# Moduli of quadrilaterals and quasiconformal reflection

Semen Nasyrov  Institute of Mathematics and Mechanics. Kazan Federal
University, Kazan, Russia  ,  Toshiyuki Sugawa  Graduate School of Information
Sciences, Tohoku University, Aoba-ku, Sendai 980-8579, Japan  and  Matti
Vuorinen  Department of Mathematics and Statistics, University of Turku,
Turku, Finland

###### Abstract.

We study the interior and exterior moduli of polygonal quadrilaterals. The
main result is a formula for a conformal mapping of the upper half plane onto
the exterior of a convex polygonal quadrilateral. We prove this by a careful
analysis of the Schwarz-Christoffel transformation and obtain the so-called
accessory parameters and then the result in terms of the Lauricella
hypergeometric function. This result enables us to understand the
dissimilarities of the exterior and interior of a convex polygonal
quadrilateral. We also give a Mathematica algorithm for the computation. In
particular, we study the special case of an isosceles trapezoidal polygon $L$
and obtain some estimates for the coefficient of quasiconformal reflection
over $L$ in terms of special functions and geometric parameters of $L$.

###### Key words and phrases:

Quasiconformal reflection, extremal length, modulus of a quadrilateral,
elliptic integrals, Gauss hypergeometric functions, Appell hypergeometric
functions

###### 2010 Mathematics Subject Classification:

30C20, 30C15, 31

##  1\. Introduction

A quadrilateral $\mbox{\boldmath$Q$}=(Q;z_{1},z_{2},z_{3},z_{4})$ is a Jordan
domain $Q$ on the Riemann sphere with four fixed points $z_{1}$, $z_{2}$,
$z_{3}$, and $z_{4}$ on its boundary. We label $z_{j}$ in such an order that
increasing of the index $j$ corresponds to the positive traverse of the
boundary $\partial Q$; we will name such quadruples
$(z_{1},z_{2},z_{3},z_{4})$ admissible.

One of the main geometric characteristics of a quadrilateral $Q$ is its
conformal modulus. There are many equivalent definitions of this concept. If
$f$ is a conformal mapping of $Q$ onto a rectangle $[0,1]\times[0,h],h>0,$
such that the points $z_{1}$, $z_{2}$, $z_{3}$, and $z_{4}$ correspond to $0$,
$1$, $1+ih$, and $ih$, then $h$ is uniquely defined and is called the
conformal modulus of $Q$ [PS, p.52, Def. 2.1.3]. In this case, we write

$h=\text{Mod}(\mbox{\boldmath$Q$}).$

Another way to define the modulus is to use the concept of the extremal length
$\lambda(\Gamma)$ of a curve-family $\Gamma$ (see, e.g. [A1]). If $\Gamma$ is
the family of curves in the domain $Q$, connecting the sides $z_{1}z_{2}$ and
$z_{3}z_{4}$ of the quadrilateral $Q$, then
$\text{Mod}(\mbox{\boldmath$Q$})=\lambda(\Gamma)$. If we consider the family
$\Gamma_{1}$ of curves connecting the sides $z_{2}z_{3}$ and $z_{1}z_{4}$ in
$Q$, then $\text{Mod}(\mbox{\boldmath$Q$})=1/\lambda(\Gamma_{1})$. At last,

$\text{Mod}(\mbox{\boldmath$Q$})=\left(\inf_{u}\int\\!\\!\\!\int_{Q}|\nabla
u|^{2}dxdy\right)^{-1}$

where the infimum is taken over all smooth functions $u$ in $Q$, continuous on
$\overline{Q}$, which are equal to $0$ on the boundary arc $z_{1}z_{2}$ and
equal to $1$ on the boundary arc $z_{3}z_{4}$. Moreover, it is known that the
infimum here is attained by a harmonic function [A1, p. 65].

Assume that $Q$ is a bounded Jordan domain in $\mathbb{C}$, $L=\partial Q$,
and $z_{1}$, $z_{2}$, $z_{3}$, $z_{4}$ are some points on $L$ satisfying the
above requirements. Consider the quadrilateral
$\mbox{\boldmath$Q$}=(Q;z_{1},z_{2},z_{3},z_{4})$. Then we will say that the
value of $\text{Mod}(Q;z_{1},z_{2},z_{3},z_{4})$ is the interior modulus. We
can also consider the quadrilateral
$\mbox{\boldmath$Q$}^{c}:=(Q^{c};z_{4},z_{3},z_{2},z_{1})$ where $Q^{c}$ is
the complement of $Q$ with respect to the Riemann sphere. In that case, we
will name $\text{Mod}(\mbox{\boldmath$Q$}^{c})$ the exterior modulus.

The main topic of this paper is to analyze the interior and exterior moduli of
polygonal quadrilaterals and to apply the results to geometric function
theory. Because the upper half-plane can be conformally mapped onto a
polygonal domain in terms of the Schwarz-Christoffel transformation, both the
interior and exterior moduli of polygonal quadrilaterals can be computed. This
transformation is semi-explicit, there are so called accessory parameters that
have to be determined separately for each case [AF, KY]. There are no
universal recipes for finding the accessory parameters, which itself leads to
numerically ill-conditioned problems. The best methods for numerical
computation of Schwarz-Christoffel type conformal mappings are those developed
and implemented by T.A. Driscoll and L.N. Trefethen [DT]. For a survey of the
available methods, see N. Papamichael and N. Stylianopoulos [PS].

We study the conformal mapping from upper half plane onto the exterior of a
convex polygonal quadrilateral with vertices $0,1,a,b$ with ${\rm
Im}\,a>0,{\rm Im}\,b>0$ from analytic point of view and our goal is to
explicitly find the Schwarz-Christoffel mapping and its parameters. Recall
first that by classical theory of elliptic functions it is known that the
upper half plane is conformally mapped onto a rectangle under the inverse of
the elliptic function $sn\,$ and the ratio of the side lengths is given by a
quotient of complete elliptic integrals [AF, KY, PS]. This fact extends to the
case of conformal mapping of the upper half plane onto parallelograms and the
mapping is given by generalized elliptic functions, now depending on the least
angle of the parallelogram, and the interior modulus is given by a quotient of
generalized complete elliptic integrals as shown in [AQVV, Section 2]. These
generalized complete elliptic integrals were introduced in [BB, p.158] and we
mention in passing that during the past decade these integrals have been
studied intensively, cf. e.g. [CQW, QMC, QMB] and the bibliographies therein.
A further extension of the above conformal mapping problem is to map the upper
half plane onto a convex polygonal quadrilateral and such a mapping is given
by the Schwarz-Christoffel transformation expressed in terms of the Gaussian
hypergeometric function ${}_{2}F_{1}(a,b;c;z)$ with parameters depending on
the angles; the previous case is a special case of this one as shown in [HVV,
Section 2]. The conformal mapping problem onto the exterior of a polygonal
quadrilateral that we study here is much more difficult. In one of our main
results, Theorem 3.22, we prove that the mapping can be expressed in terms of
the Lauricella hypergeometric function $F_{D}^{(3)}\,.$ A particular case of
this mapping was studied by P. Duren and J. Pfaltzgraff [DP], namely a
conformal mapping of the upper half plane onto the exterior of a rectangle. As
pointed out in [HVV], this mapping already appeared in the works of W.
Burnside [BU]. We also give a Mathematica function for the computation of the
exterior modulus based on Theorem 3.22 and compare its numerical precision by
a comparison to the recent numerical computation results reported in [NRRV]
and observe very good agreement of the results.

Conformal moduli play an important role in geometric function theory and
applications, in particular, they are valuable tools in the study of
quasiconformal mappings (see, e.g. [A1, AVV, D, GH, K1]). One of the
definitions of quasiconformal mappings (the so-called geometric definition)
uses the moduli. A sense-preserving homeomorphism of the Riemann sphere
$\overline{\mathbb{C}}$ is called $K$-quasiconformal ($K\geq 1$) if it
satisfies the following condition: conformal moduli are quasiinvariant under
the mapping, i.e. if $\mbox{\boldmath$Q$}=(Q;z_{1},z_{2},z_{3},z_{4})$ is an
arbitrary quadrilateral and
$f(\mbox{\boldmath$Q$})=(f(Q);f(z_{1}),f(z_{2}),f(z_{3}),f(z_{4}))$, then

(1.1)
$K^{-1}\text{Mod}(\mbox{\boldmath$Q$})\leq\text{Mod}(f(\mbox{\boldmath$Q$}))\leq
K\text{Mod}(\mbox{\boldmath$Q$}).$

A closed Jordan curve $L$ on the Riemann sphere is called a quasicircle if it
is the image of the unit circle under a quasiconformal mapping defined in the
whole plane. If we know $K$ such that (1.1) is valid, then $L$ is called a
$K$-quasicircle. An important problem is to determine, for a given Jordan
curve, whether it is a quasicircle or not, and, if the answer is positive, to
either find or estimate the minimal possible value of $K$, denoted by $K_{L}$.
The problem of finding $K_{L}$ is open also for the case of curves $L$ as
simple as the boundaries of long rectangles and we will discuss this below
[HKV, p. 455 Probl. (20)].

Ahlfors [A2] gave the following geometric characterization for quasicircles.
If $L$ is a closed Jordan curve in the plane and there is a constant $C$ such
that for every three points $z_{1}$, $z_{2}$, and $z_{3}$ on $L$ such that
$z_{3}$ lies on the subarc of $L$ with smaller diameter and with endpoints
$z_{1}$ and $z_{2}$, the inequality

(1.2) $|z_{1}-z_{3}|+|z_{2}-z_{3}|\leq C|z_{1}-z_{2}|$

holds, then $L$ is a $K$-quasicircle where $K$ depends only on $C$ [GH, p.23,
Def. 2.2.2, Thm 2.2.5]. Conversely, if $L$ is a $K$-quasicircle, then for
every appropriate triple on $L$ the inequality (1.2) holds with $C$ depending
only on $K$. The monograph of Gehring and Hag [GH] gives many more
characterizations of quasicircles and provides a survey of their many
applications.

If $L$ is a quasicircle, then there is a quasiconformal reflection with
respect to $L$, i.e. a sense-reversing quasiconformal automorphism $g$ of the
Riemann sphere which keeps every point of $L$ fixed and maps the bounded
complementary component of $L$ onto the unbounded one and vice versa. Another
important problem is to find or estimate the minimal coefficient of
quasiconformality for such a mapping $g$; further we will denote this
coefficient of quasiconformal reflection by $QR_{L}$. This is a very difficult
problem even for polygonal curves in $\mathbb{C}$ studied by R. Kühnau in a
series of papers, e.g. [K2]. A nice survey is given in [Kr, pp.525-531]. Here
we give some of these results. Since every such a curve $L$ in $\mathbb{C}$
determines its interior in a unique way, we will also say that $QR_{L}$ is the
coefficient for $Q:=\mbox{\rm int}(L)$. If there exists a circle tangent to
every side of a closed polygon $L$, then $QR_{L}=2/\alpha-1$ where $\pi\alpha$
is the least interior angle of $L$. In particular, for triangles the problem
is solved. For quadrilaterals, the problem is open even for the case of
rectangles. The value of $QR_{L}$ is known only for rectangles
$[0,a]\times[0,1]$ close to a square [W]: if $1\leq a<1.037$, then $QR_{L}=3$.
For sufficiently long rectangles with $a>2.76$ it is proven [W] that
$QR_{L}>3$. Moreover, for any $a>1$ the estimate

(1.3) $\frac{\pi}{4}\,a<QR_{L}<\pi a$

holds [W], see also [K1, Kr].

The value $QR_{L}$ is closely connected with

$M_{L}:=\sup\frac{\text{Mod}(\mbox{\boldmath$Q$})}{\text{Mod}(\mbox{\boldmath$Q$}^{c})}$

where $\mbox{\boldmath$Q$}=(Q;z_{1},z_{2},z_{3},z_{4})$, $\partial Q=L$ and
the supremum is taken over all admissible quadruples
$(z_{1},z_{2},z_{3},z_{4})$ on $L$. As Kühnau noted in [K3],

(1.4) $QR_{L}\geq M_{L},$

therefore, every lower estimation for $M_{L}$ also gives a lower estimation
for $QR_{L}$. For related resuts, see Shen [Sh].

In the present paper, we investigate the problem of estimations of $M_{L}$ and
$QR_{L}$ for isosceles trapezoidal curves. With the help of the Schwarz-
Christoffel formula, we construct conformal mappings of the upper half-plane
onto the interior and exterior of $L$. Comparison of the interior and exterior
moduli for some quadruples of points $z_{1}$, $z_{2}$, $z_{3}$, $z_{4}\in L$
allows us to obtain lower estimates for $M_{L}$, and, therefore, for $QR_{L}$.

In addition, using fairly simple methods, we get two-sided estimates of
$QR_{L}$ and $M_{L}$ for isosceles trapezoidal polygons $L$ of height $1$ in
terms of lengths of its sides and angles. In particular, our main result about
quasiconformal reflection, Theorem 6.4, states that for such $L$ with acute
angle $\pi\alpha$ and bases $c$ and $d$, $d=c+\cot(\pi\alpha)$, the following
estimates hold:

$M_{L}\geq\left\\{\begin{array}[]{rl}g(\lambda_{0})(1+C(\alpha))d,&\text{if}\
\ \frac{c}{d}\geq\lambda_{0},\\\\[5.69054pt]
g(\lambda)(1+C(\alpha))d,&\text{if}\ \
\frac{c}{d}<\lambda_{0}.\end{array}\right.$

Here $g(\lambda)=\lambda\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}\left(\sqrt{1-\lambda^{2}}\right)/\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(\lambda)$, $\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(\lambda)$ is the
complete elliptic integral of the first kind, $\lambda_{0}=0.7373921\ldots$ is
the unique point of maximum of $g$ on $(0,1),$ $g(\lambda_{0})=0.708434...$
and

$C(\alpha)=\left(\sqrt{1+\tan^{2}(\pi\alpha)/4}-\tan(\pi\alpha)/2\right)^{2},\quad
0<\alpha\leq 1/2.$

At last, in Section 7, with the help of the concept of strongly starlike
curve, we give upper bounds for $QR_{L}$ for isosceles trapezoidal polygons.

##  2\. Hypergeometric functions and their generalizations

To find some explicit formulas for the interior and exterior conformal moduli
for a trapezoidal curve we need special functions. In this section we recall
the Gaussian and Appell hypergeometric functions and some of their
generalizations [B, BF, AVV].

First we recall that, for $|z|<1$, the Gaussian hypergeometric function is
defined by the equality

${}_{2}F_{1}(a,b;c;z)=\sum_{n=1}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\,\frac{z^{n}}{n!}\,,$

for $|z|<1$, where $(q)_{n}$ denotes the Pochhammer symbol, i.e.
$(q)_{n}=q(q+1)\ldots(q+(n-1))$ for every natural $n$ and $(q)_{0}=1$. It can
be extended analytically to the domain $|z|>1$ along any path avoiding $1$ and
$\infty$. Moreover, we have

(2.1)
$\int_{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-tz)^{-b}dt=B(a,c-a)\,{}_{2}F_{1}(a,b;c;z)$

where $B(\,\cdot\,,\,\cdot\,)$ is the Euler beta function and the integral in
(2.1) converges if ${\rm Re}\,c>{\rm Re}\,a>0$. We also recall that the beta
function can be expressed via the Euler gamma function:

$B(\alpha,\beta)=\,\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\,.$

The complete elliptic integrals ${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}},{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp E}}}$ of the first and
second kinds

(2.2) ${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp
K}}}(\lambda)=\int_{0}^{1}\frac{dt}{\sqrt{(1-t^{2})(1-\lambda^{2}t^{2})}}\,,\quad{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}}(\lambda)=\int_{0}^{1}{\sqrt{\frac{1-\lambda^{2}t^{2}}{1-t^{2}}}}\,{dt}$

are, in fact, special cases of the Gaussian hypergeometric function; we have

${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda)=\frac{\pi}{2}\,\,{}_{2}F_{1}({\textstyle\frac{1}{2}},{\textstyle\frac{1}{2}};1;\lambda^{2}),\quad{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}}(\lambda)=\frac{\pi}{2}\,\,{}_{2}F_{1}(-{\textstyle\frac{1}{2}},{\textstyle\frac{1}{2}};1;\lambda^{2}).$

For the decreasing homeomorphism $\mu:(0,1)\to(0,\infty)$

$\mu(r)=\frac{\pi}{2}\frac{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\sqrt{1-r^{2}})}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(r)}\,,\quad 0<r<1,$

the following differentiation formulas hold [AVV, p.475]

(2.3) $\mu^{\prime}(r)=-\frac{\pi^{2}}{4r(1-r^{2}){\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{2}(r)},\quad\mu^{\prime\prime}(r)=\frac{\pi^{2}(2{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}}(r)-(1+r^{2}){\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(r))}{4r^{2}(1-r^{2})^{2}{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}^{3}(r)}\,.$

Let ${\mathbb{H}}^{2}=\\{z:{\rm Im}z>0\\}$ be the upper half-plane and
$0<r<1.$ Then the modulus of the quadrilateral
$\mbox{\boldmath{$H$}}^{2}_{r}=({\mathbb{H}}^{2};0,1,1/r^{2},\infty)$ can be
written as follows

(2.4)
$\text{Mod}(\mbox{\boldmath{$H$}}^{2}_{r})=\frac{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}(\sqrt{1-r^{2}})}{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(r)}\,.$

This formula follows from the definition, if we use of a canonical conformal
mapping of the the upper half plane onto a rectangle. It also follows from
[HKV, 7.12, 7.33].

The Appell hypergeometric function $F_{1}(a,b_{1},b_{2};c;z,w)$ is defined as

(2.5)
$F_{1}(a;b_{1},b_{2};c;z,w)=\sum_{m,n=0}^{\infty}\frac{(a)_{m+n}(b_{1})_{m}(b_{2})_{n}}{(c)_{m+n}}\,\frac{z^{m}w^{n}}{m!\,n!}\,,$

see, e.g. [B]. The series converges in the bidisk $B^{2}:=\\{|z|,|w|<1\\}$ and
similar to the case of hypergeometric function, can be continued analytically
outside the bidisk $B^{2}$ along any path not containing the points with $z=1$
and $w=1$. The following formula is due to Picard:

(2.6)
$\int_{0}^{1}t^{a-1}(1-t)^{\mu-1}(1-tz)^{-b_{1}}(1-tw)^{-b_{2}}dt=B(a,\mu)F_{1}(a;b_{1},b_{2};a+\mu;z,w).$

The integral converges for ${\rm Re}\,a>0$, ${\rm Re}\,\mu>0$, if $z$, $w\neq
1$. It is evident that if either $b_{2}=0$ or $w=0$, then
$F_{1}(a;b_{1},b_{2};c;z,w)={}_{2}F_{1}(a,b_{1};c;z)$.

The Lauricella hypergeometric function ([L], see also [S]) generalizes the
Appell hypergeometric function for the case of arbitrary number $n$ of
variables $z_{1},\ldots,z_{n}$:

(2.7)
$F^{(n)}_{D}(a;b_{1},\ldots,b_{n};c;z_{1},\ldots,z_{n})=\sum_{m_{1},\ldots,m_{n}\geq
0}\frac{(a)_{m_{1}+\ldots+m_{n}}(b_{1})_{m_{1}}\ldots(b_{n})_{m_{n}}}{(c)_{m_{1}+\ldots+m_{n}}m_{1}!\ldots
m_{n}!}\,z^{m_{1}}_{1}\ldots z^{m_{n}}_{n},$

$|z_{1}|,\ldots,|z_{n}|<1$. It has the integral representation

(2.8) $F^{(n)}_{D}(a;b_{1},\ldots,b_{n};c;z_{1},\ldots,z_{n})\\\
=(B(a,c-a))^{-1}\int_{0}^{1}u^{a-1}(1-u)^{c-a-1}(1-uz_{1})^{-b_{1}}\ldots(1-uz_{n})^{-b_{n}}du,$

${\rm Re}\,c>{\rm Re}\,a>0$, which gives an analytic continuation of the
Lauricella hypergeometric function outside the polydisk $|z_{k}|<1,1\leq k\leq
n$, except for the hyperplanes $z_{k}=1$.

##  3\. Conformal mappings of the half-plane onto the exterior of a polygonal
quadrilateral

Heikkala, Vamanamurthy, and Vuorinen [HVV, pp.78-82] studied the Schwarz-
Christoffel mapping of the upper half-plane onto a convex polygonal
quadrilateral. In this section we consider the conformal mapping of the upper
half-plane onto the exterior of a convex polygonal quadrilateral.

To specify the geometry, we suppose that the bounded polygonal quadrilateral
has four segments as its sides which form the exterior angles $\pi(1+\alpha)$,
$\pi(1+\beta)$, $\pi(1+\gamma)$, and $\pi(1+\delta)$ at the vertices, with the
angle parameters satisfying the constraints

(3.1) $\alpha+\beta+\gamma+\delta=2,\quad 0<\alpha,\beta,\gamma,\delta<1,$

The conformal mapping $f$ of the upper half plane onto the complement of this
quadrilateral is given by the generalized Schwarz-Christoffel formula [AF,
Section 5.6, formula (5.6.3b)]

(3.2)
$f(z)=C_{1}\int_{0}^{z}\frac{\zeta^{\alpha}(\zeta-1)^{\beta}(\zeta-t)^{\gamma}}{(\zeta-
z_{0})^{2}(\zeta-\overline{z}_{0})^{2}}\,d\zeta+C_{2}.$

The preimages of the vertices are the points $0$, $1$, $t$ ($t>1$), and
$\infty$; the point $z_{0}$ is an unknown pole. We note that from (3.1) it
follows, in particular, that the bounded complementary polygonal domain is
convex, since the interior angles at the vertices are $(1-\alpha)\pi$,
$(1-\beta)\pi$, $(1-\gamma)\pi$, and $(1-\delta)\pi$ and $\alpha$ $\beta$,
$\gamma$, $\delta>0$.

To find $z_{0}=x_{0}+iy_{0}$ we derive an equation which is obtained from the
fact that the residue of the integrand at the point $z_{0}$ vanishes. The
integrand is equal to

$\frac{g(\zeta)}{(\zeta-z_{0})^{2}}\,,\quad\text{where}\quad
g(\zeta)=\frac{\zeta^{\alpha}(\zeta-1)^{\beta}(\zeta-t)^{\gamma}}{(\zeta-\overline{z}_{0})^{2}}\,,$

and hence it has a pole of the second order at the point $z_{0}.$ Computation
yields

$\text{res}_{\,\zeta=z_{0}}\frac{g(\zeta)}{(\zeta-
z_{0})^{2}}\,=g^{\prime}(z_{0}).$

Consequently, $z_{0}$ must satisfy the equality $g^{\prime}(z_{0})=0$ which is
equivalent to $(\log g(z_{0}))^{\prime}=0$. At last,

$(\log
g(\zeta))^{\prime}=\frac{\alpha}{\zeta}+\frac{\beta}{\zeta-1}+\frac{\gamma}{\zeta-t}-\frac{2}{\zeta-\overline{z}_{0}}\,,$

and from this we obtain the desired equation for $z_{0}$:

$\frac{\alpha}{z_{0}}+\frac{\beta}{z_{0}-1}+\frac{\gamma}{z_{0}-t}=\frac{1}{iy_{0}}\,,\quad
y_{0}={\rm Im}\,z_{0}.$

######  Lemma 3.3.

In the upper half-plane $\\{z:{\rm Im}\,z>0\\}$ the equation

(3.4) $\frac{\alpha}{z}+\frac{\beta}{z-1}+\frac{\gamma}{z-t}=\frac{1}{iy},$

$z=x+iy$, has a unique solution.

###### Proof.

We can represent (3.4) in the form

$z-\overline{z}=\frac{2z(z-1)(z-t)}{Q(z)}$

where

$Q(z)=\alpha(z-1)(z-t)+\beta z(z-t)+\gamma
z(z-1)=(\alpha+\beta+\gamma)z^{2}-(\alpha(1+t)+\beta t+\gamma)z+\alpha t.$

Therefore, $\overline{z}=R(z)$ where

(3.5)
$R(z)=z-\frac{2z(z-1)(z-t)}{Q(z)}\,=\frac{(\alpha+\beta+\gamma-2)z^{3}-((\alpha-2)(1+t)+\beta
t+\gamma)z^{2}+(\alpha-2)tz}{Q(z)}.$

Since $R(z)$ is a rational function with real coefficients, we have
$\overline{R(z)}=R(\overline{z})$ and, therefore, all solutions of (3.4) are
also solutions of the equation

(3.6) $z=R(R(z))=R(z)-\frac{2R(z)(R(z)-1)(R(z)-t)}{Q(R(z))}\,.$

The function $R(R(z))$ is a rational function of degree $9$; for real $x$ it
has poles at the points where $Q(x)=0$ and $Q(R(x))=0$. For the quadratic
function $Q(x)$ we have $Q(0)$, $Q(t)>0$ and $Q(1)<1$, therefore, it has two
real zeroes $x_{1}\in(0,1)$ and $x_{2}\in(1,t)$. Moreover,

$\lim_{x\to x_{k}-}R(x)=-\infty,\quad\lim_{x\to x_{k}+}R(x)=+\infty,\quad
k=1,2,$

and, because of the inequality $\alpha+\beta+\gamma<2$, we have

$\lim_{x\to-\infty}R(x)=+\infty,\quad\lim_{x\to+\infty}R(x)=-\infty.$

This implies that on each of the intervals $(-\infty,x_{1})$, $(x_{1},x_{2})$
and $(x_{2},+\infty)$ there are exactly two points where either $R(x)=x_{1}$
or $R(x)=x_{2}$. Therefore, in addition to $x_{1}$ and $x_{2}$, we have six
points on the real axis where $R$ has poles. It is evident that all these
eight points, which are poles of $R(R(x))$, are different. Denote them by
$\tau_{j}$, $1\leq j\leq 8$; the points are labelled so that
$\tau_{1}<\tau_{2}<\ldots<\tau_{8}$.

Now we will show that $R^{\prime}(x)<0$ for real $x$ different from the points
$x_{1}$ and $x_{2}$. We have

$R(x)=x-\frac{2}{f(x)},\quad
f(x)=\frac{\alpha}{x}+\frac{\beta}{x-1}+\frac{\gamma}{x-t}\,.$

Consequently,

$R^{\prime}(x)=1+\frac{2f^{\prime}(x)}{f^{2}(x)},\quad
f^{\prime}(x)=-\frac{\alpha}{x^{2}}-\frac{\beta}{(x-1)^{2}}-\frac{\gamma}{(x-t)^{2}}\,<0.$

From the Cauchy-Schwarz inequality we obtain

$f^{2}(x)=\left(\frac{\alpha}{x}+\frac{\beta}{x-1}+\frac{\gamma}{x-t}\,\right)^{2}\leq(\alpha+\beta+\gamma)\left(\frac{\alpha}{x^{2}}+\frac{\beta}{(x-1)^{2}}+\frac{\gamma}{(x-t)^{2}}\right)<2|f^{\prime}(x)|,$

and hence, $R^{\prime}(x)<0$. It is evident that on every one of the intervals
$I_{j}:=(\tau_{j},\tau_{j+1})$, $1\leq j\leq 7$, $R(R(x))$ is strictly
increasing. Then

$\lim_{x\to\tau_{j}+}(x-R(R(x)))=+\infty,\quad\lim_{x\to\tau_{j+1}-}(x-R(R(x)))=-\infty.$

Therefore, $x-R(R(x))$ takes all real values on $I_{j}$ and there exists a
point $s_{j}\in I_{j}$ such that $s_{j}=R(R(s_{j}))$. From this we conclude
that every $s_{j}$, $1\leq j\leq 7$, satisfies the equality (3.6). From (3.5)
we conclude that $R(x)$ is a rational function of degree $3$ which is the
ratio of two polynomials of degrees $3$ and $2$. Moreover,
$R(x)\sim(1-2(\alpha+\beta+\gamma)^{-1})x$, $x\to\infty$. Therefore, $R(R(x))$
is a rational function of degree $9$ and
$R(R(x))\sim(1-2(\alpha+\beta+\gamma)^{-1})^{2}x$, $x\to\infty$. This implies
that the rational function $x-R(R(x))$ has degree $9$. Since at least $7$ of
its zeros are real, we conclude that it has at most one complex (i.e. non-
real) root in the upper half-plane. Since all solutions of (3.4) satisfy
(3.6), we see that (3.4) has at most one complex root in the upper half-plane.

To prove that such a root exists, we note that if we fix one side of the
boundary polygon $A_{1}A_{2}A_{3}A_{4}$, say $A_{1}A_{2},$ and assume that it
coincides with the segment $[0,1]$ of the real axis and change the length of
the other side, $A_{2}A_{3}$ from $0$ to $\infty$ (with fixed values of angles
$\alpha$, $\beta$, and $\gamma$), then the exterior conformal moduli of the
obtained polygons continuously increase from $0$ to $+\infty$ and, therefore,
there exists such a polygon $P$ that its modulus coincides with the modulus of
the upper half-plane with vertices at the points $0$, $1$, $t$, and $\infty$.
Then there exists a conformal mapping $f$ of the upper half-plane onto the
exterior of $P$ such that the points $0$, $1$, $t$, and $\infty$ are mapped to
the vertices of $P$. Denote $z_{0}=f^{-1}(\infty)$. Then $f$ has the form
(3.2) and $z_{0}$ satisfies (3.4). ∎

Now we will give a formula for the unique solution of (3.4). Denote

$E=\alpha+\beta+\gamma-1,$ $A=2(E-1)^{2},$
$B=(E-1)[4-3(\alpha+\gamma)+(4-3(\alpha+\beta))t],$
$C=2-3(\alpha+\gamma)+(\alpha+\gamma)^{2}+2(3-5\alpha-2\beta-2\gamma+2\alpha^{2}+2\alpha\beta+2\alpha\gamma+\beta\gamma)t$
$+(2-3(\alpha+\beta)+(\alpha+\beta)^{2})t^{2},$
$D=(1-\alpha)(\alpha+\gamma-1+(\alpha+\beta-1)t)t,$ (3.7)
$\rho(x)=\frac{\alpha tx}{(1-E)x+E(t+1)-\gamma t-\beta}.$

######  Lemma 3.8.

The unique solution of (3.4), lying in the upper half-plane, has the form
$z_{0}=x_{0}+iy_{0}$ where $x_{0}$ is the unique solution of the cubic
equation

(3.9) $Ax^{3}+Bx^{2}+Cx+D=0,$

satisfying the inequality $x^{2}<\rho(x)$ and
$y_{0}=\sqrt{\rho(x_{0})-x_{0}^{2}}$.

###### Proof.

Multiplying both sides of (3.4) by $z-t$ we obtain

$\frac{\alpha(z-t)}{z}+\frac{\beta(z-t)}{z-1}+\gamma=1+\frac{x-t}{iy}\,,$

consequently,

${\rm Re}\,\left[\frac{\alpha t}{z}+\frac{\beta(t-1)}{z-1}\right]=E$

and

$\frac{\alpha tx}{|z|^{2}}+\frac{\beta(t-1)(x-1)}{|z-1|^{2}}=E.$

If we denote $\rho=x^{2}+y^{2}$, we have

$\frac{\alpha tx}{\rho}+\frac{\beta(t-1)(x-1)}{\rho+1-2x}=E$

and

$\alpha tx(\rho+1-2x)+\rho\beta(t-1)(x-1)=E\rho(\rho+1-2x).$

Similarly, multiplying both sides of (3.4) by $z-1$ we obtain

$\frac{\alpha(z-1)}{z}+{\beta}+\frac{\gamma(z-1)}{z-t}=1+\frac{x-1}{iy}.$
${\rm Re}\,\left[\frac{\alpha}{z}+\frac{\gamma(1-t)}{z-t}\right]=E,$
$\frac{\alpha x}{\rho}+\frac{\gamma(1-t)(x-t)}{\rho+t^{2}-2tx}=E,$ $\alpha
x(\rho+t^{2}-2tx)+\rho\gamma(1-t)(x-t)=E\rho(\rho+t^{2}-2tx).$

Thus, we have a system of two equations with respect to $x$ and $\rho$:

(3.10) $\alpha tx(\rho+1-2x)+\rho\beta(t-1)(x-1)=E\rho(\rho+1-2x).$ (3.11)
$\alpha x(\rho+t^{2}-2tx)+\rho\gamma(1-t)(x-t)=E\rho(\rho+t^{2}-2tx).$

Subtracting (3.10) from (3.11) we obtain

$E\rho(t^{2}-1-2(t-1)x)=\alpha x(1-t)\rho+\alpha
x(t^{2}-t)+\rho(1-t)((\gamma+\beta)x-\gamma t-\beta)$

This is a linear equation with respect to $\rho$. Solving it, we obtain (3.7).
If we substitute this expression into any of the equations (3.10), (3.11), we
find a cubic equation (3.9) for $x_{0}$.

By Lemma 3.3, there exists only one root $x_{0}$ of (3.9) satisfying the
inequality $x^{2}<\rho(x)$. Then we find
$y_{0}=\sqrt{\rho^{2}(x_{0})-x_{0}^{2}}$. ∎

After finding the value of $z_{0}$ we can simply express the coordinates of
the vertices in terms of $\alpha$, $\beta$, $\gamma$, and $t$. Further, we
will assume that the vertices of the boundary polygon, corresponding to the
points $0$, $1$, $t$ and $\infty$, are located at the points $A_{1}=1$,
$A_{2}=0$, $A_{3}$ and $A_{4}$. Then the conformal mapping of the upper half-
plane onto the exterior of the polygonal region is given by the formula

(3.12) $f(z)=1-\frac{h(z)}{h(1)},$

where

(3.13)
$h(z)=\int_{0}^{z}\frac{x^{\alpha}(1-x)^{\beta}(1-x/t)^{\gamma}dx}{(1-x/z_{0})^{2}(1-x/\overline{z}_{0})^{2}}\,.$

(The branch of the integrand is fixed such that it takes positive values on
$(0,1)$). Here $z_{0}=z_{0}(t)$ is described in Lemma 3.8. With the help of
the Lauricella function $F^{(n)}_{D}$, we can write

(3.14)
$h(1)=B(1+\alpha,1+\beta)F_{D}^{(3)}(1+\alpha;-\gamma,2,2;2+\alpha+\beta;1/t,1/z_{0},1/\overline{z}_{0}),$
$h(z)=\frac{z^{1+\alpha}}{1+\alpha}\,F_{D}^{(4)}(1+\alpha;-\beta,-\gamma,2,2;2+\alpha;z,z/t,z/z_{0},z/\overline{z}_{0}).$

The length of the side $A_{2}A_{3}$ is given by the formula

(3.15) $l_{2}:=|A_{2}A_{3}|=\frac{I}{h(1)}\,,\quad
I=\int_{1}^{t}\frac{x^{\alpha}(x-1)^{\beta}(1-x/t)^{\gamma}dx}{(1-x/z_{0})^{2}(1-x/\overline{z}_{0})^{2}}\,.$

After the change of variables $x=1+(t-1)\tau$ we find

(3.16)
$I=\int_{1}^{t}\frac{x^{\alpha}(x-1)^{\beta}(1-x/t)^{\gamma}dx}{(1-x/z_{0})^{2}(1-x/\overline{z}_{0})^{2}}\,=\frac{(t-1)^{1+\beta+\gamma}|z_{0}|^{4}}{t^{\gamma}|z_{0}-1|^{4}}\,\int_{0}^{1}\frac{\tau^{\beta}(1-\tau)^{\gamma}(1+(t-1)\tau)^{\alpha}d\tau}{\left(1-\frac{t-1}{z_{0}-1}\tau\right)^{2}\left(1-\frac{t-1}{\overline{z}_{0}-1}\tau\right)^{2}}\\\
=\frac{(t-1)^{1+\beta+\gamma}|z_{0}|^{4}}{t^{\gamma}|z_{0}-1|^{4}}\,B(1+\beta,1+\gamma)F^{(3)}_{D}(1+\beta;-\alpha,2,2;2+\beta+\gamma;-(t-1),\textstyle\frac{t-1}{z_{0}-1},\frac{t-1}{\overline{z}_{0}-1}).$

Denote $r=1/\sqrt{t}\in(0,1)$. From the conformal invariance of the modulus we
obtain that the desired exterior conformal modulus Mod($Q$) coincides with the
conformal modulus of the quadrilateral $\mbox{\boldmath{$H$}}^{2}_{r}$ which
is the upper half-plane with vertices $z_{1}=0$, $z_{2}=1$, $z_{3}=t=1/r^{2}$,
and $z_{4}=\infty$. Applying (2.4), we have

(3.17) $\text{Mod}({\mbox{\boldmath{$Q$}}})={\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(r^{\prime})/{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(r),\quad r^{\prime}=\sqrt{1-r^{2}},$

where ${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(r)$ is the complete elliptic integral of the first kind.

We can also find the lengths of sides $l_{3}=|A_{3}A_{4}|$ and
$l_{4}=|A_{4}A_{1}|$ and the vertices $A_{3}$ and $A_{4}$. It is easy to
verify that

(3.18)
$l_{3}=\frac{\sin\pi\alpha+l_{2}\sin[\pi(\alpha+\beta)]}{\sin\pi\delta}\,,\quad
l_{4}=\frac{\sin[\pi(\beta+\gamma)]+l_{2}\sin\pi\gamma}{\sin\pi\delta}\,,$
(3.19) $A_{3}=-l_{2}e^{-\pi\beta i},\quad
A_{4}=A_{3}-l_{3}e^{-\pi(\beta+\gamma)i}=1+l_{4}e^{\pi\alpha i}.$

Therefore, we have

######  Theorem 3.20.

Let $f(z)=1-\frac{h(z)}{h(1)}$ where $h$ is given by (3.13). Then $f$ maps
conformally the upper half-plane onto the exterior of the polygonal line
$A_{1}A_{2}A_{3}A_{4}$ where $A_{1}=1$, $A_{2}=0$, $A_{3}$ and $A_{4}$ are
given by (3.19) where $l_{3}$ and $l_{4}$ are given by (3.18) and $l_{2}$;
here $l_{2}$ is defined by (3.15), $h(1)$ and $I$ are given by (3.14) and
(3.16), taking into account (3.14) and (3.16). The length of sides of
$A_{1}A_{2}A_{3}A_{4}$ are $l_{1}=1$, $l_{2}$, $l_{3}$, and $l_{4}$.

It is evident that, under the above assumptions, the exterior of the
quadrilateral is defined uniquely by the value of the length $l_{2}$ (or
$l_{4}$). From Lemma 3.8 it follows

######  Theorem 3.21.

For fixed angles $(1+\alpha)\pi$, $(1+\beta)\pi$, $(1+\gamma)\pi$, and
$(1+\delta)\pi$, the exterior conformal modulus is strictly increasing as a
function of $l_{2}$.

Actually, if it is not the case, then we can find two different (exterior)
quadrilaterals with vertices $A_{1}$, $A_{2}$, $A^{1}_{3}$, $A^{1}_{4}$ and
$A_{1}$, $A_{2}$, $A^{2}_{3}$, $A^{2}_{4}$ such that $A_{1}=1$, $A_{2}=0$, and
their conformal moduli coincide. Then the conformal mappings of the upper
half-plane onto these exterior quadrilaterals is defined by the formula

$f_{k}(z)=C_{1k}\int_{0}^{z}\frac{\zeta^{\alpha}(\zeta-1)^{\beta}(\zeta-
t_{k})^{\gamma}}{(\zeta-
z_{0k})^{2}(\zeta-\overline{z}_{0k})^{2}}\,d\zeta+1,\quad k=1,2.$

Since the moduli of the quadrilaterals are uniquely defined by $t_{k}$ and the
dependence is strictly monotone, we see that $t_{1}=t_{2}$. Then, by Lemma
3.8, we obtain that $z_{01}=z_{02}$. From the normalization $f_{k}(1)=0$,
$k=1$, $2$ it follows that $C_{11}=C_{12}$. Thus, $f_{1}\equiv f_{2}$ and,
therefore, the exterior quadrilaterals coincides, which contradicts our
assumptions.

Now we will describe how to determine the conformal modulus of the given
exterior polygonal quadrilateral with angles $(1+\alpha)\pi$, $(1+\beta)\pi$,
$(1+\gamma)\pi$, and $(1+\delta)\pi$, satisfying (3.1).

######  Theorem 3.22.

For a given exterior quadrilateral, the conformal modulus is given by the
formula (3.17) where $r=1/\sqrt{t}$ and $t$ is a unique solution to the
equation (3.15) where $h(1)$ and $I$ are defined by (3.14) and (3.16), keeping
in mind that $z_{0}=z_{0}(t)=x_{0}(t)+i\sqrt{r(x_{0}(t))-x_{0}^{2}(t)}$,
$x_{0}(t)$ is a solution to the equation (3.9) satisfying the inequality
$x^{2}<\rho(x)$ and $\rho(x)$ is given by (3.7).

######  Remark 3.23.

If we consider the general (non-convex) case, where $\alpha$, $\beta$,
$\gamma$, and $\delta$ satisfy the conditions

$\alpha+\beta+\gamma+\delta=2,\quad-1<\alpha,\beta,\gamma,\delta<1,$

instead of (3.1), then we obtain that, by Lemma 3.8, for every $M$ there are
at most three different exterior quadrilaterals of the form
$A_{1}A_{2}A_{3}A_{4}$, $A_{1}=1$, $A_{2}=0$, with angles $(1+\alpha)\pi$,
$(1+\beta)\pi$, $(1+\gamma)\pi$, and $(1+\delta)\pi$, such that their
conformal moduli equal $M$.

In connection with Remark 3.23, we can suggest

######  Conjecture 3.24.

For every $M>0$ there is only one exterior quadrilateral with given angles and
vertices $A_{1}=1$, $A_{2}=0$, $A_{3}$, $A_{4}$, conformal modulus of which
equals $M$.

With the help of Theorem 3.22, we can calculate the exterior modulus of a
sufficiently arbitrary convex polygonal line $A_{1}A_{2}A_{3}A_{4}$ with
vertices $A_{1}=1$, $A_{2}=0$, $A_{3}=A$, and $A_{4}=B$. With the help of
Wolfram Mathematica package, we created a function ExtMod$[A,B,n,wp]$ which
calculated the values of the exterior modulus and also gives the values of the
parameters $\alpha$, $\beta$, $\gamma$, $\delta$, $t$, and $z_{0}$; here $n$
and $wp$ are some additional parameters; their meaning will be explained
below. The code is contained in Appendix A. Now we briefly describe its
structure.

To find the exterior modulus with the help of (3.17), we determine the value
of $t=1/r^{2}$. For this, we use the bisection method on the segment
$[1,T_{2}]$, $T_{2}=10^{n}$, $n>0$; the number of iterations is $S=[5(n+15)]$
where $[x]$ denotes the integer part of $x$. The input consists of coordinates
of the vertices $A_{3}$ and $A_{4}$, the value of $n$ and the parameter $wp$
that specifies how many digits of precision should be maintained in internal
computations (the Wolfram Mathematica option "WorkingPrecision"). The program
calculates the values of angles $\alpha$, $\beta$, $\gamma$, and $\delta$,
which are denoted for short by $a$, $b$, $c$, and $d$. Then, on every
iteration, we solve the cubic equation (3.9) with coefficients corresponding
to the current value of $t$ and determine its unique root $x_{0}$ satisfying
$x^{2}<\rho(x)$. After this, we find $y_{0}$, $z_{0}=x_{0}+iy_{0}$ and
calculate the integrals

(3.25)
$J_{1}=\int_{0}^{1}\frac{x^{\alpha}(1-x)^{\beta}(1-x/t)^{\gamma}dx}{|1-x/z_{0}|^{4}}\,,\quad
J_{2}=\int_{1}^{t}\frac{x^{\alpha}(x-1)^{\beta}(1-x/t)^{\gamma}dx}{|1-x/z_{0}|^{4}}\,,$

corresponding to the current value of the parameter $t$, and compare its ratio
$J_{2}/J_{1}$ with $L$ which is the length of the side $|A_{2}A_{3}|$. (In
fact, $J_{2}/J_{1}$ coincides with $l_{2}$ given by (5.4).)

Now we give a suggestion how to fix the value of $n$ if we can obtain an a
priori estimate from above for the value of the desired exterior modulus $M$.
If $M$ is less than $2.3$, we can take $n=2$. For $2.3\leq M\leq 11.8$ the
value $n=(10M-6)/7$ is suitable. We note that for $M>11.8$ the program does
not work properly because of degeneration of the elliptic integrals.

Now we will give results of some numeric examples.

######  Example 3.26.

In [NRRV, subsect. 5.3] exterior polygonal quadrilaterals with the following
vertices are considered:

(1) $A_{1}=1$, $A_{2}=0$, $A_{3}=-19/25+i21/25$, $A_{4}=28/25+i69/50$;

(2) $A_{1}=1$, $A_{2}=0$, $A_{3}=-3/25+i21/25$, $A_{4}=42/25+4i$.

With the use of the boundary integral equations method, approximate values of
their exterior moduli $M_{1}$ and $M_{2}$ were found.

We calculated the moduli $M_{1}$ and $M_{2}$ by our method and the results are
given in Table 1 (see also Appendix A). To estimate the accuracy of our
calculations we also find the moduli of the conjugate quadrilaterals,
$M_{1}^{*}$ and $M_{2}^{*}$. Theoretically, the values of $M_{j}$ coincide
with $(M_{j}^{*})^{-1}$ but difference in approximate values shows how the
obtained values are distinct from the exact ones. We see that the values of
our calculations coincide with those from [NRRV] with accuracy $10^{-9}$. The
differences between $M_{j}$ and $(M_{j}^{*})^{-1}$ do not exceed $5\cdot
10^{-15}$; this gives a reason to hope that in the values of moduli we
received $14$ correct digits after the decimal dot.

On Fig. 1 we give the exterior quadrilateral and the image of a grid under the
mapping (3.12) for the case (1).

Figure 1. The exterior quadrilateral and the image of a grid under the mapping (3.12) for Example 3.26, Case (1). Table 1. The values of exterior moduli for two quadrilaterals. | $\qquad\ j=1$ | $\qquad\ j=2$
---|---|---
Approx. values of $M_{j}$ | $0.992341633097863$ | $0.959257171919002$
Approx. values of $(M_{j}^{*})^{-1}$ | $0.992341633097868$ | $0.959257171919007$
Approx. values of $M_{j}$ from [NRRV] | $0.9923416331$ | $0.9592571729$

######  Example 3.27.

Now consider the case of the exterior of a rectangle with vertices $A_{1}=1$,
$A_{2}=0$, $A_{3}=iH$, $A_{4}=1+iH$, $H>0$. The function

(3.28) $w=\frac{1}{k}\,\frac{z-\sqrt{t}}{z+\sqrt{t}}\,,\quad
k=\frac{\sqrt{t}-1}{\sqrt{t}+1}\,,$

maps the upper half-plane onto itself with the correspondence of points
$0\mapsto-1/k$, $1\mapsto-1$, $t\mapsto 1$, $\infty\mapsto 1/k$. According to
the Duren-Pfaltzgraff formula [DP, sect. (iv)],

(3.29) $H=\frac{2\mathchoice{\hbox{\,\fFt E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa
E}}{\hbox{\,\fFp E}}(k)-(1-k)\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(k)}{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}^{\prime}(k)-k\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}^{\prime}(k)}\,.$

Here $\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}(k)$ and $\mathchoice{\hbox{\,\fFt E}}{\hbox{\,\fFt
E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp E}}(k)$ are the complete elliptic integrals
given by (2.2), $\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}^{\prime}(k)=\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(k^{\prime})$, $\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}^{\prime}(k)=\mathchoice{\hbox{\,\fFt E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa
E}}{\hbox{\,\fFp E}}(k^{\prime})$, where $k^{\prime}=\sqrt{1-k^{2}}$. Using
(3.29), we can check the accuracy of our calculation. For a given $H$, we find
the approximate value of $t$, then, with the help of (3.28), determine $k$ and
by (3.29) find the value of $H_{\text{app}}$ corresponding to the found
approximate values of parameters. Comparing $H_{\text{app}}$ with the initial
value of $H$, we can estimate the accuracy of the approximate method.

In Table 2, for some $H$ we give the values of the exterior moduli $M$ and the
corresponding $H_{\text{app}}$. It is interesting that the method gives very
good results even for very large $H$. Comparing $H_{\text{app}}$ with $H$
shows that the accuracy of results for large $H$ is much better than those
obtained by considering the conjugate modulus $M^{*}$ (we can simply put
$H^{-1}$ instead of $H$) and funding, after this, the reciprocal value
$(M^{*})^{-1}$.

Table 2. The values of exterior moduli $M$ and $k$ for some rectangles. $H$ |  $H_{\text{app}}$ |  $M$ |  $k$
---|---|---|---
1 | 0.999999999999984 | 0.999999999999997 | 0.1715728752538083
2 | 1.999999999999971 | 1.154924858699707 | 0.2589511664373517
3 | 2.999999999999959 | 1.254423186704834 | 0.3183618249446048
4 | 3.999999999999940 | 1.328560829309608 | 0.3630445515606185
5 | 4.999999999999927 | 1.387897041604210 | 0.3985903936736862
10 | 9.999999999999874 | 1.580900257847724 | 0.5096661128249422
50 | 49.99999999999927 | 2.062779488244626 | 0.7306010544314864
100 | 99.99999999999962 | 2.278195883070594 | 0.7996714751224258
$\vphantom{10^{4^{3}}}10^{3}$ | 999.9999999999126 | 3.005361525457626 | 0.9312093496761309
$\vphantom{10^{4^{3}}}10^{4}$ | 10000.00000000519 | 3.737506317586474 | 0.9776888723313666
$\vphantom{10^{4^{3}}}10^{5}$ | 100000.0000021733 | 4.470341757015527 | 0.9928890530750033
$\vphantom{10^{4^{3}}}10^{6}$ | 1000000.000038778 | 5.203265238854191 | 0.997745791670292

From Table 2 we see that for $H<100$ the $|H_{\text{app}}-H|<10^{-13}$. For
large $H$, the relative error grows but even for $H=10^{6}$ it less than
$10^{-10}$ what can be considered a very good result.

##  4\. Conformal mappings of the interior and exterior of an isosceles
trapezoidal polygon onto the half-plane

######  4.1.

Interior of a trapezoidal polygon.

Further, for convenience, in investigation of the interior and exterior moduli
for considered trapezoidal lines, we will assume that one of its parallel
sides is on the real axis and, therefore, $A_{1}A_{2}$ does not coincide with
the segment $[0,1]$. This assumption does not play a significant role because
it is evident that the formulas for the corresponding conformal mappings can
be simply obtained from each other by applying conformal automorphisms of the
complex plane of the form $z\mapsto a_{0}z+b_{0}$, $a_{0}$,
$b_{0}\in\mathbb{C}$. So, let $L$ be the boundary of a trapezoid with vertices
$A_{1}(-d-i)$, $A_{2}(-c)$, $A_{3}(c)$ and $A_{4}(d-i)$. Here $d>c>0$ (Fig.
2).

Denote by $T^{+}=T^{+}(c,\alpha)$ and $T^{-}=T^{-}(c,\alpha)$ the interior and
the exterior of $L$. Let $\alpha$ be the value of the angle of $T^{+}$ at
$A_{4}$. Then the angles of $T^{+}$ at $A_{1}$, $A_{2}$, $A_{3}$, and $A_{4}$
are equal $\pi\alpha$, $\pi(1-\alpha)$, $\pi(1-\alpha)$, and $\pi\alpha$. The
corresponding angles of $T^{-}$ are equal $\pi(2-\alpha)$, $\pi(1+\alpha)$,
$\pi(1+\alpha)$, and $\pi(2-\alpha)$. Moreover, $d-c=\cot(\pi\alpha)$ and the
both non-horizontal sides are of length equal to $1/\sin(\pi\alpha)$.

Figure 2. Trapezoidal line $L$.

Let us map the lower half-plane conformally onto $T^{+}$ such that
$-(1/\lambda)\mapsto A_{1}$, $-1\mapsto A_{2}$, $1\mapsto A_{3}$, and
$1/\lambda\mapsto A_{4}$. Here $0<\lambda<1$ is some number depending on the
modulus of $T^{+}$. According to the Schwarz-Christoffel formula, the mapping
is given by the formula

$f^{+}(z)=C\int_{0}^{z}(t^{2}-1)^{-\alpha}(\lambda^{2}t^{2}-1)^{\alpha-1}\,dt$

with the constant

$C=c/I,\quad
I=\int_{0}^{1}(1-t^{2})^{-\alpha}(1-\lambda^{2}t^{2})^{\alpha-1}\,dt.$

Comparing the lengths of the bases, we obtain

(4.2)
$\frac{\int_{0}^{1}(1-t^{2})^{-\alpha}(1-\lambda^{2}t^{2})^{\alpha-1}\,dt}{\int_{1/\lambda}^{\infty}(t^{2}-1)^{-\alpha}(\lambda^{2}t^{2}-1)^{\alpha-1}\,dt}\,=\frac{c}{d}\,.$

From (4.2) we can find the parameter $\lambda$ and the conformal modulus of
$T^{+}$:

$\text{mod}\,(T^{+})=\frac{2{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)}{{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda^{\prime})}\,,\quad\lambda^{\prime}=\sqrt{1-\lambda^{2}},$

where ${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(\lambda)$ is the complete elliptic integral of the first
kind defined in (2.2).

Now we will express the above integrals (4.2) via the Gaussian hypergeometric
function.

The change of variables $\tau=t^{2}$ and (2.1) yield

(4.3)
$\int_{0}^{1}(1-t^{2})^{-\alpha}(1-\lambda^{2}t^{2})^{\alpha-1}\,dt=\frac{1}{2}\,B(1-\alpha,{\textstyle\frac{1}{2}})\,{}_{2}F_{1}({\textstyle\frac{1}{2}},1-\alpha;{\textstyle\frac{3}{2}}-\alpha;\lambda^{2}).$

Another change of variables $s=1/(\lambda t)$ and the substitution
$\tau=s^{2}$ lead to

(4.4)
$\int_{1/\lambda}^{\infty}(t^{2}-1)^{-\alpha}(\lambda^{2}t^{2}-1)^{\alpha-1}\,dt=\lambda^{2\alpha-1}\int_{0}^{1}(1-s^{2})^{\alpha-1}(1-\lambda^{2}s^{2})^{-\alpha}\,ds\\\
=\frac{1}{2}\,B(\alpha,{\textstyle\frac{1}{2}})\,\lambda^{2\alpha-1}{}_{2}F_{1}({\textstyle\frac{1}{2}},\alpha;\alpha+{\textstyle\frac{1}{2}};\lambda^{2}).$

The relations (4.3) and (4.4) allow us to write (4.2) in the form

(4.5)
$\frac{B(\alpha,{\textstyle\frac{1}{2}})}{B(1-\alpha,{\textstyle\frac{1}{2}})}\,\frac{\lambda^{2\alpha-1}{}_{2}F_{1}({\textstyle\frac{1}{2}},\alpha;\alpha+{\textstyle\frac{1}{2}};\lambda^{2})}{{}_{2}F_{1}({\textstyle\frac{1}{2}},1-\alpha;{\textstyle\frac{3}{2}}-\alpha;\lambda^{2})}=\frac{d}{c}\,.$

From the geometric reasoning we conclude that (4.5) has a unique solution
$\lambda$ on $(0,1)$.

We will also need the boundary correspondence between points of the real axis
and points on the bases of the trapezoid. For $x\in[0,1]$, with the help of
the change of variables, $t=xs$, $s^{2}=\tau$ and (2.6), we have

$\int_{0}^{x}(1-t^{2})^{-\alpha}(1-\lambda^{2}t^{2})^{\alpha-1}\,dt=x\int_{0}^{1}(1-x^{2}s^{2})^{-\alpha}(1-\lambda^{2}x^{2}s^{2})^{\alpha-1}\,ds\\\
=xF_{1}({\textstyle\frac{1}{2}};\alpha,1-\alpha;{\textstyle\frac{3}{2}};x^{2},x^{2}\lambda^{2}).$

where $F_{1}$ is the Appell hypergeometric function (2.5). Similarly, for
$x>1/\lambda$ we obtain

$\int_{x}^{\infty}(t^{2}-1)^{-\alpha}(\lambda^{2}t^{2}-1)^{\alpha-1}\,dt=x\int_{0}^{1}(x^{2}-s^{2})^{-\alpha}(\lambda^{2}x^{2}-s^{2})^{\alpha-1}\,ds\\\
=\lambda^{2(\alpha-1)}x^{-1}F_{1}({\textstyle\frac{1}{2}};\alpha,1-\alpha;{\textstyle\frac{3}{2}};x^{-2},x^{-2}\lambda^{-2}).$

######  4.6.

Exterior of a trapezoidal polygon.

Now we describe the conformal mapping of the upper half-plane onto the
exterior $T^{-}$ of a trapezoidal polygon. We will assume that the pole of the
mapping function is at the point $i$. Let $A_{1}$, $A_{2}$, $A_{3}$, and
$A_{4}$ correspond to the points $-b$, $-a$, $a$, and $b$ for some $0<a<b$.

According to the generalized Schwarz-Christoffel formula (3.2), we have

$f^{-}(z)=\widetilde{C}\int_{0}^{z}\frac{(t^{2}-a^{2})^{\alpha}(t^{2}-b^{2})^{1-\alpha}}{(1+t^{2})^{2}}\,dt$

with the constant

$\widetilde{C}=c/\widetilde{I},\quad\widetilde{I}=\int_{0}^{a}\frac{(t^{2}-a^{2})^{\alpha}(b^{2}-t^{2})^{1-\alpha}}{(1+t^{2})^{2}}\,dt.$

Because the residue of the integrand vanishes at the point $t=i$, we deduce
that the values of $a$ and $b$ are connected by the equality

(4.7) $\frac{\alpha}{1+a^{2}}\,+\,\frac{1-\alpha}{1+b^{2}}\,=\frac{1}{2}\,.$

Denote $k=a/b$, $0<k<1$.

Comparing the lengths of the sides we obtain

$\frac{\displaystyle\int_{0}^{a}\frac{(a^{2}-t^{2})^{\alpha}(b^{2}-t^{2})^{1-\alpha}}{(1+t^{2})^{2}}\,dt}{\displaystyle\int_{b}^{\infty}\frac{(t^{2}-a^{2})^{\alpha}(t^{2}-b^{2})^{1-\alpha}}{(1+t^{2})^{2}}\,dt}=\frac{c}{d}\,.$

After the change of variables $t=as$, we have

(4.8)
$\displaystyle\frac{\displaystyle\int_{0}^{1}\frac{(1-s^{2})^{\alpha}(1-k^{2}s^{2})^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,ds}{\displaystyle\int_{1/k}^{\infty}\frac{(s^{2}-1)^{\alpha}(k^{2}s^{2}-1)^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,ds}=\frac{c}{d}\,.$

From (4.7) we find that, for a fixed $\alpha$,

(4.9) $a^{2}=a^{2}(k)=\sqrt{A^{2}+k^{2}}-A,\quad
A=\textstyle(\frac{1}{2}-\alpha)(1-k^{2}).$

Solving (4.8), we find the value of $k$ and then

$\text{mod}\,(T^{-})=\frac{2{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(k)}{{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(k^{\prime})}\,,\quad
k^{\prime}=\sqrt{1-k^{2}}.$

Now we will write the integrals from (4.8) through special functions. After
the change of variable $s^{2}=\tau$, with the help of (2.6), we obtain:

$\int_{0}^{1}\frac{(1-s^{2})^{\alpha}(1-k^{2}s^{2})^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,dt=\frac{1}{2}\,B({\textstyle\frac{1}{2}},1+\alpha)F_{1}({\textstyle\frac{1}{2}};\alpha-1,2;{\textstyle\frac{3}{2}}+\alpha;k^{2},-a^{2}).$

Similarly,

$\displaystyle\int_{1/k}^{\infty}\frac{(s^{2}-1)^{\alpha}(k^{2}s^{2}-1)^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,ds=k^{3-2\alpha}\int_{0}^{1}\frac{(1-k^{2}\tau^{2})^{\alpha}(1-\tau^{2})^{1-\alpha}}{(a^{2}+k^{2}\tau^{2})^{2}}\\\
=\frac{k^{3-2\alpha}}{2a^{4}}\,B({\textstyle\frac{1}{2}},2-\alpha)F_{1}({\textstyle\frac{1}{2}};-\alpha,2;{\textstyle\frac{5}{2}}-\alpha;k^{2},-a^{2}k^{2}).$

Therefore, we have the equation to determine $k$:

(4.10)
$\frac{k^{3-2\alpha}B({\textstyle\frac{1}{2}},2-\alpha)}{2a^{4}B({\textstyle\frac{1}{2}},1+\alpha)}\frac{F_{1}({\textstyle\frac{1}{2}};-\alpha,2;{\textstyle\frac{5}{2}}-\alpha;k^{2},-a^{2}k^{2})}{F_{1}({\textstyle\frac{1}{2}};\alpha-1,2;{\textstyle\frac{3}{2}}+\alpha;k^{2},-a^{2})}=\,\frac{d}{c}$

where $a=a(k)$ (see (4.9)).

Now, as in the case of the interior modulus, we find the relations between
boundary points of the half-plane and points of the sides of $T$. We have for
$x\in(0,1)$:

$\int_{0}^{x}\frac{(1-s^{2})^{\alpha}(1-k^{2}s^{2})^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,ds=x\int_{0}^{1}\frac{(1-x^{2}s^{2})^{\alpha}(1-k^{2}x^{2}s^{2})^{1-\alpha}}{(1+a^{2}x^{2}s^{2})^{2}}\,ds\\\
=xF^{(3)}_{D}(1;-\alpha,\alpha-1,2;2;x^{2},k^{2}x^{2},a^{2}x^{2})$

where $F^{(3)}_{D}$ is the Lauricella hypergeometric function (see (2.7) and
(2.8)).

For $x>1/k$ we have

$\int_{x}^{\infty}\frac{(s^{2}-1)^{\alpha}(k^{2}s^{2}-1)^{1-\alpha}}{(1+a^{2}s^{2})^{2}}\,ds=x\int_{0}^{1}\frac{(1-x^{2}s^{2})^{\alpha}(1-k^{2}x^{2}s^{2})^{1-\alpha}}{(1+a^{2}x^{2}s^{2})^{2}}\,ds\\\
=\frac{k^{2(1-\alpha)}}{xa^{4}}\int_{0}^{1}\frac{(1-x^{-2}s^{2})^{\alpha}(1-k^{-2}x^{-2}s^{2})^{1-\alpha}}{(1+a^{2}x^{-2}s^{2})^{2}}\,ds\\\
=\frac{k^{2(1-\alpha)}}{2xa^{4}}F^{(3)}_{D}(1;-\alpha,\alpha-1,2;2;x^{-2},k^{-2}x^{-2},a^{-2}x^{-2}).$

Comparing the interior and exterior moduli we immediately obtain the following
statement.

######  Theorem 4.11.

Let $L$ be the isosceles trapezoidal curve with acute angle $\pi\alpha$ and
bases $c$ and $d$. Let $\lambda$ and $k$ be solutions of (4.5) and (4.10),
(4.9). Then the coefficient $M_{L}$ satisfies the inequality

$M_{L}\geq\max\left[\frac{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda){\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(k^{\prime})}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(\lambda^{\prime})(k)},\frac{{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda^{\prime}){\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(k)}{{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda)(k^{\prime})}\right]$

where $\lambda^{\prime}=\sqrt{1-\lambda^{2}}$, $k^{\prime}=\sqrt{1-k^{2}}$.

It is evident that the estimation from Theorem 4.11 is not sharp. Numerical
experiments with sufficiently long rectangles show that the ratio of moduli of
two quadrilaterals, external and internal, with the same vertices $z_{1}$,
$z_{2}$, $z_{3}$, and $z_{4}$ on the line is not maximal if $z_{k}$ coincides
with the ’natural’ vertices of the rectangle. The best result is for the case
where $z_{k}$ are on the bigger sides and are symmetric with respect to the
axes of symmetry of the rectangle. In the next section we will try to explain
this fact theoretically.

##  5\. The boundary of a rectangle

Consider the case $\alpha=1/2$. Let $\Pi_{d}=[-d,d]\times[-1,0]$ be a
rectangle. Denote by $\Pi^{c}_{d}$ its exterior. Now we fix a number
$\delta\in(0,d]$. Let $\bm{\Pi}_{\delta}$ be the quadrilateral
$(\Pi_{d},\delta-i,\delta,-\delta,-\delta-i,)$ and
$\bm{\Pi}_{\delta}^{c}=(\Pi_{d}^{c},-\delta-i,-\delta,\delta,\delta-i)$. We
will compare their conformal moduli. Let ${\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)$ be the
complete elliptic integral of the first kind defined by (2.2); we will write
for short ${\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}^{\prime}(\lambda)={\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda^{\prime})$
where $\lambda^{\prime}=\sqrt{1-\lambda^{2}}$.

######  Theorem 5.1.

We have

(5.2) $\frac{\text{\rm Mod}(\bm{\Pi}_{\delta})}{\text{\rm
Mod}(\bm{\Pi}_{\delta}^{c})}\geq\frac{2\lambda{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda)}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)}\,d,$

where $\lambda=\delta/d$.

###### Proof.

It is obvious that

(5.3) $\text{\rm Mod}(\bm{\Pi}_{\delta})=2\delta.$

Now we will estimate $\text{\rm Mod}(\bm{\Pi}_{\delta}^{c})$.

Figure 3. Rectangle with shifted vertices.

Let ${\Pi}_{d}^{c+}$ be the part of ${\Pi}_{d}^{c}$, lying in the quarter of
the plane $\\{z:{\rm Re}\,z\geq 0,{\rm Im}\,z\geq-1/2\\}$. Consider the
quadrilateral $\bm{\Pi}_{\delta}^{c+}:=(\Pi^{c+}_{d};0,\delta,d-i/2,\infty)$.
By the symmetry principle,

(5.4) $\text{\rm Mod}(\bm{\Pi}_{\delta}^{c+})=\text{\rm
Mod}^{-1}(\bm{\Pi}_{\delta}^{c}).$

On the other side, the modulus is equal to the extremal length of the family
of curves, $\Gamma$, connecting in ${\Pi}_{d}^{c+}$ the sides $[0,\delta]$ and
$[d-i/2,\infty]$. Now consider the subdomain $G$ of ${\Pi}_{d}^{c+}$ which is
the first quarter of the plane. The modulus of the quadrilateral
$\bm{G}:=(G;0,\delta,d,\infty)$ is equal to the extremal length of the family
of curves, $\Gamma_{1}$, connecting in $G$ the sides $[0,\delta]$ and
$[d,\infty]$. Since $\Gamma_{1}<\Gamma$, we obtain

(5.5) $\text{\rm Mod}(\bm{\Pi}_{\delta}^{c+})\geq\text{\rm Mod}(\bm{G}).$

Under the conformal automorphism $w=(1/\delta)z$ of $G$, the vertices of the
quadrilateral $\bm{G}$ are mapped to the points $0$, $1$, $1/\lambda$,
$\infty$. The modulus of the obtained quadrilateral is well-known; it is
expressed via elliptic integrals. Since conformal modulus is invariant under
similarity mappings, we find

(5.6) $\text{\rm Mod}(\bm{G})=\frac{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda)}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)}.$

From (5.3)–(5.6) we obtain (5.2). ∎

Consider the function

$g(\lambda):=\frac{\lambda{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda)}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)}.$

The graph of $g(\lambda)$ is given on the Fig. 4. It is evident that
$\lim_{\lambda\to 0+}g(\lambda)=0$.

Figure 4. The graph of the function $g(\lambda)$.

######  Lemma 5.7.

The function $g$ is concave on $(0,1)$ and has a unique maximum point
$\lambda_{0}=0.7373921\ldots$ which is a unique root of the equation

(5.8) $(\lambda^{\prime})^{2}{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda){\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda)=\pi/2,$

$\lambda^{\prime}=\sqrt{1-\lambda^{2}}$, on the interval $(0,1)$. The maximal
value of $g$ is equal to $g(\lambda_{0})=0.708434\ldots$

###### Proof.

In terms of the function $\mu$ we see that
$g(\lambda)=2\lambda\mu(\lambda)/\pi$ and hence

$\frac{\pi}{2}g^{\prime}(\lambda)=\mu(\lambda)+\lambda\mu^{\prime}(\lambda),\quad\frac{\pi}{2}g^{\prime\prime}(\lambda)=2\mu^{\prime}(\lambda)+\lambda\mu^{\prime\prime}(\lambda).$

By (2.3) we obtain

(5.9)
$g^{\prime}(\lambda)=\frac{g(\lambda)}{\lambda}\,-\frac{\pi/2}{(\lambda^{\prime})^{2}{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{2}(\lambda)}\,=\frac{(\lambda^{\prime})^{2}{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda){\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda)-\pi/2}{(\lambda^{\prime})^{2}{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}^{2}(\lambda)}\,,$

and also, after simplification,

$g^{\prime\prime}(\lambda)=-\,\frac{\pi/2}{\lambda(1-\lambda^{2})^{2}{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{3}(\lambda)}((3-\lambda^{2}){\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\lambda)-2{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp E}}}(\lambda))<0.$

For the last inequality note that $(3-\lambda^{2}){\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda)>2{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(\lambda)>2{\mathchoice{\hbox{\,\fFt E}}{\hbox{\,\fFt
E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp E}}}(\lambda)$, $0<\lambda<1$. Therefore,
$g$ is concave on $(0,1)$ and the statement on the maximum follows from (5.9).
∎

######  Corollary 5.10.

For $\lambda\in(0,\lambda_{0})$ we have $g(\lambda)>g(\lambda_{0})\lambda$.

######  Corollary 5.11.

For the coefficient of quasiconformal reflection $M_{\partial\Pi_{d}}$ we have
the estimation $M_{\partial\Pi_{d}}\geq\gamma d$ where
$\gamma=2g(\lambda_{0})=1.4168687\ldots$

Actually, it is easy to show that the maximal value of the function
$g(\lambda)$ is attained at a unique point $\lambda_{0}\in(0,1)$ which is the
unique root of the equation (5.8) on $(0,1)$.

######  Remark 5.12.

From Corollary 5.11 and (1.4) it follows that $QR_{\partial\Pi_{d}}\geq\gamma
d$. A sharper estimate for $QR_{\Pi_{d}}$ follows from (1.3):

$QR_{\partial\Pi_{d}}\geq(\pi/2)d,\quad\pi/2=1.5707963\ldots$

But the method of the proof of Theorem 5.1, which is rather simple, can be
used to obtain a similar estimate for the case of isosceles trapezoids.

##  6\. Estimation of the coefficient $M_{L}$ for isosceles trapezoidal
polygon $L$

Now we apply the same method to obtain a lower estimate for the coefficients
$M_{L}$ and $QR_{L}$ of arbitrary isosceles trapezoidal polygon $L$.

First we will prove

######  Lemma 6.1.

Let $D^{+}$ be the part of $T^{-}$ lying in the half-plane $\\{y\geq-1\\}$ and
let $0<\delta<c$. Denote by $M_{1}$ the conformal modulus of
$(D;-\delta,\delta,d-i,-d-i)$ and by $M_{2}$ the conformal modulus of the
quadrilateral which is the upper half-plane with vertices $-\delta$, $\delta$,
$d$, $-d$. Then

(6.2) $M_{1}\geq C(\alpha)M_{2}$

where

$C(\alpha)=\left(\sqrt{1+\tan^{2}(\pi\alpha)/4}-\tan(\pi\alpha)/2\right)^{2},\quad
0<\alpha\leq 1/2.$

###### Proof.

Consider the piecewise-linear mapping $F(x,y)=x+iv(x,y)$ where

$v(x,y)=\left\\{\begin{array}[]{cr}y,&|x|\leq c.\\\
y+(|x|-c)\tan(\pi\alpha),&c\leq|x|\leq d,\\\ y+1,&|x|\geq d.\\\
\end{array}\right.$

It is easy to verify that $F$ is a homeomorphism of $\mathbb{C}$, mapping
$(D;-\delta,\delta,d-i,-d-i)$ onto $(H;-\delta,\delta,d,-d)$. We will show
that $F$ is a $K$-quasiconformal mapping with

(6.3) $K=(\sqrt{4+\tan^{2}(\pi\alpha)}+\tan(\pi\alpha))^{2}/4.$

Actually, $F$ is conformal in $\\{|x|\leq c\\}$ and $\\{|x|\geq d\\}$. Since
$F(x,y)$ is even with respect to $x$, we only need to show that $F$ is
$K$-quasiconformal in the strip $\\{c\leq x\leq d\\}$ where it has the form

$F(z)=z+i(x-c)\tan(\pi\alpha)=\frac{1}{2}\,[(2+i\tan(\pi\alpha))z-i\tan(\pi\alpha)\overline{z}]-ic\tan(\pi\alpha),$

therefore,

$\left|\frac{F_{\overline{z}}}{F_{z}}\right|\leq
k:=\frac{\tan(\pi\alpha)}{\sqrt{4+\tan^{2}(\pi\alpha)}}.$

and this implies that $F$ is $\frac{1+k}{1-k}$ -quasiconformal mapping. But
$\frac{1+k}{1-k}=K$ where $K$ is given by (6.3).

At last, because of quasiinvariance of conformal modulus under quasiconformal
mapping, we obtain that $M_{1}\geq K^{-1}M_{2}$ where
$K^{-1}=(\sqrt{4+\tan^{2}(\pi\alpha)}-\tan(\pi\alpha))^{2}/4=C(\alpha)$. ∎

Now, with the help of Lemma 6.1 we will estimate $M_{L}$.

######  Theorem 6.4.

Let $L$ be the isosceles trapezoidal polygon with acute angle $\pi\alpha$ and
bases $c$ and $d$, $c<d$.

1) If $\frac{c}{d}\geq\lambda_{0}$, where $\lambda_{0}$ is described in Lemma
5.7, then

(6.5) $M_{L}\geq g(\lambda_{0})(1+C(\alpha))d.$

2) If $\frac{c}{d}<\lambda_{0}$, then

$M_{L}\geq g(\lambda)(1+C(\alpha))d,\quad\lambda=c/d.$

###### Proof.

1) Let $\frac{c}{d}\geq\lambda_{0}$. Then we put $\delta=\lambda_{0}d$,
$0<\delta<c$. Denote

$\mbox{\boldmath$T$}_{\delta}^{+}=(T^{+};\delta-i,\delta,-\delta,-\delta-i),\quad\mbox{\boldmath$T$}_{\delta}^{-}=(T^{-};-\delta-i,-\delta,\delta,\delta-i),$

then

(6.6) $\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{+})\geq
2\delta=2\lambda_{0}d.$

The line $y=-1$ separates $T^{-}$ into two domains. One of them is $D^{+}$,
denote by $D^{-}$ the other one. Let
$\mbox{\boldmath$D$}^{+}=(D^{+};-\delta,\delta,d-i,-d-i)$,
$\mbox{\boldmath$D$}^{-}=(D^{-};-\delta-i,\delta-i,d-i,-d-i)$. Then, with the
use of Lemma 6.1, we obtain

$\text{Mod}(\mbox{\boldmath$D$}^{-})=\frac{{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}^{\prime}(\lambda_{0})}{2{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\lambda_{0})},\quad\text{Mod}(\mbox{\boldmath$D$}^{+})\geq
C(\alpha)\text{Mod}(\mbox{\boldmath$D$}^{-})$

Now we note that $\mbox{\boldmath$D$}^{+}$, $\mbox{\boldmath$D$}^{-}$ and the
quadrilateral
$(\mbox{\boldmath$T$}_{\delta}^{-})^{*}=(T^{-};-\delta,\delta,d-i,-d-i)$,
conjugate to $\mbox{\boldmath$T$}_{\delta}^{-}$, are symmetric with respect to
the imaginary axis. Applying the Grötzsch’ lemma [VA, pp.13-15] to the right
halves of the considered quadrilaterals and taking into mind that, by the
symmetry principle, their moduli are half as small as the initial moduli, we
obtain

(6.7)
$(\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{-}))^{-1}=\text{Mod}((\mbox{\boldmath$T$}_{\delta}^{-})^{*})\geq\text{Mod}(\mbox{\boldmath$D$}^{+})+\text{Mod}(\mbox{\boldmath$D$}^{-})$

Then, multiplying (6.6) and (6.7), we obtain

$\frac{\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{+})}{\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{-})}\geq
g(\lambda_{0})(1+C(\alpha))d.$

and this implies (6.5).

2) Now let $\frac{c}{d}<\lambda_{0}$. Then, by similar reasoning as above, we
obtain

$\frac{\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{+})}{\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{-})}\geq
g(\lambda)(1+C(\alpha))d$

where $\lambda=\frac{c}{d}$. ∎

######  Remark 6.8.

The estimation obtained in Theorem 6.4 is good for small $\alpha$ because then
$C(\alpha)$ is close to $1$. For $\alpha$, close to $\pi/2$, the value of
$C(\alpha)$ is close to zero and probably can be essentially improved.

Using (1.4), we can estimate $QR_{L}$ for trapezoidal curves.

######  Corollary 6.9.

Under the assumption of Theorem 6.4, if $\frac{c}{d}\geq\lambda_{0}$, then
$QR_{L}\geq g(\lambda_{0})(1+C(\alpha))d$. If $\frac{c}{d}<\lambda_{0}$, then
$QR_{L}\geq g(\lambda)(1+C(\alpha))d$ where $\lambda=c/d$.

We can also enhance the estimations given in Theorem 6.4 and Corollary 6.9, if
we calculate $\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{+})$ and
$\text{Mod}(\mbox{\boldmath$T$}_{\delta}^{+})$ with the help of the formulas
given in Section 4.

Now we will describe the algorithm in more detail.

1) Let $\frac{c}{d}\geq\lambda_{0}$. First we find the preimages of $\delta$
and $\delta-i$, $\delta=\lambda_{0}d$, under the conformal mapping of the
lower half-plane onto $T^{+}$ described in Subsection 4.1.

For this, we find a unique $x_{*}\in[0,1]$ from the equation

$x_{*}F_{1}({\textstyle\frac{1}{2}};\alpha,1-\alpha;{\textstyle\frac{3}{2}};x_{*}^{2},\lambda^{2}x_{*}^{2})=\frac{\lambda_{0}d}{c}\,B(1-\alpha,\textstyle\frac{1}{2})\,{}_{2}F_{1}(\textstyle\frac{1}{2},1-\alpha,\textstyle\frac{3}{2}-\alpha,\lambda^{2}).$

and a unique $x_{**}\in(1/\lambda,+\infty)$, satisfying

$x_{**}^{-1}\lambda^{2(1-\alpha)}F_{1}({\textstyle\frac{1}{2}};\alpha,1-\alpha;{\textstyle\frac{3}{2}};x_{**}^{-2},\lambda^{-2}x_{**}^{-2})=\frac{\lambda_{0}d}{c}\,B(1-\alpha,\textstyle\frac{1}{2})\,{}_{2}F_{1}(\textstyle\frac{1}{2},1-\alpha,\textstyle\frac{3}{2}-\alpha,\lambda^{2}).$

Now we find $\widetilde{\lambda}=x_{*}/x_{**}$ and the modulus of
$\mbox{\boldmath$T$}^{+}$:

(6.10) $\text{Mod}(\mbox{\boldmath$T$}^{+})=\frac{2{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{\lambda})}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{\lambda}^{\prime})},\quad\widetilde{\lambda}^{\prime}=\sqrt{1-\widetilde{\lambda}^{2}}.$

Then we find the preimages of $\delta$ and $\delta-i$ under the conformal
mapping of the upper half-plane onto $T^{-}$ described in Subsection 4.6.

We find $y_{*}\in(0,1)$ from the equation

$y_{*}F_{D}^{(3)}({\textstyle\frac{1}{2}};-\alpha,\alpha-1,2;{\textstyle\frac{3}{2}};y_{*}^{2},y_{*}^{2}k^{2},-y_{*}^{2}a^{2})=\frac{\lambda_{0}d}{c}F_{1}({\textstyle\frac{1}{2}};\alpha-1,{\textstyle\frac{3}{2}}+\alpha;k^{2}m-a^{2}),$

and $y_{**}\in(1/k,+\infty)$ from the equation

$y_{**}^{-1}k^{2(1-\alpha)}a^{-4}F_{D}^{(3)}({\textstyle\frac{1}{2}};-\alpha,\alpha-1,2;{\textstyle\frac{3}{2}};y_{**}^{-2},y_{**}^{-2}k^{-2},-y_{**}^{-2}a^{-2}).$

Then we put $\widetilde{k}=y_{*}/y_{**}$ and obtain

(6.11) $\text{Mod}(\mbox{\boldmath$T$}^{-})=\frac{2{\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{k})}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{k}^{\prime})},\quad\widetilde{k}^{\prime}=\sqrt{1-\widetilde{k}^{2}}.$

From (6.10) and (6.11) we deduce that

$M_{L}\geq\frac{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa
K}}{\hbox{\,\fFp K}}}(\widetilde{\lambda}){\mathchoice{\hbox{\,\fFt
K}}{\hbox{\,\fFt K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{k}^{\prime})}{{\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp
K}}}(\widetilde{\lambda}^{\prime}){\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}}(\widetilde{k})}\,.$

2) If $\frac{c}{d}<\lambda_{0}$, then we replace $x_{*}$ and $y_{*}$ with $1$
and in the second equation we put $c/d$ instead of $\lambda_{0}$.

######  Remark 6.12.

If the base $c$ is sufficiently large, then the choice of quadrilateral with
vertices $\pm\delta$, $\pm\delta-i$ for estimation of $M_{L}$ and $QR_{L}$ is
rather good because of a result by W. Hayman (see, e.g. [PS, thrm. 2.3.8]). It
states that if two sides of a quadrilateral are segments on the vertical lines
$\\{x=0\\}$ and $\\{x=1\\}$ and the other two are graphs of two continuous on
$[0,1]$ functions $y=\varphi(x)$ and $y=\psi(x)$, and
$h:=\min\psi-\max\varphi>0$, then the modulus $M$ of the quadrilateral
satisfies the inequality

$h\leq M\leq h+1.$

In the symmetric case, the estimation could be improved.

##  7\. Upper estimate of $QR_{L}$ for isosceles trapezoidal polygons $L$

Let $\alpha$ be a real number with $0<\alpha<1.$ An analytic function $f$ on
the unit disk $\mathbb{D}$ is said to be strongly starlike of order $\alpha$
if $f^{\prime}(0)\neq 0$ and if $f$ satisfies the inequality

$\left|\arg\frac{zf^{\prime}(z)}{f(z)-f(0)}\right|<\frac{\pi\alpha}{2}$

for $0<|z|<1.$ In particular, ${\rm Re}\,[zf^{\prime}(z)/(f(z)-f(0))]>0$ and
thus $f$ is a starlike univalent function on the unit disk. A simply connected
domain $\Omega$ in $\mathbb{C}$ is said to be strongly starlike of order
$\alpha$ with respect to $w_{0}$ if the conformal homeomorphism
$f:\mathbb{D}\to\Omega$ with $f(0)=w_{0}$ and $f^{\prime}(0)>0$ is strongly
starlike of order $\alpha.$ The following result is due to Fait, Krzyż and
Zygmunt [FKZ]. Here and hereafter, we set

(7.1) $K(\alpha)=\frac{1+\sin(\pi\alpha/2)}{1-\sin(\pi\alpha/2)}$

for $0<\alpha<1.$

######  Lemma 7.2.

Let $0<\alpha<1.$ A strongly starlike function $f$ of order $\alpha$ on
$\mathbb{D}$ extends to a $K(\alpha)$-quasiconformal mapping of $\mathbb{C}.$

In particular, we see that the boundary of a strongly starlike domain is a
Jordan curve in $\mathbb{C}.$ As a consequence, we obtain the following
result.

######  Corollary 7.3.

Let $L$ be the boundary of a strongly starlike domain $\Omega$ of order
$\alpha.$ Then $QR_{L}\leq K(\alpha),$ where $K(\alpha)$ is given in (7.1).

###### Proof.

We follow Ahlfors’ construction [A2]. Let $j$ be the inversion in the unit
circle, $j(z)=1/\overline{z}.$ By Lemma 7.2, a conformal mapping
$f:\mathbb{D}\to\Omega$ extends to a $K(\alpha)$-quasiconformal mapping of the
Riemann sphere $\mathbb{C}\cup\\{\infty\\},$ which is denoted by the same
symbol $f.$ Then $F=f\circ j\circ f^{(-1)}$ is a $K(\alpha)$-quasiconformal
reflection across $L.$ ∎

To check strong starlikeness, it is convenient to look at the following
quantity. For a domain $\Omega$ and $w_{0}\in\Omega,$ we define

$R_{\Omega,w_{0}}(\theta)=\sup\\{r>0:w_{0}+te^{i\theta}\in\Omega\leavevmode\nobreak\
\text{for all}\leavevmode\nobreak\ t\in[0,r)\\}$

for $\theta\in\mathbb{R}.$ The following result is contained in [Su].

######  Lemma 7.4.

Let $\Omega$ be a domain in $\mathbb{C}$ containing a point $w_{0}$ and let
$0<\alpha<1.$ The domain $\Omega$ is strongly starlike of order $\alpha$ with
respect to $w_{0}$ if and only if $R(\theta)=R_{\Omega,w_{0}}(\theta)$ is
absolutely continuous and satisfies the inequality
$|R^{\prime}(\theta)|/R(\theta)\leq\tan(\alpha\pi/2)$ for almost all
$\theta\in\mathbb{R}.$

We consider the isosceles trapezoid $L$ described in Section 4. Let $\Omega$
be the domain bounded by $L.$ We now show the following for $\Omega.$

######  Lemma 7.5.

Let $0<s<1.$ Then $\Omega$ is strongly starlike of order $\alpha(s)$ with
respect $-is,$ where $\alpha(s)\in(0,1)$ is determined by

$\tan\frac{\pi\alpha(s)}{2}=\max\left\\{\frac{c}{s}\,,\,\frac{d}{1-s}\,,\,\frac{1-s+(d-c)d}{c+(d-c)s}\,,\,\,\frac{s-(d-c)c}{c+(d-c)s}\right\\}.$

###### Proof.

Since $\Omega$ is symmetric in the imaginary axis, it is enough to consider
the function $R(\theta)=R_{\Omega,-is}(\theta)$ for $-\pi/2<\theta<\pi/2.$ We
define $\theta_{1}$ and $\theta_{2}$ in $(0,\pi/2)$ by requiring

$\tan\theta_{1}=\frac{s}{c}\,,\quad\tan\theta_{2}=\frac{1-s}{d}\,.$

Then the function $R(\theta)$ is described by

$R(\theta)=\begin{cases}\vskip 6.0pt plus 2.0pt minus
2.0pt\dfrac{s}{\sin\theta}\,,&\theta_{1}<\theta<\pi/2,\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\dfrac{(1-s)c+sd}{\cos\theta+(d-c)\sin\theta}\,,&-\theta_{2}\leq\theta\leq\theta_{1},\\\
\dfrac{1-s}{-\sin\theta}\,,&-\pi/2<\theta<-\theta_{2}.\end{cases}$

In the first case $\theta_{1}<\theta<\pi/2,$ we have
$|R^{\prime}(\theta)|/R(\theta)=1/\tan\theta\leq 1/\tan\theta_{1}=c/s.$
Similarly, we have $|R^{\prime}(\theta)|/R(\theta)|\leq
1/\tan\theta_{2}=d/(1-s)$ in the third case. When
$-\theta_{2}<\theta<\theta_{1},$ we have

$\frac{R^{\prime}(\theta)}{R(\theta)}=\frac{\sin\theta-(d-c)\cos\theta}{\cos\theta+(d-c)\sin\theta}\,,\quad\left(\frac{R^{\prime}(\theta)}{R(\theta)}\right)^{\prime}=\frac{1+(d-c)^{2}}{(\cos\theta+(d-c)\sin\theta)^{2}}>0.$

Hence, $R^{\prime}/R$ is increasing in this interval and, in particular,

$-\frac{1-s+(d-c)d}{d-(d-c)(1-s)}=-\frac{1-s+(d-c)d}{c+(d-c)s}\leq\frac{R^{\prime}(\theta)}{R(\theta)}\leq\frac{s-(d-c)c}{c+(d-c)s}$

for $\theta\in(-\theta_{2},\theta_{1}).$ Therefore, by Lemma 7.4, the required
formula follows. ∎

We are now able to show the following result.

######  Theorem 7.6.

Let $L$ be an isosceles trapezoidal polygon of height $1$ and bases $c$ and
$d$ with $c\leq d.$ Then

$QR_{L}\leq(\sqrt{1+\tau^{2}}+\tau)^{2},$

where

$\tau=\max\left\\{c+d,\,\,\frac{1-c^{2}+d^{2}}{2c}\right\\}.$

###### Proof.

Let $\Omega$ be the domain bounded by $L.$ If $s=c/(c+d)$, then $c/s=d/(1-s)$
and we apply the previous lemma to show that $\Omega$ is strongly starlike of
order $\alpha$ with $\tan(\alpha\pi/2)=\tau.$ Thus the required assertion
follows with the help of the formula
$\sin\theta=\tan\theta/\sqrt{1+\tan^{2}\theta}.$ ∎

######  Remark 7.7.

When $c=d,$ we have a rectangle $L$ of height 1 and width $a=2d.$ We may
assume that $a\geq 1.$ Then we have $\tau=a$ and thus

$QR_{L}\leq(\sqrt{1+a^{2}}+a)^{2}\leq(3+2\sqrt{2})a^{2}.$

We recall that Werner’s estimation (1.3) gives us $QR_{L}\leq\pi a.$
Therefore, the last theorem yields only a poor estimate.

##  8\. Appendix A. The Wolfram Mathematica code for calculation of exterior
modulus for polygonal quadrilateral

Here we give a code which defines the function ExtMod($A,B,n,wp$) described in
Section 3.

    ExtMod[A_?NumberQ, B_?NumberQ, n_, wp_] := Module[{m = n, A1 = 1., A2 = 0.,
    A3 = A, A4 = B, a, b, c, d, i, r1, r2, s, t, t1, t2, x, x0, x1, x2, x3, x4,
    y0, z0, A0, B0, C0, D0, sol, J1, J2, K, L, L2, M, S, T2},
    T2 = 10^m; S = IntegerPart[5 (m + 15)]; a = Arg[A4 - A1]/Pi;
    b = 1 - Arg[A3 - A2]/Pi; c = 1 - b - Arg[A4 - A3]/Pi; d = 2 - a - b - c;
    L = Abs[A3 - A2]; t1 = 1; t2 = T2;
       Do[t = (t1 + t2)/2; K = a + b + c - 1;
       A0 = 2 (K - 1)^2; B0 = (K - 1) (4 - 3 (a + c) + (4 - 3 (a + b)) t);
       C0 = 2 - 3 (a + c) + (a + c)^2 + 2 (3 - 5*a - 2*b - 2*c + 2*a^2 + 2*a*b
       + 2*a*c + b*c) t + (2 - 3 (a + b) + (a + b)^2) t^2;
       D0 = (1 - a) (a + c - 1 + (a + b - 1) t) t;
       sol = Solve[A0*x^3 + B0*x^2 + C0*x + D0 == 0, x];
       x1 = x /. sol[[1]]; x2 = x /. sol[[2]]; x3 = x /. sol[[3]];
       r1[x_] = a*t*x; r2[x_] = ((1 - K) x + K (t + 1) - c*t - b);
       x4 = If[r2[x1]^2 x1^2 < r1[x1] r2[x1], x1, x2];
       x0 = If[r2[x3]^2 x3^2 < r1[x3] r2[x3], x3, x4];
       y0 = Sqrt[r1[x0]/r2[x0] - x0^2]; z0 = x0 + I*y0;
       Quiet[J1 = Re[NIntegrate[s^a (1 - s)^b (1 - s/t)^c/Abs[1 - s/z0]^4,
       {s, 0, 1}, WorkingPrecision -> wp]]];
       Quiet[J2 = Re[NIntegrate[s^a (s - 1)^b (1 - s/t)^c/Abs[1 - s/z0]^4,
       {s, 1, t}, WorkingPrecision -> wp]]];
       L2 = J2/J1; If[L2 < L, t1 = t, t2 = t], {i, S}];
    t = N[t]; M = EllipticK[(1. - 1/t)]/EllipticK[1/t]; {M, a, b, c, d, t, z0}];

Now we will give the results of calculating of the exterior moduli of the
polygons from Example 3.26. The working precision is equal $16$ and the values
of $n$ is equal $2$ because the values of the moduli are sufficiently small.

Polygon 1.

    A = 28/25 + I*69/50; B = -19/25 + I*21/25; sol = ExtMod[B, A, 2, 16];
    Print["ExtMod(", A, ", ", B, ",", 16, ") = " NumberForm[sol[[1]], 16], ","];
    Print["alpha = " NumberForm[sol[[2]], 16], "," " beta = "
    NumberForm[sol[[3]], 16], "," " gamma = " NumberForm[sol[[4]], 16], "," "
    delta = " NumberForm[sol[[5]], 16], ","]; Print ["t = " NumberForm[sol[[6]],
    16], "," " z0 = " NumberForm[sol[[7]], 16], "."]

    ExtMod(28/25+(69 I)/50, -(19/25)+(21 I)/25,16) =  0.992341633097864,
    alpha =  0.4723903292882761,  beta =  0.2659022512561763,
    gamma =  0.6450651518079917,  delta =  0.6166422676475559,
    t =  1.966910456214164,  z0 =  (1.215406699779183+1.315084271771535 I).

Polygon 2.

    A = 42/25 + I*4; B = -3/25 + I*21/25; sol = ExtMod[B, A, 2, 16];
    Print["ExtMod(", A, ", ", B, ",", 16, ") = " NumberForm[sol[[1]], 16], ","];
    Print["alpha = " NumberForm[sol[[2]], 16], "," " beta =
    " NumberForm[sol[[3]], 16], "," " gamma = " NumberForm[sol[[4]], 16], "," "
    delta = " NumberForm[sol[[5]], 16], ","]; Print ["t = " NumberForm[sol[[6]],
    16], "," " z0 = " NumberForm[sol[[7]], 16], "."]

    ExtMod(42/25+4 I, -(3/25)+(21 I)/25,16) =  0.959257171919002,
    alpha =  0.4463997482438991,  beta =  0.4548327646991334,
    gamma =  0.2099823197839025, delta =  0.888785167273065,
    t =  1.83346758954612,  z0 =  (0.7429152683728336+1.983082728044083 I).

Table 3. The values of exterior moduli for some polygonal quadrilaterals. A | B | ExtMod[B,A]
---|---|---
7 + 5 I |  $-1+2$ I | 1.158095606321043
8 + 3 I |  $-1$ \+ I | 1.130410084465672
5 + 5 I |  $-3$ \+ I | 1.233703270301942
7 + 4 I |  $-3$ \+ 3 I | 1.274708414007269
5 + 5 I |  $-1$ \+ 2 I | 1.140576491710462
7 + 5 I |  I | 1.015468479689712
7 + 3 I | 1 + 2 I | 1.135151674872884
4 + 5 I |  $-2$ \+ I | 1.157883901548636
1 + I |  I | 0.999999999999995

Table 3 was computed with the function ExtMod.

######  Remark 8.1.

There is also another method to validate the results of the function ExtMod.
We can compare the results in the case of a rectangle with vertices
$0,1,1+i*h,i*h,h>0,$ to the analytic formula given by Duren and Pfaltzgraff
[DP], see also [HRV1] for further bibliographic references. By this formula,
defining

$\psi(r)=\frac{2(\mathchoice{\hbox{\,\fFt E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa
E}}{\hbox{\,\fFp E}}(r)-(1-r)\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(r))}{\mathchoice{\hbox{\,\fFt
E}}{\hbox{\,\fFt E}}{\hbox{\,\fFa E}}{\hbox{\,\fFp
E}}(\sqrt{1-r^{2}})-r\mathchoice{\hbox{\,\fFt K}}{\hbox{\,\fFt
K}}{\hbox{\,\fFa K}}{\hbox{\,\fFp K}}(\sqrt{1-r^{2}})},$ ${\rm
DP}(h)=\mu(\psi^{-1}(h))/\pi,$

we have

${\rm ExtMod}[I*h,1+I*h,2,16][[1]]={\rm DP}(1/h).$

For the range $h\in[0.5,10]$ this last identity holds with approximate error
$10^{-14}.$

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