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Can a pole of an analytic function have a rational number as its order?

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1 Answer 1

No. Isolated singularities of analytic functions are either removable, poles (of integer order) or essential.

You may wonder: what about $1/\sqrt{z}\,$? Could not we say that $z=0$ is a pole of order $1/2$? No, because $z=0$ is not an isolated singularity of $f$; it is a branching point. It is impossible to define $1/\sqrt{z}$ in such a way that it is analytic in a punctured neighbourhood of $z=0$.

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Well, strictly speaking, $1/z$ is not analytic in any neighbourhood of $0$ either. Perhaps you mean a punctured disc? –  Zhen Lin Nov 6 '11 at 21:55
    
@Zhen Lin Yes, of course. –  Julián Aguirre Nov 6 '11 at 22:19
    
Sir, Mr. julian aguirre, as u have mentioned that poles can have integer order. i think that the order of pole can be positive integer only. if it will be of negative order then it will turn into zero of the function. Am i right sir? –  GAURAV SHARMA Nov 7 '11 at 16:38
    
Yes of course, I meant positive integer. –  Julián Aguirre Nov 7 '11 at 16:46
    
Please explain 'punctured' neighbourhood. –  GAURAV SHARMA Nov 9 '11 at 15:55

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