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If $h(z)$ is analytic on the disk centered at 0 of radius r, by the Cauchy Residue formula \[ \int \int_D h(z)\, dx dy = \pi r^2 h(0) \] The disk is the simplest example of a quadrature domain since the integral of a holomorphic function over the domain is determined by the value at a single point.

How about the next simplest cases? What are connected quadrature domains whose integrals only depend on a few points (e.g. 2 or 3)?

\[ \int \int_D h(z)\, dx dy = c_1 h(z_1) + c_2 h(z_2) + c_3 h(z_3) \]

Probably these will all be close to the union of a few circles (with jumps in the coefficients as the radius changes).

alt text http://s16.postimage.org/vzkdfcqnn/domain.gif

It looks exact quadrature domains can be constructed using meromorphic functions on Riemann surfaces and is related to uniformation of surfaces. There are relations to Laplacian growth, Random Matrix Theory & Integrable Hierarchies.

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A bounded simply connected domain with analytic boundary is a quadrature domain if and only if the inverse of the Riemann mapping function (mapping the disc onto the domain) is rational.

See, for example, P. J. Davis, The Schwarz function and its applications, The Mathematical Association of America, Bualo, N. Y., 1974. The Carus Mathematical Monographs, No. 17. or A. Varchenko and P. Etingof, Why the boundary of a round drop becomes a curve of order 4, AMS 1991.

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  • $\begingroup$ Can you define quadrature domains by $\partial f(\mathbb{D})$ for some rational function $f = p/q$ ? $\endgroup$
    – john mangual
    Commented Aug 1, 2013 at 22:12
  • $\begingroup$ If $f$ is univalent in $D$, then $f(D)$ will be a quadrature domain. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 16, 2018 at 3:10

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