# Can the pole of a analytic function be of rational order

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

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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 $z=0$.
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