# What is a conformal mapping with $\infty$ in its image?

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 $U\setminus\{0\}$?)

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$f$ 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 $\{(\mathbf C,z\mapsto z), (\mathbf (\mathbf C\cup\{\infty\})\setminus\{0\}, z\mapsto 1/z)\}$).
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.