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I am learning about conformal field theories and have a question about poles of order > 1.

If a conformal transformation acts as

[tex]z \rightarrow f(z)[/tex],

f(z) must be both invertable and well-defined globally. I want to show that f(z) must have the form

[tex]f(z) = \frac{az+b}{cz+d}[/tex].

Here is what I have so far. A general complex function f(z) can have branch cuts, essential singularities and multiple poles of varying order.

If f(z) has a branch cut then it is not well-defined in the vicinity of the branch point---i.e. it is multi-valued.

If f(z) has an essential singularity, then the transformaiton is not invertable.

So f(z) must look like

[tex]f(z) = \frac{P(z)}{Q(z)}[/tex],

where P(z) and Q(z) are power series in z. If f(z) has multiple poles (of any order), then it is not well-defined at those points. That is, if f(z) has poles at z0, z1, ..., then it maps all of these points to 0.

The case I am having trouble with is when f(z) has higher order poles, i.e. poles of order n > 1. The textbook I am using ("Conformal Field Theories" by Di Francesco, Mathieu, and Senechal) says "the image of a small neighborhood of z0 (the location of the pole) is wrapped n times around z0, and thus is not invertable". Can someone explain this statement to me? In what sense is there "wrapping" about the higher order poles?

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# Complex analysis question

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