Talk:Meromorphic function

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must the set of poles be finite? -- Tarquin

Nope. 1/sin(z) is meromorphic. AxelBoldt

how big can the set of poles be

I don't think I completely understand the definition.

Can there be an infinite amount of poles? And if so, do they have to be countable? --anon

Yes, they can be an infinite set. Hopefully my rewrite shows that.

They really do have to be isolated? --anon Yes, by definition. If they are not isolated, it is impossible to prove that a meromorphic function is a ratio of two holomorphic functions. Oleg Alexandrov 17:28, 11 August 2005 (UTC)[reply]

the amount of poles of a meromorphic function must not be countable?

http://mathworld.wolfram.com/MeromorphicFunction.html

here they speak of

if this a different definition or is it some non trivial theorem that if your definition is true, the number of poles is countable

thanks --anon

I don't see any contradiction between our page and mathworld. The poles of a meromorphic function must be isolated, by definition. Now, one can prove a theorem saying that a set of isolated points is finite or countable. So, is it the proof of this theorem that you are interested in? Oleg Alexandrov 20:44, 11 August 2005 (UTC)[reply]

Poles of sin(1/z)

Outside of every neighbourhood of origin the function sin(1/z) is bounded in bounded sets. Thus origin cannot be it's accumulation point of poles. Perhaps here is a typo and it should be 1/sin(1/z)? —Preceding unsigned comment added by J Kataja (talk • contribs) 10:01, 9 January 2008 (UTC)[reply]

You're right. The function

\sin(1/z)

is actually not meromorphic in the origin, but for a different reason. -- EJ (talk) 12:00, 9 January 2008 (UTC)[reply]