Can a pole of an analytic function have a rational number as its order?
Can the pole of a analytic function be of rational order
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complex-analysis
1 Answers
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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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0Well, strictly speaking, $1/z$ is not analytic in any neighbourhood of $0$ either. Perhaps you mean a punctured disc? – 2011-11-06
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0@Zhen Lin Yes, of course. – 2011-11-06
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0Sir, 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? – 2011-11-07
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0Yes of course, I meant positive integer. – 2011-11-07
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0Please explain 'punctured' neighbourhood. – 2011-11-09
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0A punctured neighbourhood of $z_0\in\mathbb{C}$ is a neighbourhood of $z_0$ with the point $z_0$ deleted, like $\{z\in\mathbb{C}:0<|z-z_0|
0$. – 2011-11-09