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I was wondering when a function was conformal at a pole? In class, when learning about Möbius transformations we put down a definition saying $f$ is conformal at a pole $z$ if $1/f$ is conformal at the zero $z$. However, I'm not sure whether this was given as a general statement, or whether it only applies to Möbius transformations.

Also, does it matter whether the function that has a pole is meromorphic, say, on the entire Riemann sphere, or does the answer change if it also has an essential singularity, such as, for example, $f(z) = \frac {1}{sinz}$ (pole at $z = 0$, but an essential singularity at $z = \infty$)?

I've looked online a bit, and found some sources saying that even Möbius transformations aren't conformal at their poles, i.e. at $z =-\frac{d}{c}$, which directly contradicts what I've learned. Unfortunately, the professor that is teaching the class is out of town for a week, so I can't ask him, and figured I'd turn to this community. It also seems it's somewhat of a general question that I haven't found answered anywhere else, so it might be good to hear more about.

For full disclosure, this is related to homework, but I feel the question is stated in general terms, not in terms of me trying to solve a specific problem (i.e. the problem on the homework isn't to answer this question directly), so I didn't attach a homework tag to it.

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1 Answer 1

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You will likely find some sources where conformality is defined only at the points where $f$ is holomorphic, and other sources where the value of $\infty$ is allowed. Many definitions are not (and probably will never be) consistent across the entire mathematical literature.

But some general principles are universal:

  1. Conformality is a local property. Looking at an arbitrarily small neighborhood of point $a$ is enough to tell whether $f$ is conformal at $a$. Singularities away from $a$ do not matter.
  2. Conformality is preserved under composition: if $f$ is conformal at $a$ and $g$ is conformal at $f(a)$, then $g\circ f$ is conformal at $a$.
  3. Conformality is preserved under taking inverse: if $f$ is conformal at $a$, then $f^{-1}$ is conformal at $f(a)$.

Conclusion: if we accept that the inversion $z\mapsto 1/z$ is conformal everywhere including $0$, then (by composition) we must also accept that $1/f$ is conformal at $a$ if and only if $f$ is conformal at $a$.

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