conformal radius

Definition.

Let $G\subset ℂ$ be a simply connected region that is not the whole plane and let $a\in G$ be any point. The Riemann mapping theorem tells us that there exists a unique one-to-one and onto holomorphic map $f:𝔻\to G$ (where $𝔻$ is the unit disc) such that $f(0)=a$ and $f^{\prime }(0)>0$. Then define the conformal radius $r(G,a)=f^{\prime }(0)$.

Example.

For example, take $G\mathrm{=}B\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}\delta \mathrm{)}$ (the open ball of radius $\delta$ around 0) for some $\delta \mathrm{>}\mathrm{0}$, then $r\mathit{}\mathrm{(}G\mathrm{,}\mathrm{0}\mathrm{)}\mathrm{=}\delta$ because we have a map $f\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}\delta \mathrm{\cdot }z$ as our unique map. And thus this definition coincides with our definition of radius for this special case.

Example.

For another example we look at how the conformal radius is affected by the choice of the point $a$. So suppose that we take $G$ to be the unit disc ($\mathrm{D}$) itself and we take some point $a\mathrm{\in }\mathrm{D}$. The unique map that takes 0 to $a$ is the map $f\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}\frac{z\mathrm{+}a}{\mathrm{1}\mathrm{+}\overline{a}\mathit{}z}$ (where $\overline{a}$ is the complex conjugate of $a$) and by the quotient rule we get that $f^{\mathrm{\prime }}\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}\frac{\mathrm{1}\mathrm{-}{\mathrm{|}a\mathrm{|}}^{\mathrm{2}}}{{\mathrm{(}\mathrm{1}\mathrm{+}\overline{a}\mathit{}z\mathrm{)}}^{\mathrm{2}}}$. And so $r\mathit{}\mathrm{(}\mathrm{D}\mathrm{,}a\mathrm{)}\mathrm{=}{f}^{\mathrm{\prime }}\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{1}\mathrm{-}{\mathrm{|}a\mathrm{|}}^{\mathrm{2}}$, so the conformal radius of the unit disc goes to 0 as we move the point $a$ towards the boundary of the disc, and it is largest (equal to 1) when $a\mathrm{=}\mathrm{0}$.

From the first example we can now see another way of characterizing the conformal radius. Take the inverse map (inverses of holomorphic one-to-one functions are also always holomorphic) and call it $\phi :G\to 𝔻$ (the map such that $\phi (f(z))=z$). We take the derivative (see the entry on univalent functions (http://planetmath.org/UnivalentFunction)) we get ${\phi }^{\prime }(f(0))=\frac{1}{f^{\prime }(0)}$, that is ${\phi }^{\prime }(a)=\frac{1}{r}$ (where we call $r=r(G,a)$ for brevity now). If we multiply the map by the conformal radius we get a map $\gamma :G\to B(0,r)$ such that $\gamma (z)=r\cdot \phi (z)$ and ${\gamma }^{\prime }(a)=1$. By uniqueness of the map arising from the Riemann mapping theorem we can see that $\gamma$ is also unique. Thus we could define the conformal radius as follows.

Definition.

Let $G\subset ℂ$ be a region and let $a\in G$ be any point. By application of Riemann mapping theorem there exists a unique map $\gamma :G\to B(0,r)$ for some $r>0$, such that $\gamma (a)=0$ and ${\gamma }^{\prime }(a)=1$. The conformal radius is then defined as $r(G,a)=r$.

This definition gives more of an intuitive understanding of why we’d call this the conformal radius of $G$. We look at the unique map with ${\gamma }^{\prime }(a)=1$, that is, the map that doesn’t “stretch” the set. So the radius of $G$ with respect to $a$ is really the radius of the unique ball around zero to which $G$ is conformally equivalent without any “stretching” needed.

Title	conformal radius
Canonical name	ConformalRadius
Date of creation	2013-03-22 14:18:33
Last modified on	2013-03-22 14:18:33
Owner	jirka (4157)
Last modified by	jirka (4157)
Numerical id	9
Author	jirka (4157)
Entry type	Definition
Classification	msc 30C55
Related topic	RiemannMappingTheorem