A modulus obstruction to quasiconformal equivalence of round annuli

Assume the Axiom of Choice. Write $\lambda$ for extremal length and $\mu$ for its reciprocal curve-family modulus (Extremal length and the curve-family modulus of a path family). Call two domains $K$-quasiconformally equivalent if a $K$-quasiconformal homeomorphism carries one onto the other, using the equivalent geometric and analytic definitions on this page (Orientation-preserving homeomorphisms and the geometric definition of quasiconformality, The ACL and Sobolev analytic definition of quasiconformality, The geometric and analytic definitions of quasiconformality agree). Then:

(a) If the round annuli $A(1,2)$ and $A(1,R)$ (Annuli in the complex plane) are $K$-quasiconformally equivalent for finite $R>1$, then $\frac{1}{2\pi }\log R\le K\frac{1}{2\pi }\log 2,$ so $\log R\le K\log 2$. In particular, for every finite $K\ge 1$, the annuli $A(1,2)$ and $A(1,2^{K+1})$ are not $K$-quasiconformally equivalent. For example, $A(1,2)$ and $A(1,8)$ are not $2$-quasiconformally equivalent, though this estimate does not exclude equivalence for $K\ge 3$.

(b) The inverse direction gives the lower bound as well: any such equivalence satisfies $\frac{1}{K}\log 2\le \log R\le K\log 2,$ equivalently the ratio $M(A(1,R))\mathrm{/}M(A(1,2))$ of their conformal parameters from The conformal parameter of a round annulus is a complete invariant lies in $[1\mathrm{/}K,K]$.

(c) No finite round annulus $A(r,R)$ with $0<r<R<\mathrm{\infty }$ is $K$-quasiconformally equivalent to the punctured disc $D^{\ast }=\{0<\mathrm{\mid }z\mathrm{\mid }<1\}$ for any finite $K\ge 1$ (The punctured disc has infinite conformal parameter, unlike every finite annulus).