The conformal parameter of a round annulus is a complete invariant

The radial deformation retraction in [F6], based at $s_0$, induces inverse homomorphisms on the domain and circle fundamental groups by [F4]. Scaling that circle to the unit circle identifies its group with $\mathbb{Z}$ by [F5]; the scaling leaves $\int dz\mathrm{/}z$ unchanged because the integrand and coordinate integrators acquire reciprocal factors. Changing basepoint along a radial/circular path also preserves the winding integer, since the path and its reversal contribute opposite contour integrals. Thus winding identifies the fundamental group of each radial domain with $\mathbb{Z}$. A biholomorphism between two such domains and its inverse induce inverse group isomorphisms, so it acts on winding numbers by an automorphism of $\mathbb{Z}$, necessarily multiplication by a sign $\epsilon \in \{+1,-1\}$. Since $\mathbb{Z}$ is abelian, the conclusion is independent of the basepoint paths; it sends the winding-one closed-loop family onto the winding-$\epsilon$ family. Rectifiability is preserved in both directions by the conformal path transport in Conformal invariance, monotonicity, and the series and parallel laws for extremal length.

For any rectifiable closed $\gamma$, complex conjugation $c(z)=\bar{z}$ satisfies ${\int }_{c\circ \gamma }dw\mathrm{/}w=\overline{{\int }_{\gamma }dz\mathrm{/}z}$, by expanding the componentwise Riemann–Stieltjes definition, so it interchanges winding $+1$ and $-1$. If $\tilde{\gamma }$ is an arc-length parametrization of $\gamma$, then $c\circ \tilde{\gamma }$ is one for $c\circ \gamma$ because $c$ is a Euclidean isometry. The arc-length integral formula in [F1] gives ${\mathrm{\ell }}_{\rho }(c\circ \gamma )={\mathrm{\ell }}_{\rho \circ c}(\gamma )$. Also $A(\rho \circ c)=A(\rho )$ by the Borel change-of-variables formula in [F7]. Since $\rho \mapsto \rho \circ c$ is a bijection on finite-positive-area Borel densities, the two sign families have equal extremal length.

If $R\mathrm{/}r=R^{\mathrm{\prime }}\mathrm{/}{r}^{\mathrm{\prime }}$, the map $f(z)=(r^{\mathrm{\prime }}\mathrm{/}r)z$ is a bijective holomorphic map $A(r,R)\to A(r^{\mathrm{\prime }},R^{\mathrm{\prime }})$ with holomorphic inverse $w\mapsto (r\mathrm{/}{r}^{\mathrm{\prime }})w$, so the annuli are conformally equivalent.

Fix $n\ge 1$ and put ${\epsilon }_n=2^{-n}$. Given $\gamma \in {\mathrm{\Gamma }}_{D^{\ast }}$, continuity and $\mathrm{\mid }\gamma (0)\mathrm{\mid }=0<{\epsilon }_n<1=\mathrm{\mid }\gamma (1)\mathrm{\mid }$ give a nonempty compact level set $\{t:\mathrm{\mid }\gamma (t)\mathrm{\mid }={\epsilon }_n\}$; compactness of $[0,1]$ gives its largest member $t_n$. For $t_n<t<1$, one has ${\epsilon }_n<\mathrm{\mid }\gamma (t)\mathrm{\mid }<1$, since another value at or below ${\epsilon }_n$ would force a later hit of that level, so $\gamma {\mathrm{\mid }}_{[t_n,1]}$ belongs to ${\mathrm{\Gamma }}_{{\epsilon }_n,1}$. If $\gamma$ is nonrectifiable, its assigned length is already $+\mathrm{\infty }$. If it is rectifiable, subpath additivity in [F8] gives the length comparison below.

If a biholomorphism $h:D^{\ast }\to \mathbb{C}$ existed, its inverse $h^{-1}:\mathbb{C}\to D^{\ast }$ would be entire and bounded by $1$. Liouville's theorem [F11] would make it constant, contradicting bijectivity.

If a biholomorphism $h:\mathbb{D}\to D^{\ast }$ existed, then $h$ would be nowhere zero. By [F12], the disc is homologically simply connected, so [F10] supplies a holomorphic $g:\mathbb{D}\to \mathbb{C}$ with $e^{g(z)}=h(z)$. Set $L(w)=g(h^{-1}(w))$ on $D^{\ast }$. Then $e^{L(w)}=w$, and [F10] gives $L^{\mathrm{\prime }}(w)=1\mathrm{/}w$. Integrating around the positively oriented circle $\mathrm{\mid }w\mathrm{\mid }=1\mathrm{/}2\subset D^{\ast }$, [F11] gives $0=\int L^{\mathrm{\prime }}(w)dw=\int dw\mathrm{/}w=2\pi i$, a contradiction.

Conversely, let $h:A(r,R)\to A(r^{\mathrm{\prime }},R^{\mathrm{\prime }})$ be a biholomorphism. By step 1.1, it maps ${\mathrm{\Theta }}_{r,R}^{+}$ onto one of the two target sign families. Conformal invariance [F3] and the sign equality in step 1.2 give ${\displaystyle \lambda ({\mathrm{\Theta }}_{r,R}^{+})=\lambda ({\mathrm{\Theta }}_{r^{\mathrm{\prime }},R^{\mathrm{\prime }}}^{+})\mathrm{.}}$ Using [F2], this is ${\displaystyle \frac{2\pi }{\log (R\mathrm{/}r)}=\frac{2\pi }{\log (R^{\mathrm{\prime }}\mathrm{/}{r}^{\mathrm{\prime }})}\mathrm{.}}$ Both logarithms are positive by [F9]; cancellation and strict monotonicity in [F9] give $R\mathrm{/}r=R^{\mathrm{\prime }}\mathrm{/}{r}^{\mathrm{\prime }}$. This proves (i).

Let $\rho$ be any Borel density on $A({\epsilon }_n,1)$ with $0<A(\rho )<\mathrm{\infty }$, and extend it by zero to a Borel density $\tilde{\rho }$ on $D^{\ast }$. Its area is unchanged. Step 1.4 and [F8] show ${\displaystyle {\mathrm{\ell }}_{\tilde{\rho }}({\mathrm{\Gamma }}_{D^{\ast }})\ge {\mathrm{\ell }}_{\rho }({\mathrm{\Gamma }}_{{\epsilon }_n,1}),}$ because every path contains the annular subpath and the endpoint values carry no length mass. Thus the extremal-length quotient of $\tilde{\rho }$ on ${\mathrm{\Gamma }}_{D^{\ast }}$ is at least the quotient of $\rho$ on ${\mathrm{\Gamma }}_{{\epsilon }_n,1}$. Taking suprema and applying [F2] yields ${\displaystyle \lambda ({\mathrm{\Gamma }}_{D^{\ast }})\ge \lambda ({\mathrm{\Gamma }}_{{\epsilon }_n,1})=\frac{\log (2^n)}{2\pi }=\frac{n\log 2}{2\pi }\mathrm{.}}$ As $n$ is unbounded and $\log 2>0$, this proves $M(D^{\ast })=+\mathrm{\infty }$.

Define ${\mathrm{\Theta }}_{D^{\ast }}^{+}$ and ${\mathrm{\Theta }}_{{\mathbb{C}}^{\ast }}^{+}$ as the families of rectifiable closed paths of winding $+1$ in the indicated domains. For every $n\ge 1$, ${\mathrm{\Theta }}_{A(2^{-n},1)}^{+}\subseteq {\mathrm{\Theta }}_{D^{\ast }}^{+}$ and ${\mathrm{\Theta }}_{A(2^{-n},2^n)}^{+}\subseteq {\mathrm{\Theta }}_{{\mathbb{C}}^{\ast }}^{+}$. These subannular families have full traces in the larger domains, so [F1] and monotonicity [F3] give ${\displaystyle 0\le \lambda ({\mathrm{\Theta }}_{D^{\ast }}^{+})\le \frac{2\pi }{n\log 2},0\le \lambda ({\mathrm{\Theta }}_{{\mathbb{C}}^{\ast }}^{+})\le \frac{2\pi }{2n\log 2}\mathrm{.}}$ Letting $n$ grow proves both extremal lengths are zero. By step 1.2 the corresponding winding-$-1$ families also have extremal length zero.

If either $\mathrm{\Omega }=D^{\ast }$ or $\mathrm{\Omega }={\mathbb{C}}^{\ast }$ were conformally equivalent to a finite round annulus $A(r,R)$, step 1.1 would map its winding-one family onto one of the two target sign families. Conformal invariance [F3] and step 1.2 would then equate its zero extremal length from step 2.3 with the strictly positive value $2\pi \mathrm{/}\log (R\mathrm{/}r)$ in [F2], a contradiction. This proves both finite-annulus exclusions in (iii).

Steps 1.3 and 2.1 prove (i), step 2.2 proves (ii), and steps 1.5, 1.6, and 3.1 prove every exclusion in (iii).