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Let $D$ be a circular quadrilateral (that is a Jordan region whose boundary consists of 4 arcs of circles all orthogonal to the unit circle) whose interior angles are all equal to 0, the vertices lie on the unit circle, and $D$ is inside the unit disc. Suppose also that $D$ is symmetric with respect to the real and imaginary axes, and has one vertex $z_1=\exp(i\theta)$ where $\theta\in (0,\pi/2)$. Then this number $\theta$ determines such a $D$ completely.

Let $f$ be the inverse of the Riemann map, so that $f$ is the conformal map from the unit disc onto $D$, $f(0)=0$ and $f'(0)>0$.

Is it true that maximum $f'(0)$ is achieved when $\theta=\pi/4$ ?

There is a strong computer evidence for this, as well as the general considerations (where else can the maximum be?).

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  • $\begingroup$ Alexandre, do you mean to say that $D$ is an ideal hyperbolic quadrilateral in the Poincare disk? (In other words, the arcs of $D$ are orthogonal to the unit circle). Otherwise, there is another parameter needed to define $D$, namely the angle that the arcs of $D$ make with the unit circle. $\endgroup$
    – Jeremy Kahn
    Commented Oct 7, 2012 at 1:13
  • $\begingroup$ Alexandre, have you looked at Chapter V of Nehari's book Conformal Mapping? He considers the general problem of finding the "Schwarz-Christoffel" mapping for hyperbolic polygons. It's complicated in general, but in your case, you should be able to write it down. $\endgroup$
    – Autumn Kent
    Commented Oct 7, 2012 at 3:08
  • $\begingroup$ See also mathoverflow.net/questions/46102/… $\endgroup$
    – Autumn Kent
    Commented Oct 7, 2012 at 3:09
  • $\begingroup$ Jeremy, you are right, I consider an ideal hyperbolic quasrilateral, and I made a correction in my description. $\endgroup$
    – Alexandre Eremenko
    Commented Oct 7, 2012 at 14:06
  • $\begingroup$ Richard, I know the literature, at least the classical books. This is probably the simplest problem on conformal mapping where there is no explicit solution. You can look in my preprint arXiv:1110.2696, where I wrote all I know abut this, and computed the thing numerically. $\endgroup$
    – Alexandre Eremenko
    Commented Oct 7, 2012 at 14:10

2 Answers 2

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Dear Alexandre,

These ideal hyperbolic quadrilaterals are sometimes called Lambert quadrilaterals. See e.g. pointers to the literature in arXiv:1203.6494v2 [math.MG] 10 Apr 2012 .Matti Vuorinen

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  • $\begingroup$ Matti, it is interesting to know that they have a name. Do you know why Lambert? Lambert was an XVIII century mathematician who is remembered for Lambert series. Why is his name attached to these quadrilaterals? (Unless there was some other Lambert). $\endgroup$
    – Alexandre Eremenko
    Commented Feb 4, 2013 at 5:54
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Alexandr Solynin told me that he solved this problem (even the more general one, for hyperbolic n-gons with all zero angles) in 1993. A. Solynin, Some extremal problems for circular polygons, (Russian) Zapiski Nauchnyh Seminarov POMI 206, 1993, 127-136. English translation is in Journal of Math Sci. 80 4, 1996, 1956-1961.

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