conformal radius

Definition.

Let $G\subset ℂ$ be a simply connected region that is not the wholeplane and let $a\in G$ be any point.The Riemann mapping theorem tells us that there existsa unique one-to-one and onto holomorphic map $f:𝔻\to G$ (where$𝔻$ is the unit disc) such that $f(0)=a$ and $f^{\prime }(0)>0$. Thendefine 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 $r\mathit{}\mathrm{(}G\mathrm{,}\mathrm{0}\mathrm{)}\mathrm{=}\delta$ becausewe have a map $f\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}\delta \mathrm{\cdot }z$ as our unique map.And thus this definitioncoincides with our definition of radius for this special case.

Example.

For another example we look at how the conformal radius is affected bythe 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 $a$ isthe 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 radiusof the unit disc goes to 0 as we move the point $a$ towards the boundaryof the disc, andit is largest (equal to 1) when $a\mathrm{=}\mathrm{0}$.

From the first example we can now see another way of characterizing the conformalradius. 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 bythe 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 themap arising from the Riemann mapping theorem we can see that $\gamma$ isalso unique. Thus we could define the conformal radius as follows.

Definition.

Let $G\subset ℂ$ be a region and let $a\in G$ be anypoint. By application of Riemann mappingtheorem there exists a unique map $\gamma :G\to B(0,r)$ forsome $r>0$, such that $\gamma (a)=0$ and ${\gamma }^{\prime }(a)=1$. Theconformal 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.