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Can i have an example of connected Reinhardt domain in $C^n$ which contains zero. But it is not complete.

Complete means: For $w= (w_1,..w_n)\in D$, if $z$ is such that $|z_j|\leq |w_j$ for all $j$ implies $z\in D$.

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$\{(z_1,\ldots,z_n) \mid |z_i| < 1 \} \cup \{(z_1,\ldots,z_n) \mid 2 < |z_i| < 3 \}$ would be an example –  Jacob Ikabruob Apr 11 '12 at 8:11
    
Sorry but i cant see: is this connected?? –  zapkm Apr 11 '12 at 8:31

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up vote 2 down vote accepted

Something like $\{ (z,w) \in \mathbb{C}^2 : |z| < 1, |w| < 1 \} \setminus \{ (z,w) \in \mathbb{C}^2 : 1/3 \le |z| \le 2/3, |w| \le 1/2 \}$ would work. Draw a picture in the $(|z|, |w|)$-plane to see what it looks like.

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I think: $(0,0)\in \mathbb C^2$ doesn't lie in this. –  zapkm Apr 11 '12 at 17:27
    
How so? The origin certainly is inside the bidisc (the first set) and is not inside the piece that is removed. –  mrf Apr 11 '12 at 19:08
    
Yeah, sorry my mistake.. thanks –  zapkm Apr 11 '12 at 20:07
    
You might consider upvoting as well as accepting the answer. I don't think the downvote was warranted. –  mrf Apr 12 '12 at 7:09
    
I tried, but i can't upvote..Please edit something[even you can change fullstop and and bracket at the last] in the answer, then upvote will be allowed for me... Thnx. –  zapkm May 5 '12 at 13:19

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