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

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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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    A 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| for any r>0.2011-11-09