- #36

#### Svein

Science Advisor

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I did not say that ##1/z## was continuous. I said that in complex analysis we do not have a problem with ##1/z##. It is called aOffice_Shredder said:@Svein I'm not sure how that addresses the question of whether ##1/z## is a continuous function though, which is the analogous question to ##1/x## in ##\mathbb{R}##

*pole*.