \right| = \left|z \right|^{2} < 2[/tex]

    Can you complete it ?
     
  8. Sep 19, 2010 #7
    Well your last line implies [tex]\vert z \vert < \sqrt{2} [/tex] but how can we get from [tex]\vert 1-z^2 \vert <1 [/tex] to [tex] -1 < 1-z^2 <1[/tex]? The modulus of complex numbers and the absolute value of real numbers do not share this property.
     
  9. Sep 19, 2010 #8

    Dick

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    You are quite right about that. Write z=r*exp(i*theta) and try to write out what |1-z^2|<1 means in terms of r and theta.
     
  10. Sep 19, 2010 #9
    I need to stop browsing for the day. I am posting nonsense. Please, forgive my incorrect solution.

    I do not know what possessed me to make such a claim.

    I was even ordering complex numbers :surprised.
     
  11. Sep 20, 2010 #10
    Right. So I have [tex]\vert 1-r^2e^{2i\theta} \vert < 1[/tex]. This is where I get stuck. In real calculus one can easily get rid of those pesky vertical bars, but here, I don't know where to go. Like I said above, I've tried invoking the triangle inequality but I come to something which doesn't help since it is a lower bound. I've also tried expanding the inside of the modulus into sine's and cosine's but that gets messier, then a little nicer, then messier.
     
  12. Sep 20, 2010 #11
    Don't worry about it; I appreciate your help. And after all, it's almost the same error I made to begin with!
     
  13. Sep 20, 2010 #12

    Dick

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    Yes, expand the inside in terms of sin's and cos's. You want to use that if a and b are real then |a+bi|=sqrt(a^2+b^2).
     
  14. Sep 21, 2010 #13
    Having solved the problem (finally), I see now my main (very, very silly) mistake: [tex]r^2e^{2i\theta} = r^2e^{i(2\theta)} = r^2(cos(2\theta) + isin(2\theta)) \neq r^2(cos^2(\theta) + isin^2(\theta))[/tex] which is the expansion I had been doing. So, it turns out the set is the interior of a lemniscate whose equation is immediate upon *properly* expanding into sine's and cosine's. Thanks for the help everyone.
     
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