wu :: forums - Print Page


    
      
        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 4^th, 2009, 5:14pm)

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, (z-a_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))/(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 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) = 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.