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Complex analysis

by matheater
Tags: analysis, complex
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matheater
#1
Jun5-08, 10:00 AM
P: 7
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?
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nicksauce
#2
Jun5-08, 10:21 AM
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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.
nicksauce
#3
Jun5-08, 10:24 AM
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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.

matheater
#4
Jun5-08, 01:56 PM
P: 7
Complex analysis

Quote Quote by nicksauce View Post
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 ?
nicksauce
#5
Jun5-08, 02:07 PM
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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.
matheater
#6
Jun5-08, 02:11 PM
P: 7
Quote Quote by nicksauce View Post
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?
nicksauce
#7
Jun5-08, 02:16 PM
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Hmm this is true... I retract my answer, and hope someone else can help you.
Sqens
#8
Jun13-08, 06:03 PM
P: 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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