Conformal radius

Remarks

Equivalently, $\operatorname{rad}(a, \Omega) = 1/\abs{\varphi'(a)}$ for any conformal map $\varphi$ of $\Omega$ onto $\mathbb{D}$ with $\varphi(a) = 0$. Example: let $\Omega = \mathbb{D}$ and $a \in \mathbb{D}$. The map $f(w) = (w + a)/(1 + \bar a w)$ sends $\mathbb{D}$ onto $\mathbb{D}$ with $f(0) = a$ and $f'(0) = 1 - \abs{a}^2 > 0$, so $\operatorname{rad}(a, \mathbb{D}) = 1 - \abs{a}^2$. Non-example: $\Omega = \C$ is simply connected, but no conformal map of $\mathbb{D}$ onto $\C$ exists, so the definition does not apply. With $d = \operatorname{dist}(a, \partial\Omega)$ one has $d \le \operatorname{rad}(a, \Omega) \le 4d$. The lower bound follows from the Schwarz lemma applied to $f^{-1}$ on the disc of centre $a$ and radius $d$; the upper bound is the Koebe quarter theorem. The Green's function of $\Omega$ satisfies $g(z, a) = -\log\abs{z - a} + \log \operatorname{rad}(a, \Omega) + o(1)$ as $z \to a$. Under a conformal map $F$ of $\Omega$ onto $\Omega'$, $\operatorname{rad}(F(a), \Omega') = \abs{F'(a)} \operatorname{rad}(a, \Omega)$.