

Title: analytic constant complex functions Post by fightnu on Mar 4^{th}, 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 4^{th}, 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 a_{i}, but that at each point, (za_{i}) 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 a_{i}, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.giff(z)dz = 0 along any closed curve in D \ {a_{i}}. Let F(z,w) = (f(z)  f(w))/(zw). 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)/(zw) 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 5^{th}, 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 6^{th}, 2009, 3:26pm To show that h(z) = z^{2} 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/cgibin/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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