
#1
Jun508, 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/√(z1),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)/(z1)^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
Jun508, 10:21 AM

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P: 1,276

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) / (za)^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
Jun508, 10:24 AM

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P: 1,276

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
Jun508, 01:56 PM

P: 7

Complex analysis 



#5
Jun508, 02:07 PM

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P: 1,276

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)}{(xa)^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
Jun508, 02:11 PM

P: 7





#7
Jun508, 02:16 PM

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P: 1,276

Hmm this is true... I retract my answer, and hope someone else can help you.




#8
Jun1308, 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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