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$f(z_1,z_2,z_3)= \frac{1}{z_1+z_2}$ it's an holomorphic function in $$\mathbb{C}^3\setminus\{(0,0,z_3)\mid z_3 \in \mathbb{C}\} $$ Since the set where is not defined is analytic of codimension 2, by the Riemann extension theorem it's possible to extended even if is not locally bounded. Can I see explicitly how it is the extension? For example at a point (0,0,1).

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That should be $\mathbb{C}^3 \setminus \{(z_1, z_2, z_3) \mid z_1+z_2=0 \}$. The singular locus has co-dimension one, not two. –  WimC Feb 24 '12 at 16:41

1 Answer 1

Recording a CW-answer: $f$ does not admit holomorphic continuation beyond its original domain of definition. The extension theorem which OP wanted to apply is not applicable here, as pointed out by @WimC.

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