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Topic: analytic constant complex functions (Read 9836 times) 

fightnu
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analytic constant complex functions
« on: Mar 4^{th}, 2009, 5:14pm » 
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hello .... could anyone give me a hint on this problem " if g(z) is a bounded analytic function on the complex plane except at z=0, I need to prove that g should be constant " ? I tried to prove that a new function,say h(z)=(z^2 )g(z) has the form h= c z^2, so g should be constant,,, but I couldn't prove this;; can any one help me in that


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Eigenray
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Re: analytic constant complex functions
« Reply #1 on: Mar 4^{th}, 2009, 10:38pm » 
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What do you know about isolated singularities? You need to show that z=0 is neither a pole nor an essential singularity. The general result is that if f(z) is analytic and bounded in a neighborhood of a, then z=a is a removable singularity. There is a version of Cauchy's theorem that says: Suppose f(z) is analytic in some disk D, with the exception of a finite set of a points a_{i}, but that at each point, (za_{i}) f(z) 0 as z a_{i}, then f(z)dz = 0 along any closed curve in D \ {a_{i}}. Let F(z,w) = (f(z)  f(w))/(zw). For fixed w 0, this function satisfies the conditions, so if we fix a circle around 0, f(w) = f(z)/(zw) dz holds for w 0. But the RHS is an analytic function of w inside the circle, so the singularity has been removed.


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MonicaMath
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Re: analytic constant complex functions
« Reply #2 on: Mar 5^{th}, 2009, 12:34pm » 
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thenk you for replay I t is cleare using the idea of removable singularities. But I'm trying to prove it without using this idea as I mentioned it in the first post of the problem. anyway thank you


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Eigenray
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Re: analytic constant complex functions
« Reply #3 on: Mar 6^{th}, 2009, 3:26pm » 
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To show that h(z) = z^{2} g(z) is a polynomial of degree 2, you can use the generalized Liouville's theorem you asked about before. And since h(0) = h'(0) = 0, g(z) must be a constant.


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