Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have a bijection $f$ between two open sets in $\mathbb{C}\cup\{\infty\}$, for example $U=\{|z|<1\}$ and $\Omega$, with $\infty\in\Omega$. Let $f(0)=\infty$.
What do we mean by saying that "$f$ is a conformal mapping" in this case? What conditions are imposed on the behaviour of $f$ near $0$? (Maybe continuity and analyticity in $U\setminus\{0\}$?)

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

$f$ is conformal if it is holomorphic and has a nonzero derivative.

In your case, it means that $z\mapsto \frac{1}{f(z)}$ is holomorphic and has nonzero derivative on $U$ (because the usual atlas for Riemannian sphere is $\{(\mathbf C,z\mapsto z), (\mathbf (\mathbf C\cup\{\infty\})\setminus\{0\}, z\mapsto 1/z)\} $).

share|improve this answer
add comment

The map $g(z) = 1/z$ takes a neighbourhood of $\infty$ to a neighbourhood of $0$. $f$ with $f(0) = \infty$ is conformal on a neighbourhood of $0$ iff $g \circ f = 1/f$ (defined to be $0$ at $0$) is conformal there.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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