$f(z_1,z_2,z_3)= \frac{1}{z_1+z_2}$ 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).
Explicit extension of a function that is not holomorphic only on an analytic set of codimension 2.
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complex-analysis
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5That 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. – 2012-02-24
1 Answers
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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.