I suppose this is the proper place for this question:)(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Complex analysis question

**Physics Forums | Science Articles, Homework Help, Discussion**