Re: question and help please
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Fri, 31 Mar 2006 15:47:12 +0100
On 31-03-2006 15:31, David C. Ullrich wrote:
if f : G -------> C is analytic except for poles , show that the set ofLet z_0 be a point of G. Then either _f_ is defined at z_0 or z_0 is a
poles of f cannot have a limit point in G ..
pole of _f_. If _f_ is defined at z_0, then there's a power series
sum_n a_n(z - z_0)^n which converges to f(z) in some disk centered at
z_0. So, that disk has no poles of _f_. If z_0 is a pole of _f_, then
there's some natural number _m_ and some power series
sum_n a_n(z - z_0)^n such that, for some r > 0, the disk D(z_0,r) is
contained in G and that, for each z in D(z_0,r)\{z_0},
f(z) = (sum_n a_n(z - z_0)^n)/(z - z_0)^m.
So, z_0 is the only pole of _f_ in D(z_0,r). This proves that there's
some neighborhood of z_0 such that it either has no poles or that z_0
itself is the only pole there.
Actually this doesn't really require proof. A pole is _by definition_
a certain sort of _isolated singularity_, so by definition if z_0
is a pole then there exists a neighborhood of z_0 in which there
are no other poles.
That's the definition that I use too, but I assumed that the OP's
definition of pole of a function _f_ was a point z_0 such that, for some
natural number _m_ and some power series sum_n a_n(z - z_0)^n,
f(z) = (sum_n a_n(z - z_0)^n)/(z - z_0)^m
near z_0. Of course, this *implies* that poles are isolated.
Best regards,
Jose Carlos Santos
.
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