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general >> complex analysis >> analytic constant complex functions
(Message started by: fightnu on Mar 4th, 2009, 5:14pm)

Title: analytic constant complex functions
Post by fightnu on Mar 4th, 2009, 5:14pm
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

:(

Title: Re: analytic constant complex functions
Post by Eigenray on Mar 4th, 2009, 10:38pm
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) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0 as z http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif ai, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.giff(z)dz = 0 along any closed curve in D \ {ai}.

Let F(z,w) = (f(z) - f(w))/(z-w).  For fixed w http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0, this function satisfies the conditions, so if we fix a circle around 0,
f(w) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif f(z)/(z-w) dz
holds for w http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0.  But the RHS is an analytic function of w inside the circle, so the singularity has been removed.

Title: Re: analytic constant complex functions
Post by MonicaMath on Mar 5th, 2009, 12:34pm
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

Title: Re: analytic constant complex functions
Post by Eigenray on Mar 6th, 2009, 3:26pm
To show that h(z) = z2 g(z) is a polynomial of degree http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 2, you can use the generalized Liouville's theorem you asked about [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=complex;action=display;num=1235678200]before[/link].  And since h(0) = h'(0) = 0, g(z) must be a constant.



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