Properties of Grötzsch and Teichmüller Moduli

Lemma
For any number $$R>1$$, denote the modulus of the Gr&ouml;tzsch annulus by $$M(R)$$. Then
 * $$M(R) = 2M\left(\frac{1+R}{2\sqrt{R}}\right).$$

Similarly, for any $$R>0$$, denote the modulus of the Teichm&uuml;ller annulus by $$\Lambda(R)$$. Then
 * $$\Lambda(R)\cdot \Lambda\left(\frac{1}{R}\right) = \frac{1}{4}.$$

(In particular, $$\Lambda(1)=\frac{1}{2}$$.)

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

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

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

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
 * $$ z\mapsto \frac{1+\sqrt{R}z}{z + \sqrt{R}}$$

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

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

To prove the first relation between the Teichm&uuml;ller and Gr&ouml;tzsch moduli, observe that the Koebe Function $$ z\mapsto \frac{z}{(1+z)^2} $$ maps $$G(R)$$ conformally onto the set
 * $$ V := \C\setminus \left( \left[\frac{1}{4},\infty\right)\cup \left[0,\frac{R}{(1+R)^2}\right] \right)$$.

On the other hand, the M&ouml;bius transformation
 * $$z\mapsto z\cdot \frac{(1+R)^2}{R} - 1$$

takes $$V$$ conformally onto the Teichm&uuml;ller domain for
 * $$ \frac{1}{4}\cdot \frac{(1+R)^2}{R} - 1 = \frac{(1-R)^2}{4R}. $$

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
 * $$R = \frac{1+Q}{2\sqrt{Q}}.$$

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

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

Alternatively, we can also prove the last equality directly: reflection of the Gr&ouml;tzsch 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 Teichm&uuml;ller domain for $$R^2-1$$.