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.


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 $a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.