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Complex analysis

  1. Jun 5, 2008 #1
    I have the following problems
    (1)Consider the series ∑z^n,|z|<1 z is in C
    I thik this series is absolutely and uniformly comvergent since the series ∑|z|^n is con vergent for |z|<1,but I have a book saying that it is absolutely convergent,not uniformly.........i am confused...
    (2)for the function f(z)=1/√(z-1),z=1 is a (a)pole (b)an essential singularity ?
    I think it is an essential singularity since if it is a pole,say of order m then m is a positive integer and we can write f(z)=g(z)/(z-1)^m, where g(z) is analytic at z=1 and g(1)≠0,
    but the given function cannot be written in this way,but the answer is given pole,i am again confused...
    Can anybody help me?
     
  2. jcsd
  3. Jun 5, 2008 #2

    nicksauce

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    1) It is indeed not uniformly convergent. You really just have to apply the definition of uniform convergence to see it.
    2) It is a fact that a singularity is only an essential singularity if and only if the limit does not exist and it is not infinity. The definition of a pole is we can write f(z) = g(z) / (z-a)^n, for some holomorphic g, and the order of the pole is the smallest n for which this is true. If f(z) = z^-1/2, then this is satisfied by taking g(z) = z^1/2 and n = 1.
     
  4. Jun 5, 2008 #3

    nicksauce

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    For #1 we had a similar question on a homework set...
    You can see the solution here, http://www.math.mcgill.ca/jaksic/MATH249.html
    go to Assignment 5 solutions, and it's the first question.
     
  5. Jun 5, 2008 #4
    Thank u very much for your kind help,but I couldn't understand the second answer,will u please explain it a bit more ?
     
  6. Jun 5, 2008 #5

    nicksauce

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    I don't know what more I can do but restate what I wrote before. I don't have my complex analysis book with me right now, but wikipedia define a pole as follows:

    Suppose f has a singularity at x = a... If there exists a holomorphic function g(z) so that we can write
    [tex]f(z) = \frac{g(z)}{(x-a)^n}[/tex], then a is a pole of order n, where n is the smallest number for which f can be written like this.

    Now consider [tex]f(z) = z^{-\frac{1}{2}}[/tex], which has a singularity at z = 0. We can write
    [tex]f(z) = \frac{z^ {\frac{1}{2}}}{z}[/tex]
    Therefore z=0 is a pole of order 1.
     
  7. Jun 5, 2008 #6
    but g(0)=0,how is this satisfying the criteria for being a pole?
     
  8. Jun 5, 2008 #7

    nicksauce

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    Hmm this is true... I retract my answer, and hope someone else can help you.
     
  9. Jun 13, 2008 #8
    Neither. Both poles and essential singularities require the relevant function to be holomorphic on a deleted neighborhood of the singularity. [tex]z^{-\frac{1}{2}}[/tex] isn't even continuous on one of these neighborhoods. In descriptive terms, however, it would look like half of a simple pole stretched around.
     
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