# Complex analysis

by matheater
Tags: analysis, complex
 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?
 HW Helper Sci Advisor 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) / (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.
 HW Helper Sci Advisor 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.
P: 7

## Complex analysis

 Quote by nicksauce 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 ?
 HW Helper Sci Advisor 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 $$f(z) = \frac{g(z)}{(x-a)^n}$$, then a is a pole of order n, where n is the smallest number for which f can be written like this. Now consider $$f(z) = z^{-\frac{1}{2}}$$, which has a singularity at z = 0. We can write $$f(z) = \frac{z^ {\frac{1}{2}}}{z}$$ Therefore z=0 is a pole of order 1.
P: 7
 Quote by nicksauce 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 $$f(z) = \frac{g(z)}{(x-a)^n}$$, then a is a pole of order n, where n is the smallest number for which f can be written like this. Now consider $$f(z) = z^{-\frac{1}{2}}$$, which has a singularity at z = 0. We can write $$f(z) = \frac{z^ {\frac{1}{2}}}{z}$$ Therefore z=0 is a pole of order 1.
but g(0)=0,how is this satisfying the criteria for being a pole?
 P: 8 Neither. Both poles and essential singularities require the relevant function to be holomorphic on a deleted neighborhood of the singularity. $$z^{-\frac{1}{2}}$$ isn't even continuous on one of these neighborhoods. In descriptive terms, however, it would look like half of a simple pole stretched around.