Talk:Liouville's theorem (complex analysis)

Latest comment: 14 years ago by 130.235.35.193 in topic Domination

Correction Aug 2007 edit

In the remarks at the bottom of the page, shouldn't it be:

.. So sum{k=0,n} a_k z^k for |z| sufficiently large? Not sum{k=n+1,inf} a_k z^k 70.23.234.252 15:36, 19 August 2007 (UTC)Reply

yes, you're right. my bad. Mct mht 17:36, 19 August 2007 (UTC)Reply

Correction, June 2008 edit

In the section about the image of an entire function being dense in C it is not "absurd" that g is constant. It simply implies that f is also constant and thus shows that the image of an entire function f is dense if and only if f is constant. This should be reformulated. —Preceding unsigned comment added by Pere Callahan (talkcontribs) 00:03, 16 June 2008 (UTC)Reply

Domination edit

It is not clear to me why "if g(z)=0, then f(z)=0" implies "f(z)/g(z) can be analytically extendend to zeroes of g". This is not true in general: g could have zeroes of higher order than f. Of course, in that case f would not be bounded by g, by elementary calculations, but that would make the proof less elegant. --Aleph4 (talk) 12:36, 21 April 2009 (UTC)Reply

There is a simple correction which I have added. Simply use the fact that h is bounded by hypothesis and thus all singularities must be removable. —Preceding unsigned comment added by 130.235.35.193 (talk) 11:50, 17 June 2009 (UTC)Reply