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Topic: analytic constant complex functions (Read 10067 times) |
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fightnu
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analytic constant complex functions
« on: Mar 4th, 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 4th, 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 ai, but that at each point, (z-ai) f(z)  0 as z ai, then  f(z)dz = 0 along any closed curve in D \ {ai}.
Let F(z,w) = (f(z) - f(w))/(z-w). For fixed w  0, this function satisfies the conditions, so if we fix a circle around 0,
f(w) = f(z)/(z-w) 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 5th, 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 6th, 2009, 3:26pm » |
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To show that h(z) = z2 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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