Properties of Grötzsch and Teichmüller Moduli

Lemma
For any number $R>1$, denote the modulus of the Grötzsch annulus by $M(R)$. Then:
 * $M(R) = 2M\left(\dfrac{1+R}{2 \sqrt R}\right)$.

Similarly, for any $R>0$, denote the modulus of the Teichmüller annulus by $\Lambda(R)$. Then:
 * $\Lambda(R)\cdot \Lambda \left({\dfrac 1 R}\right) = \dfrac 1 4$.

(In particular, $\Lambda(1) = \dfrac 1 2$.)

Furthermore, these two quantities are related by
 * $ M(R)= \Lambda\left(\dfrac{(1-R)^2}{4R}\right)$

and
 * $2M(R) = \Lambda(R^2-1)$.

Proof
We begin by proving the equation relating $\Lambda(R)$ and $\Lambda(1/R)$.

To do so, we consider the quadrilateral $Q(R)$ given by the upper half plane with the boundary arcs $[-1,0]$ and $[R,\infty)$.

Then
 * $\operatorname{mod}(Q(R)) = 2\Lambda(R)$.

Now consider the quadrilateral $Q'$ that again consists of the upper half plane, but now with the boundary arcs $(-\infty,0]$ and $[0,R]$.

Then
 * $\operatorname{mod}(Q')= \dfrac{1}{\operatorname{mod}(Q(R))} = \dfrac{1}{2\Lambda(R)}$.

On the other hand, the function $z\mapsto \dfrac{-z}{R}$ takes $Q'$ conformally to $Q(1/R)$. Hence, by Invariance of Extremal Length under Conformal Mappings, we have
 * $2\Lambda\left(\dfrac{1}{R}\right) = \operatorname{mod}(Q') = \dfrac{1}{2\Lambda(R)}$.

Rearranging yields the desired identity.

To prove the first equality (regarding $M(R)$), let $G(R)$ denote the Grötzsch annulus, and consider the set
 * $U := \{z\in\C: |z|>1, z\notin [\sqrt{R},\infty)\text{ and }z\notin (-\infty,-\sqrt{R}]\}$.

Then $z\mapsto z^2$ maps $U$ to $G(R)$ as a covering map of degree $2$.

Hence
 * $M(R) = \operatorname{mod}(G(R)) = 2\operatorname{mod}(U)$.

On the other hand, the Möbius transformation
 * $ z\mapsto \dfrac{1+\sqrt{R}z}{z + \sqrt{R}}$

is a conformal isomorphism between $U$ and $G\left(\dfrac{1+R}{2\sqrt{R}}\right)$.

The claim now follows again from Invariance of Extremal Length under Conformal Mappings.

To prove the first relation between the Teichmüller and Grötzsch moduli, observe that the Koebe Function:
 * $ z\mapsto \dfrac{z}{(1+z)^2} $

maps $G(R)$ conformally onto the set:
 * $ V := \C\setminus \left( \left[\dfrac{1}{4},\infty\right)\cup \left[0,\dfrac{R}{(1+R)^2}\right] \right)$.

On the other hand, the Möbius transformation:
 * $z\mapsto z\cdot \dfrac{(1+R)^2}{R} - 1$

takes $V$ conformally onto the Teichmüller domain for:
 * $ \dfrac{1}{4}\cdot \dfrac{(1+R)^2}{R} - 1 = \dfrac{(1-R)^2}{4R}$.

So $ M(R) =\Lambda((1-R)^2/4R) $, as claimed.

The second relation can be proved from the first, together with the property of $M(R)$ that we proved above. Indeed, choose $Q$ such that:
 * $R = \dfrac{1+Q}{2\sqrt{Q}}$.

(This is possible because the right-hand side is a strictly-increasing function from the interval $[1,\infty)$ to itself.)

Then:
 * $\Lambda(R^2-1)=\Lambda\left(\dfrac{(1+Q)^2}{4Q}-1\right) = \Lambda\left(\dfrac{(1-Q)^2}{4Q}\right) = M(Q)=2M(R)$.

Alternatively, we can also prove the last equality directly: reflection of the Grötzsch annulus in the unit circle yields the set
 * $ W := \C\setminus (\,[0,1/R]\cup [R,\infty)\,)$.

It follows that $\operatorname{mod}(W) = 2M(R)$.

On the other hand, the map
 * $z\mapsto Rz - 1$

takes $W$ to the Teichmüller domain for $R^2-1$.