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I Complex Analysis Radius of Convergence.

  1. Jun 21, 2016 #1
    Hello, I have two questions regarding the Radius of convergence.

    1. What should we do at the interval (R-eps, R)
    2. It used definition to prove radius of convergence, but I am not sure if it is necessary-sufficient condition of convergence. I get that this can be a sufficient condition but not sure of the necessity

    Attached Files:

  2. jcsd
  3. Jun 22, 2016 #2


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    If I understand the question, this may help -- The ε was picked arbitrarily small so the proof shows convergence all the way to ε = 0. So there really is no specific interval (R-ε, R) to worry about.
  4. Jun 22, 2016 #3

    hm the author used (L+eps)|Z|<1 to argue that L^n*|Z|^n < (L+eps)^n|Z|^n < r^n whereas r<1

    but the thing is as I showed at the uploaded png file (if the derivation is correct), it is clear that |Z|= R-eps satisfies the condition |Z| < R and if |Z|= R-eps, r = 1.

    which implies (L+eps)|Z|<1 cannot be true. for all |Z|<R
    Ah, also I made a mistake. It should be R+R^2*eps-eps-R*eps^2

    hmm do you agree?
    Last edited: Jun 22, 2016
  5. Jun 22, 2016 #4

    maybe I screwed up at somewhere.

    Attached Files:

  6. Jun 22, 2016 #5


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    Maybe not. But the point is that ε >0 is arbitrarily small (or r is arbitrarily near 1) so the the interval you are worried about shrinks down to nothing.
  7. Jun 22, 2016 #6
    Although we ignore the interval, the argument that (L+eps)|Z|<1 does not hold true if we take |Z| = R - eps. (I guess) It just bugs me and I am stuck at that page ;(

    and my last conclusion R<1 doesn't seem right at all too.
  8. Jun 23, 2016 #7

    Stephen Tashi

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    It isn't clear to me what is to be proven. We've only been shown the last part of the text that states "the theorem".
  9. Jun 23, 2016 #8
    I am sorry.

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