# Modulus of an annulus

The value reciprocal to the extremal length of the family of closed curves in the annulus $ r _ {1} \leq | z | \leq r _ {2} $
which separate the boundary circles; the modulus is equal to

$$ \frac{1}{2 \pi } \mathop{\rm ln} \frac{r _ {2} }{r _ {1} } . $$

By a conformal mapping onto an associated annulus $ K $, the modulus $ m _ {G} $ of an annular domain $ G $ can be obtained. It turns out that $ m _ {G} = D ( u) / 2 \pi $, where $ D ( u) $ is the Dirichlet integral of the real part of the function $ u $ mapping $ G $ onto $ K $: $ \{ 1 < | w | < e ^ {m _ {G} } \} $. (Thus, a given annular domain is mapped onto an annulus with a fixed ratio of the radii of the boundary circles. This fact can be taken as another definition of the modulus of an annulus, its generalization leads to the idea of the modulus of a plane domain.)

A generalization of the modulus of an annular domain is the modulus $ m _ \gamma $ of a prime end (cf. Cluster set; Limit elements) $ \gamma $ of an open Riemann surface $ R $ relative to a neighbourhood. Depending on whether $ m _ \gamma $ is finite or infinite, the prime end has hyperbolic or parabolic type and $ R $ either does or does not have a Green function.

For a simply-connected domain $ D $ of hyperbolic type the so-called reduced modulus $ m _ {z _ {0} } $ relative to $ z _ {0} \in D $ is defined as the limit

$$ \lim\limits _ {r \rightarrow 0 } \left ( m ( r) + \frac{1}{2 \pi } \mathop{\rm ln} r \right ) , $$

where $ m ( r) $ is the modulus of the annular domain $ D ( r) = D \cap \{ | z - z _ {0} | > r \} $. It turns out that $ m _ {z _ {0} } = ( \mathop{\rm ln} R ) / 2 \pi $, where $ R $ is the conformal radius (cf. Conformal radius of a domain) of $ D $ relative to $ z _ {0} \in D $.

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#### References

[a1] | Z. Nehari, "Conformal mapping" , Dover, reprint (1975) |

**How to Cite This Entry:**

Modulus of an annulus.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Modulus_of_an_annulus&oldid=47877