Complex analysis

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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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.
 
  • #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.
 
  • #4
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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.
Thank u very much for your kind help,but I couldn't understand the second answer,will u please explain it a bit more ?
 
  • #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.
 
  • #6
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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.
but g(0)=0,how is this satisfying the criteria for being a pole?
 
  • #7
nicksauce
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Hmm this is true... I retract my answer, and hope someone else can help you.
 
  • #8
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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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