``` wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) 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) dzholds 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 replayI 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. Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board