The contents of the Reinhardt domain page were merged into Annuity on 2020-11-20. For the contribution history and old versions of the merged article please see its history.
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Example in 1 dimensional case
It seems to me that the following sentence is not exact:
"Note that in one complex variable, a logarithmically convex Reinhardt domain is simply a disc."
I think it should be anulus instead of disc. But if we add the condition of containing the origin then it has to be a disc.
What do you think? That is actually the first time I study Reinhardt domain, so I don't want to do any change myself for the moment.
Bongilles (talk) 22:15, 30 January 2013 (UTC)[reply]
I basically thought the same thing. Based on the definition presented here, the Reinhardt domain appears to be up to n open annuluses containing the zn. This could include a disc if one of the zn is the origin. It might be useful to mention annuluses (annuli?) in the article, if this interpretaion is correct (I've first encountered Reinhardt domains just now by reading this article). It may be that to be "logarithmically convex" a 1d Reinhardt domain must be a disk, which would make the article correct (I don't know what logarithmically convex means).--Wikimedes (talk) 19:56, 8 August 2014 (UTC)[reply]
I see the same issue, but I also don't know this subject well. Looking at the referenced EoM article, it looks like what's missing the concept of a "complete" Reinhardt domain. Also, Reinhardt domains are "domains of holomorphy" for functions, which seems to include Laurent series. Taylor series (I guess) have complete Reinhardt domains as their domain of convergence, and the statement in question should read (I think) "...a logarithmically convex, complete Reinhardt domain is simply a disc." Baccala@freesoft.org (talk) 18:30, 2 February 2020 (UTC)[reply]
