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can we say that essential singularities has order like poles ? I mean ;

we know e^(1/z) has an essential singularity at z=0 then do I need to say e^(1/z^3) has essential singularity at z=0 with order 3 ?

thanks

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    No. The term order only applies to removable singularities. Your functions both have "infinite order" poles; $e^{1/z^3}$ certainly does not have a singularity of order 3 at 0.2012-06-03
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    it does not really give any computational advantages, as the behaviour of the function remains equally "bad" in the neighbourhood of the essential singularity2012-06-04

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