# Minimal surfaces over harmonic shears

Simran Bedi Department of Mathematics, University of Delhi.
<EMAIL_ADDRESS>Sanjay Kumar Deen Dyal Upadhyaya College, University of
Delhi.
<EMAIL_ADDRESS>
###### Abstract

Harmonic mappings have long intrigued researchers due to their intrinsic
connection with minimal surfaces. In this paper, we investigate shearing of
two distinct classes of univalent conformal mappings which are convex in
horizontal direction with appropriate dilatations. Subsequently, we present a
family of minimal surfaces constructed by lifting the harmonic mappings
obtained through shear construction method given by Clunie-Sheil. Furthermore,
we contribute to addressing an open problem partially, proposed by Boyd and
Dorff, by identifying the resulting minimal surfaces for certain values of the
parameters in one of the classes of mappings. Notably, this family of minimal
surfaces transforms from the well-established Enneper surface to a Helicoid.

Keywords: Harmonic univalent mappings, Harmonic shear, Convex along real
directions, Minimal Surfaces

##  1 Introduction

The study of minimal surfaces has fascinated differential geometers, not
solely because of the exotic structures that they exhibit but due to their
inherent relationship with harmonic mappings. Results and properties of
harmonic univalent functions are generally used to investigate the properties
of minimal surfaces. The reason why we emphasize their univalency is because,
geometrically, it means that the image of this map does not show any
overlapping or self-intersection. Further, when we lift this univalent map
from the complex plane to $\mathbb{R}^{3}$, the resulting surface inherently
does not self-intersect itself, called a “minimal graph”. Construction of such
rather special functions is an intricate task.

One of the approach is to use the shearing technique, introduced by Clunie and
sheil-small in 1984 [1], whose fundamental building blocks are: a conformal
mapping and a dilatation. It proved to be a revolutionary paper in the study
of harmonic mapping which gave rise to a lot of research in this field.
Another revolutionary discovery was the Weierstrass-Enneper representation,
which allows us to take a harmonic univalent function with an appropriate
dilatation and lift it to a minimal graph. Several researchers have used this
technique [2], however, it is often difficult to identify the resulting
minimal graphs.

One approach to recognize the minimal surface is to use a change of variables.
In [3], Dorff and Muir have constructed and identified a family of minimal
graphs associated with the generalized Koebe function using this approach. In
[4], Boyd and Dorff proposed the following open problem, which is the prime
object of our investigation:

######  Problem.

Determine the minimal graphs formed by lifting harmonic univalent mappings in
the paper “Gauss curvature estimates for minimal graphs” by Nowak and
Woloszkiewicz [5].

We have contributed to answering this open problem partially, by recognizing
the surfaces formed by lifting the harmonic maps for certain values, obtained
through the shearing method applied on the conformal univalent mapping
$F(z)=z/(1+c\,z+z^{2})$, given in the aforementioned paper. We have also
constructed a two-parameter family of harmonic mappings by shearing the same
map with a different dilatation, $\omega(z)=z(z+a)/(1+az)$.

Also, We have chosen another family of conformal univalent maps which are
convex in horizontal direction(CHD) given by,
$F(z)=z-(1/n^{2})z^{n};n=2,3,4,\ldots$ [6]. We have investigated its shearing
for $\omega(z)=z^{n}$ and then have lifted the resulting harmonic maps to
construct corresponding minimal graphs for even $n.$ We have plotted images of
open unit disk under the conformal mappings, images of harmonic mappings, and
the resulting minimal surfaces they give rise to. The projections of these
minimal surfaces can be observed in the images showing their corresponding
harmonic mappings.

##  2 Background

Now that we have some essential ideas, we need the background for the theory
of Harmonic mappings. This section will cover important definitions and
theorems, which we will utilize throughout the rest of the paper.

Let $\mathbb{D}=\\{z:|z|<1\\}$ be the open unit disk in the complex plane.

######  Definition 1.

Let $S$ denotes the family of analytic functions on $\mathbb{D}$ that are
normalized and univalent; that is,

$S=\\{f:\mathbb{D}\rightarrow\mathbb{C}\mid f\mbox{ is analytic and univalent
with }f(0)=0,f^{\prime}(0)=1\\}.$

The next theorem tells us that a harmonic function defined on $\mathbb{D}$ can
be expressed in terms of analytic functions, known as its canonical
decomposition [6].

######  Theorem 1.

Let $f=u+\dot{\iota}v$ be a complex-valued harmonic function such that
$f:D\rightarrow\mathbb{C}$, where $D$ is a simply-connected domain, then there
exist analytic functions $h$ and $g$ such that $f=h+\bar{g}$.

######  Definition 2.

Let $S_{H}^{0}$ be the family of complex-valued harmonic mappings on
$\mathbb{D}$ that are univalent and normalized; that is,

$S_{H}^{0}=\\{f:\mathbb{D}\rightarrow\mathbb{C}\mid f\mbox{ is harmonic,
univalent with }h(0)=0,g(0)=0,h^{\prime}(0)=1,g^{\prime}(0)=0\\}.$

Note that the harmonic mapping $f=h+\bar{g}$ can also be expressed as
$f=\operatorname{Re}{(h+g)}+\dot{\iota}\operatorname{Im}{(h-g)}$. By a result
of Lewy [7], $f=h+\bar{g}$ is locally univalent and sense-preserving in
$\mathbb{D}$ if and only if the Jacobian of $f$,
$J_{F}(z)=|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}>0$ for all $z\in\mathbb{D}.$
The dilatation of $f$ is defined as $\omega(z)=g^{\prime}(z)/h^{\prime}(z)$.
Jacobian being positive is equivalent to the condition $|\omega(z)|<1$ for all
$z\in\mathbb{D}$.

######  Definition 3.

A domain $\Omega$ is said to be convex in the horizontal direction(CHD) if it
has a connected intersection with every line parallel to the real axis. It is
also called convex in the direction of real axis(CRA).

The following theorem [1] establishes the construction of a harmonic map with
a specified dilatation which is a crucial tool for construction of minimal
surfaces:

######  Theorem 2 (Clunie and Sheil-Small ).

A harmonic map $f=h+\bar{g}$ locally univalent in $\mathbb{D}$ is a univalent
mapping of $\mathbb{D}$ such that $f(\mathbb{D})$ is a CHD Domain if and only
if , h-g is a conformal univalent mapping of $\mathbb{D}$ onto a CHD Domain.

This process is known as shearing or shear method and can be presented step by
step as:

  1. (i)

Choose a conformal univalent map $F$ and decompose it as $F=h-g.$

  2. (ii)

Choose an appropriate dilatation $\omega$ and write it as
$\omega=g^{\prime}/h^{\prime}.$

  3. (iii)

Solve for $h$ and $g$ using equations obtained in the preceding two steps.

  4. (iv)

Write $f=h+\bar{g}$, which is the required harmonic univalent map convex in
horizontal direction obtained by shearing a conformal map along parallel
lines.

Now, we lay some background about minimal surfaces before discussing their
connection with harmonic mappings.

######  Definition 4.

A minimal surface is a surface whose mean curvature vanishes at all its
points.

At each point, the bending upward in one direction is matched with the bending
downward in the orthogonal direction. Every point on the surface is a saddle
point. We can also define minimal surfaces in terms of first fundamental form,
second fundamental form, area function and isothermal coordinates [8]. In this
paper, we will come across two well-known minimal surfaces which are:
Enneper’s surface and helicoid. The helicoid is parametrized on
$\mathbb{D}\setminus(-1,0)$ as:

$Y_{0}(z)=\left(\operatorname{Re}\left(z-\frac{1}{z}\right),\operatorname{Im}\left(z+\frac{1}{z}\right),2\,\operatorname{Im}(\log
z)\right)$

Enneper’s surface is parametrized on $\mathbb{D}$ as:

$Y_{2}(z)=\left(\operatorname{Re}\left(z-\frac{z^{3}}{3}\right),\operatorname{Im}\left(z+\frac{z^{3}}{3}\right),2\,\operatorname{Re}(-z^{2})\right)$

Note that the negative sign in the third component function has the effect of
reflecting the surface through xy-plane. Moreover, scalings by a factor,
substitution by a $m\ddot{o}bius$ transformation and reflection across planes
containing two axes does not affect the geometry of the surface.
Karl Weierstrass and Alfred Enneper discovered a way to easily construct
minimal surfaces by demonstrating their connection to harmonic mappings in the
next theorem [9].

######  Theorem 3 (Weierstrass-Enneper representation).

Let $\Omega\subset\mathbb{C}$ be a simply connected domain containing the
origin. If a minimal graph

$\\{(u,v,F(u,v)):u+iv\in\Omega\\}$

is parameterized by sense-preserving isothermal parameters
$z=x+iy\in\mathbb{D}$, the projection onto its base plane defines $a$ harmonic
mapping $w=u+iv=f(z)$ of $\mathbb{D}$ onto $\Omega$ whose dilatation is the
square of an analytic function. Conversely, if $f=h+\bar{g}$ is a harmonic
univalent mapping of $\mathbb{D}$ onto $\Omega$ with dilatation
$\omega=g^{\prime}/h^{\prime}$ being the square of an analytic function, then
with $z=x+iy\in\mathbb{D}$, the parameterization

$\displaystyle\mathbf{X}(z)=\left(\operatorname{Re}\\{h(z)+g(z)\\},\operatorname{Im}\\{h(z)-g(z)\\},2\operatorname{Im}\left\\{\int_{0}^{z}\sqrt{g^{\prime}(\zeta)h^{\prime}(\zeta)}d\zeta\right\\}\right)$

defines a minimal graph whose projection into the complex plane is
$f(\mathbb{D})$. Except for the choice of sign and an arbitrary additive
constant in the third coordinate function, this is the only such surface.

##  3 Shearing of one-slit and two-slit conformal mappings

(a) $c=-2$

(b) $c=0$

(c) $c=1$

(d) $c=2$

Figure 1: Image of the conformal map $F_{c}(|z|=0.999)$

In this section, we apply shearing technique on a family of conformal mappings
given in [5] denoted by $F_{c}(z)$ with two different dilatations
$\omega_{a}(z)=z(z+a)(1+az)$ and $\omega(z)=z^{n}$. We further lift the
obtained two-parameter family of harmonic mappings $f_{c,a}$ from complex
plane to minimal surfaces in $\mathbb{R}^{3}$ for $a=0$ and even $n$, using
Theorem 3. Consider the family of univalent conformal map $F_{c}$ defined by,

$\displaystyle F_{c}(z)=\dfrac{z}{1+cz+z^{2}}$ (1)

where $c\in(-2,2)$. It maps the unit disk univalently onto a domain convex in
the horizontal direction [10].
For special case of $c=-2$, the image of under $F_{c}(\mathbb{D})$ is a single
slit domain like Koebe domain depicted in Fig: 1(a), which is entire complex
plane except for a slit along the negative real axis represented as:

$\mathbb{C}\setminus\\{x:x\in(-\infty,-1/4)\\}$

and similarly a slit along positive real axis for $c=2$ shown in Fig: 1(d)
given by:

$\mathbb{C}\setminus\\{x:x\in(1/4,\infty)\\}.$

For $c=0$, it maps the unit disk onto a two slit domain represented in Fig:
1(b), that is, entire complex plane except for two half-lines given by:

$\mathbb{C}\setminus\\{x:x\in(-\infty,-1/2)\cup(1/2,\infty)\\}.$

Let us consider the harmonic shear of $F_{c}$ defined in (1), with the
dilatation

$\omega_{a}(z)=\frac{z(z+a)}{(1+az)},$

such that $-1\leq a\leq 1$. In particular, for $a=-1$, $a=1$ and $a=0$,
$\,\omega(z)$ takes value $-z,z$ and $z^{2}$ respectively. This requires
solution of the pair of differential equations given by:

$\displaystyle
h_{c,a}^{\prime}(z)-g_{c,a}^{\prime}(z)=F_{c}^{\prime}(z)\,\,\,\mbox{and}\,\,\,g_{c,a}^{\prime}(z)=z\frac{(z+a)}{(1+az)}h_{c,a}^{\prime}(z).$

After straightforward but tedious computations for normalized $h_{c,a}$ and
$g_{c,a}$, the solution is expressed as:

$\displaystyle h_{c,a}(z)=-\frac{p(z)}{q(z)},$ (2)
$\displaystyle\text{where}\,\,\,\,p(z)=$ $\displaystyle
16\sigma_{2}+16\sigma_{1}-4c^{2}\sigma_{2}+16z^{2}\sigma_{2}-4c^{2}\sigma_{1}+16z^{2}\sigma_{1}+2z\sigma_{3}\sigma_{4}-4c^{2}z^{2}\sigma_{2}$
$\displaystyle+2ac^{3}\sigma_{1}-4c^{3}z\sigma_{1}-8ac\sigma_{2}+16cz\sigma_{2}-4c^{2}z^{2}\sigma_{1}+2ac^{3}\sigma_{2}-4c^{3}z\sigma_{2}$
$\displaystyle-8ac\sigma_{1}+16cz\sigma_{1}+2ac^{3}z^{2}\sigma_{1}-8acz^{2}\sigma_{2}-8ac^{2}z\sigma_{2}$
$\displaystyle+2ac^{4}z\sigma_{2}+2ac^{3}z^{2}\sigma_{2}-8acz^{2}\sigma_{1}-8ac^{2}z\sigma_{1}$
$\displaystyle+2ac^{4}z\sigma_{1}+2az^{2}\sigma_{3}\sigma_{4}-cz^{2}\sigma_{3}\sigma_{4}$
$\displaystyle-c^{2}z\sigma_{3}\sigma_{4}+acz\sigma_{3}\sigma_{4},$
$\displaystyle q(z)=$
$\displaystyle(c^{2}-4)\sigma_{3}\,\sigma_{4}(1+cz+z^{2}),$

such
that$\,\,\sigma_{1}=\mathrm{atanh}\left(\frac{{\left(c^{2}-4\right)}\,{\left(c+2\,z\right)}}{\sigma_{3}\,\sigma_{4}}\right),\;\sigma_{2}=\mathrm{atanh}\left(\frac{4\,c-c^{3}}{\sigma_{3}\,\sigma_{4}}\right),\;\sigma_{3}={{\left(c-2\right)}}^{3/2}\text{and}\;\sigma_{4}={{\left(c+2\right)}}^{3/2}.$
Similary, solving for $g_{c,a}(z)$ gives,

$g_{c,a}(z)=-r(z)/q(z),$ (3) $\displaystyle\text{where}\,\,\,\,\,\,r(z)$
$\displaystyle=16\sigma_{2}+16\sigma_{1}-4c^{2}\sigma_{2}+16z^{2}\sigma_{2}-4c^{2}\sigma_{1}+16z^{2}\sigma_{1}-2z\sigma_{3}\sigma_{4}-4c^{2}z^{2}\sigma_{2}$
$\displaystyle+2ac^{3}\sigma_{1}-4c^{3}z\sigma_{1}-8ac\sigma_{2}+16cz\sigma_{2}-4c^{2}z^{2}\sigma_{1}+2ac^{3}\sigma_{2}-4c^{3}z\sigma_{2}$
$\displaystyle-8ac\sigma_{1}+16cz\sigma_{1}+2ac^{3}z^{2}\sigma_{1}-8acz^{2}\sigma_{2}-8ac^{2}z\sigma_{2}$
$\displaystyle+2ac^{4}z\sigma_{2}+2ac^{3}z^{2}\sigma_{2}-8acz^{2}\sigma_{1}-8ac^{2}z\sigma_{1}$
$\displaystyle+2ac^{4}z\sigma_{1}+2az^{2}\sigma_{3}\sigma_{4}-cz^{2}\sigma_{3}\sigma_{4}$
$\displaystyle+a\,c\,z\sigma_{3}\sigma_{4}$

and $q(z),\,\sigma_{1},\,\sigma_{2},\,\sigma_{3}$ are as above.

So, $f_{c,a}(z)=h_{c,a}(z)+\overline{g_{c,a}(z)}$ is the desired map
corresponding to conformal map $F_{c,a}$. By Theorem 2, we have $f_{c,a}\in
S_{H}^{0}$ and is convex in direction of real axis. So, $f_{c,a}$ is the
required harmonic map obtained by applying shear construction.
Next we discuss special cases by taking $c=-2$ and $c=2$ in (1). By shearing

$F_{-2}(z)=\frac{z}{(1-z)^{2}}\,\,\text{with
dilatation}\,\,\,\omega_{a}(z)=\frac{z(z+a)}{1+az},\,\,\,\,$

(a) $c=-2,a=-1$

(b) $c=-2,a=0$

(c) $c=-2,a=1$

(d) $c=0,a=-1$

(e) $c=0,a=0$

(f) $c=0,a=1$

(g) $c=2,a=-1$

(h) $c=2,a=0$

(i) $c=2,a=1$

Figure 2: Images of harmonic shear $f_{c,a}(|z|=0.999)$ having dilatation
$\omega(z)=\frac{z(z+a)}{1+az}$.

and solving for $h$ and $g$, we get

$h(z)=-\frac{6\,z+3\,a\,z^{2}-a\,z^{3}-6\,z^{2}+2\,z^{3}}{6\,{{\left(z-1\right)}}^{3}}\,\,\,\text{and}\,\,\,g(z)=-\frac{3\,a\,z^{2}-a\,z^{3}+2\,z^{3}}{6\,{{\left(z-1\right)}}^{3}}$

So, the required map $f_{-2,a}$ can be expressed as
$f_{-2,a}(z)=\operatorname{Re}\\{h(z)+g(z)\\}+\dot{\iota}\operatorname{Im}\\{h(z)-g(z)\\}$,
i.e,

$\displaystyle
f_{-2,a}(z)=\operatorname{Re}\left\\{\frac{a}{3}-\frac{z^{2}+(a-1)z-\frac{a}{3}+\frac{2}{3}}{(z-1)^{3}}-\frac{2}{3}\right\\}+\frac{\dot{\iota}}{4}\operatorname{Im}\left\\{\left(\frac{1+z}{1-z}\right)^{2}-1\right\\}$
(4)

By Theorem 2, we see that $f_{-2,a}\in S_{H}^{0}.$ For $a=-1$, $f(\mathbb{D})$
is a right half plane. In order to study mapping properties of $f_{-2,a}$, let
$w=(1+z)/(1-z)$, i.e, $z=(w-1)/(w+1)$, which leads us to:

$\displaystyle f_{-2,a}(z)$
$\displaystyle=\operatorname{Re}\left\\{\frac{1}{4}\left(\frac{1+a}{3}w^{3}+(1-a)w-\frac{2(2-a)}{3}\right)\right\\}+\dot{\iota}\operatorname{Im}\left\\{\frac{1}{4}(w^{2}-1)\right\\}$
$\displaystyle=\frac{1}{4}\left(\frac{1+a}{3}(x^{3}-3xy^{2})+(1-a)x-\frac{2(2-a)}{3}\right)+\dot{\iota}\frac{1}{2}xy,\,\,x>0.$

Note that each point $z\neq 1$ on the unit circle is carried onto a point $w$
on the imaginary axis so that $x=0$ and $F_{-2,a}=-(2-a)/6.$ Similar
discussion as in the case of harmonic koebe function K(z) in [9] shows that
for $-1<a\leq 1$, $f(\mathbb{D})$ is a slit domain i.e, it maps onto the
entire plane minus the interval $(-\infty,-(2-a)/6)$ on the negative real
axis. For each $a$, the tip of the slit is located at $-(2-a)/6$, which we can
visualize in Fig: 2(a), 2(b) and 2(c).
Similarly, proceeding the same way for $F_{2}(z)=z/(1+z)^{2}$, where the
harmonic shear is given by:

$\displaystyle
f_{2,a}(z)=\operatorname{Re}\left\\{\frac{a}{3}-\frac{z^{2}+(a+1)z+\frac{a}{3}+\frac{2}{3}}{(z+1)^{3}}+\frac{2}{3}\right\\}+\dot{\iota}\operatorname{Im}\left\\{\frac{z}{(1+z)^{2}}\right\\}.$
(5)

$f(\mathbb{D})$ is the left half plane for a=1 and a slit domain where the tip
of the slit is located at $(a+2)/6$ for each $-1\leq a<1$, as can be seen in
Fig: 2(g), 2(h) and 2(i). Notice that the images of harmonic shear in fig: 2,
has been changed only horizontally, which is the direction of the shear. We
also observe that the shear for $c=2,a=1$ in fig: 2(i) has collapsed the image
of unit disk onto the point $1/2$. Therefore, we get the following result:

######  Theorem 4.

Consider the conformal univalent map $F_{c}$ of the unit disk $\mathbb{D}$
onto a domain convex in direction of real axis given by (1), where
$c\in[-2,2]$. Let the dilatation function be given by
$\omega_{a}(z)=z(z+a)/(1+az)$, where $-1\leq a\leq 1$, Then the horizontal
shear of $F_{c}$ with $c\in(-2,2)$ and dilatation $\omega_{a}$ is given by a
two parameter family of harmonic mappings
$f_{c,a}(z)=h_{c,a}(z)+\overline{g_{c,a}(z)}$ where $h_{c,a}(z)$ and
$g_{c,a}(z)$ are given by (2) and (3) such that $f_{c,a}\in S_{H}^{0}$. For
special cases of c=-2 and c=2 with $-1\leq a\leq 1$, the shear is given by (4)
and (5). Additionally, we have $f(\mathbb{D})$ is a slit domain where the tip
of the slit is located at $-(2-a)/6$ for $c=-2$ and at $(a+2)/6$ for $c=2$
such that $f_{-2,a},f_{2,a}\in S_{H}^{0}$.

###  3.1 Minimal surfaces for $a=0$

For the special case $a=0$, we have $\omega(z)=z^{2}$ which is a square
analytic dilatation. Then we can apply Weierstrass-Enneper formula to lift the
harmonic mapping to a minimal graph on $\mathbb{D}$. Thus, we have the
following result:

######  Theorem 5.

For $c\in(-2,2)$, define
$f_{c}=h_{c}+\bar{g_{c}}:\mathbb{D}\rightarrow\mathbb{C}$ to be the harmonic
mapping satisfying $h_{c}(z)-g_{c}(z)=F_{c}(z)$ and
$g_{c}^{\prime}(z)=\omega(z)h_{c}^{\prime}(z)$ normalized by
$h_{c}(0)=g_{c}(0)=g_{c}^{\prime}(0)=h_{c}^{\prime}(0)-1=0$, where $F_{c}$ is
given by

$F_{c}(z)=\frac{z}{1+cz+z^{2}},$ $\omega(z)=z^{2}.$

Then, $f_{c}\in S_{H}^{0}$ and $f_{c}(\mathbb{D})$ is convex in the horizontal
direction. As $c$ varies from 0 to 2, $f_{c}(\mathbb{D})$ transforms from a
infinite vertical strip mapping to a single slit mapping along positive real
axis, and as c varies from $-2$ to 0, it transforms from a single slit mapping
along negative real axis to a strip mapping. Furthermore, since $w$ is a
square analytic function, $f_{c}$ lifts to a minimal graph $X_{c}$ on
$\mathbb{D}$ for each $c\in(-2,2)$ . For some special cases, $c=-2,c=0$ and
$c=2,X_{-2}(\mathbb{D})$ is part of Enneper’s surface, $X_{0}(\mathbb{D})$ is
part of helicoid and $X_{2}(\mathbb{D})$ is again part of Enneper’s surface.

(a) $c=-2$
(Identified as Enneper’s surface)

(b) $c=0$
(Identified as part of Helicoid)

(c) $c=1$

(d) $c=2$
(Identified as Enneper’s surface)

Figure 3: Minimal surfaces over harmonic map $f_{c},\,c=0,2,-2,1$ with
dilatation $z^{2}.$

###### Proof.

By Theorem 2, $f_{c}\in S_{H}^{0}$ and $f_{c}(\mathbb{D})$ is convex in the
horizontal direction.Then, the solution of the pair of differential equations

$h_{c}^{\prime}-g_{c}^{\prime}=F_{c}^{\prime}$ $\omega
h^{\prime}-g^{\prime}=0$

defines a family of harmonic mappings
$f_{c}=\operatorname{Re}\\{h_{c}+g_{c}\\}+\dot{\iota}\operatorname{Im}\\{h_{c}-g_{c}\\}.$
For $c\in(-2,2)$,

$\displaystyle u=\operatorname{Re}\\{h_{c}+g_{c}\\}$
$\displaystyle=\operatorname{Re}\left\\{\frac{z}{(c-2)(z+1)}+\frac{z}{(c+2)(z-1)}-\frac{8\left(\operatorname{atanh}\left(\frac{c\left(c^{2}-4\right)}{\sigma_{1}}\right)-\operatorname{atanh}\left(\frac{\left(c^{2}-4\right)(c+2z)}{\sigma_{1}}\right)\right)}{\sigma_{1}}\right\\}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,-\operatorname{Re}\left\\{\frac{z\left(z^{2}+1\right)}{\left(z^{2}-1\right)\left(z^{2}+cz+1\right)}\right\\},$
$\displaystyle\mbox{where}\,\,\sigma_{1}=(c-2)^{3/2}(c+2)^{3/2}$
$\displaystyle
v=\operatorname{Im}\\{h_{c}-g_{c}\\}=\operatorname{Im}\left\\{\frac{z}{z^{2}+cz+1}\right\\}.$

Since $\omega(z)=z^{2}$, by Theorem 2, $f_{c}$ lifts to a minimal graph
corresponding to each $c\in(2,2)$. Applying this theorem yields the following
representations of minimal graphs for $c\in(-2,2)$:

$X_{c}(z)=(u,v,F(u,v)),$

where $x_{1}$ and $x_{2}$ are as above, and

$\displaystyle
F(u,v)=2\operatorname{Im}\left\\{\frac{\frac{2}{c^{2}-4}+\frac{cz}{c^{2}-4}}{z^{2}+cz+1}-\frac{2}{c^{2}-4}+\frac{2c\operatorname{atanh}\left(\frac{c\left(c^{2}-4\right)}{\sigma_{1}}\right)}{\sigma_{1}}-\frac{2c\operatorname{atanh}\left(\frac{\left(c^{2}-4\right)(c+2z)}{\sigma_{1}}\right)}{\sigma_{1}}\right\\}$

where $\sigma_{1}=(c-2)^{3/2}(c+2)^{3/2}$. We can observe how
$f_{c}(\mathbb{D})$ transforms from an infinite strip to a slit when $c$ goes
from 0 to 2 in Fig: 2(e) and 2(h) and vice-versa when $c$ goes from -2 to 0
can be seen in Fig: 2(b) and Fig: 2(e).

We establish the claim of identifying the surfaces as helicoid and Enneper’s
surface by considering the following cases:

######  Case 1:

$c=-2$, the analytic function becomes $F_{-2}(z)=\dfrac{z}{(z-1)^{2}}.$
Applying the shearing technique with dilatation $\omega(z)=z^{2}$ and then
lifitng the resulting harmonic map, we have represetation of the corresponding
minimal graph as $X_{-2}(\mathbb{D})=(u,v,F(u,v))$ as:

$\displaystyle
X_{-2}(z)=\left(\operatorname{Re}\left\\{-\frac{z^{2}-z+\frac{2}{3}}{(z-1)^{3}}-\frac{2}{3}\right\\},\operatorname{Im}\left\\{\frac{z}{(z-1)^{2}}\right\\},2\,\operatorname{Im}\left\\{\frac{1}{6}-\frac{3z-1}{6(z-1)^{3}}\right\\}\right).$

Applying the change of variables technique to recognize the surface, and
substituting $w=(1+z)/(1-z)$ and again expressing as $z=(w-1)/(w+1)$, the
surface representation becomes:

$\displaystyle
X_{-2}(z)=\left(\operatorname{Re}\left\\{\frac{w^{3}}{12}+\frac{w}{4}-\frac{1}{3}\right\\},\operatorname{Im}\left\\{\frac{w^{2}}{4}-\frac{1}{4}\right\\},2\,\operatorname{Im}\left\\{\frac{w^{3}}{24}-\frac{w}{8}+\frac{1}{12}\right\\}\right).$

Interchanging the second and third coordinates, Translation by 1/3, and
scaling by a factor of 4, we get

$\displaystyle
X_{-2}(\mathbb{D})=\left(\operatorname{Re}\left\\{\frac{w^{3}}{3}+w\right\\},\operatorname{Im}\left\\{\frac{w^{3}}{3}-w\right\\},\operatorname{Im}\left\\{w^{2}\right\\}\right).$

We see that our original surface $X_{-2}(\mathbb{D})$ is part of the enneper
surface formed by using right half plane as domain instead of the standard
unit disk.

######  Case 2:

For c=0, the function $F_{c}(z)$ becomes $F(z)=z/(1+z^{2})$ with dilatation
$\omega(z)=z^{2}.$ Solving for $h$ and $g$, we get
$h(z)=(z+\mathrm{atan}\left(z\right)+z^{2}\,\mathrm{atan}\left(z\right))/(2\,{\left(z^{2}+1\right)})$
and
$g(z)=(\mathrm{atan}\left(z\right)-z+z^{2}\,\mathrm{atan}\left(z\right))/(2\,{\left(z^{2}+1\right)})$,
such that $f=h+\overline{g}$ is the harmonic map required to lift to a minimal
graph. So, we have

$\displaystyle X_{0}(z)=(x_{1},x_{2},x_{3})$
$\displaystyle=\left(\operatorname{Re}\\{\mathrm{atan}\left(z\right)\\},\operatorname{Im}\\{\frac{z}{z^{2}+1}\\},2\operatorname{Im}\left\\{\frac{1}{2}-\frac{1}{2\,{\left(z^{2}+1\right)}}\right\\}\right)$
$\displaystyle=\left(\operatorname{Re}\left\\{\frac{\dot{\iota}}{2}\,\log\left(\frac{\dot{\iota}+z}{\dot{\iota}-z}\right)\right\\},\operatorname{Im}\left\\{\frac{z}{z^{2}+1}\right\\},2\operatorname{Im}\left\\{\frac{1}{2}-\frac{1}{2\,{\left(z^{2}+1\right)}}\right\\}\right),$

which are hardly recognizable as representations of part of the helicoid.
However, applying the change of variables $w=(\dot{\iota}+z)/(\dot{\iota}-z)$
and again writing $z=\dot{\iota}(w-1)/(w+1)$,

$\displaystyle X_{0}(z)$
$\displaystyle=\left(\operatorname{Re}\left\\{\frac{\dot{\iota}}{2}\,\log\left(w\right)\right\\},\operatorname{Im}\left\\{\frac{\dot{\iota}}{4}\left(w-\frac{1}{w}\right)\right\\},2\operatorname{Im}\left\\{-\frac{1}{8}\left(w+\frac{1}{w}-2\right)\right\\}\right).$

Scaling by 4 gives,

$\displaystyle X_{0}(z)$
$\displaystyle=\left(\operatorname{Im}\\{2\log\left(w\right)\\},\operatorname{Re}\left\\{w-\frac{1}{w}\right\\},\operatorname{Im}\left\\{-\left(w+\frac{1}{w}\right)\right\\}\right).$

Since, negative sign over $x_{3}$, interchanging of the second and third
coordinates, and then first and third coordinates won’t alter the geometry and
gives us the formula of helicoid. So we see $X_{0}(\mathbb{D})$ is the same
surface as $Y_{0}({z\in\mathbb{C}:\operatorname{Re}{z}>0}))$, i.e.,
$X_{0}(\mathbb{D})$ is part of the helicoid.

######  Case 3:

For c=2, the conformal map takes the form $F_{2}(z)=\dfrac{z}{(1+z)^{2}}.$
After similar computations we get,

$\displaystyle
X_{2}(z)=\left(\operatorname{Re}\left\\{\frac{2}{3}-\frac{z^{2}+z+\frac{2}{3}}{(z+1)^{3}}\right\\},\operatorname{Im}\left\\{\frac{z}{(1+z)^{2}}\right\\},2\,\operatorname{Im}\left\\{\frac{1}{6}-\frac{\frac{z}{2}+\frac{1}{6}}{(z+1)^{3}}\right\\}\right).$

Substituting $w=(1-z)/(1+z)$ and expressing in terms of $z$ as
$z=(1-w)/(1+w)$,then translating by $-1/3$, scaling by a factor of $-1/4$ and
finally adjusting the negative over $x_{3}$ the surface representation
becomes:

$\displaystyle
X_{2}(z)=\left(\operatorname{Re}\left\\{\frac{w^{3}}{3}+w\right\\},\operatorname{Im}\left\\{w^{2}\right\\},\operatorname{Im}\left\\{\frac{w^{3}}{3}-w\right\\}\right).$

We see interchanging the coordinates suitably gives the parameterization of
Enneper’s surface.

We have plotted the minimal surfaces formed for $c=-2,0,1$ and $2$ in Fig:
3(a), 3(b), 3(c) and 3(d). Their projection onto $x$-$y$ plane can be observed
and matched with images of the corresponding harmonic maps in Fig: 2(b), 2(e)
and 2(h). ∎

Further, this idea of lifting can be generalized by changing the dilatation to
$\omega(z)=z^{2n}$, $n\in\mathbb{N}$. In 2014, Ponnusamy et al. have
constructed a two-parameter family of harmonic mappings $f_{n,c}$ with
dilatation $z^{n}$ by shearing the four slit conformal mapping defined by
$\phi(z)$ in [11]. We will further lift the resulting shears to a family of
minimal surfaces for even $n$ and $A=0$.

$\displaystyle\phi(z)=A\,\log\left(\frac{1+z}{1-z}\right)+B\,\frac{z}{1+cz+z^{2}}.$

Taking $A=0,B=1$ serves our purpose. By writing $c=-2\,\cos\gamma$ where
$\gamma\in(0,\pi)$. Applying shearing technique gives,

$h(z)=-\frac{\dot{\iota}}{2\,\sin\gamma}(e^{-\dot{\iota}\gamma}\textit{I}_{2}-e^{\dot{\iota}\gamma}\textit{I}_{3}),$
$\displaystyle\text{where}\,\,\textit{I}_{2}=\int_{0}^{z}\frac{d\zeta}{(\zeta-e^{-\dot{\iota}\gamma})^{2}(1-\zeta^{n})}\,\,\,\mbox{and}\,\,\,\textit{I}_{3}=\int_{0}^{z}\frac{d\zeta}{(\zeta-e^{\dot{\iota}\gamma})^{2}(1-\zeta^{n})}.$

Using
$\dfrac{1}{1-z^{n}}=-\frac{1}{n}\sum\limits_{k=0}^{n-1}\dfrac{z_{k}}{z-z_{k}}$,
where $z_{k}=e^{\frac{2\pi\dot{\iota}k}{n}}$ for $k=0,1,2,...n-1$ and assuming
that $\eta=e^{\dot{\iota}\gamma}\neq z_{k}$, consider

$\displaystyle\textit{I}_{\eta}$
$\displaystyle=\int_{0}^{z}\frac{d\zeta}{(\zeta-\eta)^{2}(1-\zeta^{n})}$
$\displaystyle=-\frac{1}{n}\sum\limits_{k=0}^{n-1}\left\\{\frac{1}{\eta-
z_{k}}\left(\frac{1}{\eta-z}-\frac{1}{\eta}\right)-\frac{1}{(\eta-
z_{k})^{2}}\left[\log\left(\frac{\eta-z}{\eta}\right)-\log\left(\frac{z_{k}-z}{z_{k}}\right)\right]\right\\}.$

(a) c=0, n=4

(b) $c=0,n=6$

(c) $c=1,n=4$

(d) $c=1,n=6$

(e) $c=2,n=4$

(f) $c=2,n=6$

Figure 4: Minimal surfaces over harmonic shears of $F(z)=z/(1+cz+z^{2})$ with
dilatation $\omega(z)=z^{n}$ for even n.

Let $\mathbf{N}=\\{0,1,\dots,n-1\\}$ be an index set. Suppose that
$a\in\mathbf{N}$, then we define $\mathbf{N}_{a}=\mathbf{N}\setminus\\{a\\}$.
In the case of $\gamma=2\pi m/n$, where $m=0,1,2,\dots,n-1$,

$\displaystyle\textit{I}_{3,m}$
$\displaystyle=\int_{0}^{z}\frac{d\zeta}{(\zeta-z_{m})^{2}(1-\zeta^{n})}$
$\displaystyle=\frac{1}{n}\sum\limits_{k\in\mathbf{N}_{m}}\int_{0}^{z}\frac{z_{k}\,d\zeta}{(\zeta-
z_{m})^{2}(\zeta-z_{k})}+\frac{1}{n}\int_{0}^{z}\frac{z_{m}\,d\zeta}{(\zeta-
z_{m})^{3}}.$

The above sum can be computed as above and the last integral is,

$\displaystyle\frac{1}{n}\int_{0}^{z}\frac{d\zeta}{(\zeta-
z_{m})^{3}}=\frac{1}{2n}\left[\frac{1}{(z-z_{m})^{2}}-\frac{1}{z_{m}^{2}}\right].$

We can readily find $g$ from $h-\phi$ for $A=0$. Thus, we get the harmonic map
$h=f+\overline{g}$ that we can lift to a minimal surface for square dilatation
by taking $\omega(z)=z^{2n}$. Since,
$x_{3}=2\operatorname{Im}\left\\{\int_{0}^{z}\sqrt{\omega(\zeta)}\,h^{\prime}(\zeta)\,d\zeta\right\\}$,
we compute

$\int_{0}^{z}\sqrt{\omega(\zeta)}\,h^{\prime}(\zeta)\,d\zeta=-\frac{\dot{\iota}}{2\,\sin\gamma}\left(e^{-\dot{\iota}\gamma}\int_{0}^{z}\frac{\zeta^{n}}{(\zeta-e^{-\dot{\iota}\gamma})^{2}(1-\zeta^{2n})}d\zeta-e^{\dot{\iota}\gamma}\int_{0}^{z}\frac{\zeta^{n}}{(\zeta-e^{\dot{\iota}\gamma})^{2}(1-\zeta^{2n})}d\zeta\right).$

Using,
$\dfrac{z^{n}}{1-z^{2n}}=\dfrac{z^{n}}{(1-z^{n})(1+z^{n})}=\dfrac{1}{2}\left[\dfrac{1}{1-z^{n}}-\dfrac{1}{1+z^{n}}\right],$
the above expression becomes:

$\displaystyle=\frac{-\dot{\iota}}{2\,\sin\gamma}\left(e^{-\dot{\iota}\gamma}\int_{0}^{z}\frac{1}{2(\zeta-e^{-\dot{\iota}\gamma})^{2}(1-\zeta^{n})}-e^{-\dot{\iota}\gamma}\int_{0}^{z}\frac{1}{2(\zeta-e^{-\dot{\iota}\gamma})^{2}(1+\zeta^{n})}\right)d\zeta$
$\displaystyle\,\,\,\,\,\,-\frac{-\dot{\iota}}{2\,\sin\gamma}\left(e^{\dot{\iota}\gamma}\int_{0}^{z}\frac{1}{2(\zeta-e^{\dot{\iota}\gamma})^{2}(1-\zeta^{n})}-e^{\dot{\iota}\gamma}\int_{0}^{z}\frac{1}{2(\zeta-e^{\dot{\iota}\gamma})^{2}(1+\zeta^{n})}\right)d\zeta.$

We observe that the first and third integral in above expression are
$\textit{I}_{2}$ and $\textit{I}_{3}.$ For the second (say $\textit{I}_{4}$)
and fourth integral ($\textit{I}_{5}$), again by partial fractions we can
express:

$\frac{1}{1+z^{n}}=-\frac{1}{n}\sum\limits_{l=0}^{n-1}\frac{z_{l}}{z-z_{l}},$

where $z_{l}$’s are $n^{th}$ roots of equation $z^{n}+1=0$ such that
$z_{l}=e^{\frac{2\pi\dot{\iota}l}{n}+\frac{\pi\dot{\iota}}{n}}$, where
$l=0,1,2,\dots,n-1$, then

$\int_{0}^{z}\sqrt{\omega(\zeta)}\,h^{\prime}(\zeta)\,d\zeta=\frac{-\dot{\iota}}{2\,\sin\gamma}\left(\frac{e^{-\dot{\iota}\gamma}}{2}\textit{I}_{2}-\frac{e^{-\dot{\iota}\gamma}}{2}\textit{I}_{4}-\frac{e^{\dot{\iota}\gamma}}{2}\textit{I}_{3}+\frac{e^{\dot{\iota}\gamma}}{2}\textit{I}_{5}\right),$

where,

$\displaystyle\textit{I}_{4}=\int_{0}^{z}\frac{1}{(\zeta-e^{-\dot{\iota}\gamma})^{2}(1+\zeta^{n})}d\zeta\,\,\,\mbox{and}\,\,\,\textit{I}_{5}=\int_{0}^{z}\frac{1}{(\zeta-e^{\dot{\iota}\gamma})^{2}(1+\zeta^{n})}d\zeta.$

Assuming that $\rho=e^{\dot{\iota}\gamma}\neq z_{l}$, i.e,
$\gamma\neq(2l+1)\frac{\pi}{n}$, and computing in a similar manner just like
$\textit{I}_{\eta}$, we get

$\displaystyle\textit{I}_{\rho}$
$\displaystyle=\int_{0}^{z}\frac{d\zeta}{(\zeta-\rho)^{2}(1-\zeta^{n})}$
$\displaystyle=-\frac{1}{n}\sum\limits_{k=0}^{n-1}\left\\{\frac{1}{\rho-
z_{l}}\left(\frac{1}{\rho-z}-\frac{1}{\rho}\right)-\frac{1}{(\rho-
z_{l})^{2}}\left[\log\left(\frac{\rho-z}{\rho}\right)-\log\left(\frac{z_{l}-z}{z_{l}}\right)\right]\right\\}.$

In case of $\gamma=(2s+1)\frac{\pi}{n}$ for $s=0,1,2,\dots,n-1$ and computing
in similar manner just like $\textit{I}_{3,m}$ above, we get
$\textit{I}_{3,p}=\textit{I}_{5}$.

$\displaystyle\textit{I}_{3,s}$
$\displaystyle=\int_{0}^{z}\frac{d\zeta}{(\zeta-
z_{s})^{2}(1-\zeta^{n})}=-\frac{1}{n}\sum\limits_{l\in\mathbf{N}_{s}}\int_{0}^{z}\frac{z_{l}\,d\zeta}{(\zeta-
z_{s})^{2}(\zeta-z_{l})}+\frac{1}{n}\int_{0}^{z}\frac{z_{s}\,d\zeta}{(\zeta-
z_{s})^{3}}.$

This last integral can again be computed as done for $I_{3,m}$ above. In Fig:
4, we have plotted the minimal surfaces obtained after calculating $x_{3}$
coordinate, for different values of $c$ and even $n$.

Table 1: The integrals $I_{2}$ and $I_{3}$ for the analytic part $h=\frac{1}{2\sin\gamma}i\left(e^{-i\gamma}I_{2}-e^{i\gamma}I_{3}\right)$ of the harmonic shear $f$ with a dilatation $\omega(z)=z^{n}$. $\gamma$ | $I_{2}$ | $I_{3}$
---|---|---
is not $2\pi m/n$ | $I_{\bar{\eta}}$ | $I_{\eta}$
is $2\pi m/n$ | $I_{3,n-m}$ | $I_{3,\mathrm{\leavevmode\nobreak\ m}}$
Table 2: The integrals $I_{4}$ and $I_{5}$ for the $x_{3}=\operatorname{Im}\left\\{\frac{-\dot{\iota}}{2\,\sin\gamma}\left(e^{-\dot{\iota}\gamma}2\textit{I}_{2}-e^{-\dot{\iota}\gamma}\textit{I}_{4}-e^{\dot{\iota}\gamma}\textit{I}_{3}+e^{\dot{\iota}\gamma}\textit{I}_{5}\right)\right\\}$ coordinate for the minimal surfaces. $\gamma$ | $I_{4}$ | $I_{5}$
---|---|---
is not $(2s+1)\pi/n\,\,\,\,\,$ | $I_{\bar{\rho}}$ | $I_{\rho}$
is $(2s+1)\pi/n$ | $I_{3,n-s}$ | $I_{3,s}$

##  4 Harmonic shears of inner region of an epicycloid

In this section, we will apply shear construction method on a new class of
conformal mappings denoted by $F_{n}(z)$ with dilatation $\omega(z)=z^{n}$ and
will further lift to a family of minimal surfaces using Theorem 3. We have
created images of these minimal surfaces for $n=2,3,4$ in Fig: 6. We have also
plotted images of conformal mappings $F_{n}(z)$ alongside images of
corresponding harmonic mappings $f_{n}(z)$ for $n=2,3,4$ in Fig: 5.
let

$\displaystyle F_{n}(z)=h_{n}(z)-g_{n}(z)=z-\frac{1}{n^{2}}\,z^{n},$ (6)

be the class of conformal univalent mappings of the unit disk $\mathbb{D}$
onto a domain convex in horizontal direction and $n$ is any natural number
greater than equal to $2$. We observed that images of the open unit disk under
conformal maps form the interior region of an epicycloid characterized by
$n-1$ cusps. Let

$\displaystyle\omega(z)=\frac{g^{\prime}(z)}{h^{\prime}(z)}=z^{n},$

be the dilatation. Differentiating $(1)$ gives us,

$\displaystyle h_{n}^{\prime}(z)-g_{n}^{\prime}(z)=1-\frac{1}{n}\cdot
z^{n-1}.$

Considering the pair of these differential equations above, we get

$h_{n}^{\prime}(z)=\frac{1-\frac{1}{n}\cdot z^{n-1}}{1-z^{n}}.$

(a) $n=2$

(b) $n=3$

(c) $n=4$

Figure 5: Conformal mapping $F_{n}$ of the unit disk onto inner region of an
epicycloid with $n-1$ cusps, alongside its harmonic shear $f_{n}$ on right
with $n+2$ concave boundary arcs for the dilatation $z^{n},\,n=2,3,4$.

Integrating from $0$ to $z$ and normalizing such that $h(0)=0$ and similarly,
solving for $g(z)$ gives:

$\displaystyle h_{n}(z)=z\hskip
2.27626pt_{2}F_{1}\left(1,\frac{1}{n};\frac{1}{n}+1;z^{n}\right)+\frac{\log\left(z^{n}-1\right)}{n^{2}}-\frac{\pi\dot{\iota}}{n^{2}},$
(7) $\displaystyle g_{n}(z)=z\hskip
2.27626pt_{2}F_{1}\left(1,\frac{1}{n};\frac{n+1}{n};z^{n}\right)-z+\frac{\log\left(z^{n}-1\right)+z^{n}}{n^{2}}-\frac{\pi\dot{\iota}}{n^{2}},$
(8)

where ${}_{2}F_{1}(a,b;c;z)$ represents a hypergeometric function, which is a
power series such that,

${}_{2}F_{1}(a,b;c;z)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!},$

where $a,b$, and $c\in\mathbb{C}$ and $(x)_{n}=x(x+1)\cdots(x+n-1)$ is the
Pochhammer symbol. So, the desired map corresponding to analytic map $F_{n}$
is $f_{n}=h_{n}+\overline{g_{n}}$. By Theorem 2, $f_{n}\in S_{H}^{0}$ and
$f_{n}$ is convex in horizontal direction.

We see that image of equally spaced radial segments and concentric circles
under sense-preserving harmonic univalent map $f_{n}$ is a radial plot with
n+2 concave boundary arcs. Notice that, although the image of $F_{n}$ is
bounded, the image of $f_{n}$ is unbounded.The harmonic map $f_{n}$ can also
be expressed as $f_{n}(z)=x_{1}(z)+\dot{\iota}x_{2}(z)$ where

$\displaystyle x_{1}(z)$
$\displaystyle=\operatorname{Re}\left\\{h(z)+g(z)\right\\}$
$\displaystyle=\operatorname{Re}\left\\{2z\hskip
2.27626pt_{2}F_{1}\left(1,\frac{1}{n};\frac{n+1}{n};z^{n}\right)-z+\frac{2\log\left(z^{n}-1\right)+z^{n}}{n^{2}}-\frac{2\pi\dot{\iota}}{n^{2}}\right\\},$
$\displaystyle x_{2}(z)$ $\displaystyle=\operatorname{Im}\\{h(z)-g(z)\\}$
$\displaystyle=z-\frac{z^{n}}{n^{2}}.$

For $n=2m$, $f_{n}$ lifts to a minimal graph where

$\displaystyle x_{3}$
$\displaystyle=2\,\operatorname{Im}\left\\{\int_{0}^{z}\sqrt{\omega(\zeta)}h_{n}^{\prime}(\zeta)\,d\zeta\right\\}$
$\displaystyle=2\,\operatorname{Im}\left\\{\int_{0}^{z}\zeta^{m}\left(\frac{1-\frac{1}{2m}\cdot\zeta^{2m-1}}{1-\zeta^{2m}}\right)\,d\zeta\right\\}$
$\displaystyle=2\,\operatorname{Im}\left\\{\int_{0}^{z}\left(\frac{\zeta^{m}}{1-\zeta^{2m}}-\frac{1}{2m}\cdot\frac{\zeta^{3m}}{\zeta(1-\zeta^{2m})}\right)\,d\zeta\right\\}$
$\displaystyle=2\,\operatorname{Im}\left\\{\int_{0}^{z}\left(\frac{1}{2}\left(\frac{1}{1-\zeta^{m}}-\frac{1}{1+\zeta^{m}}\right)-\frac{1}{2m}\cdot\left(\frac{\zeta^{m-1}}{2(\zeta^{m}+1)}-\frac{\zeta^{m-1}}{2(\zeta^{m}-1)}-\zeta^{m-1}\right)\right)\,d\zeta\right\\}$
$\displaystyle=2\,\operatorname{Im}\left\\{z\hskip
2.27626pt_{2}F_{1}\left(1,\frac{1}{m};\frac{1}{m}+1;z^{m}\right)-z\hskip
2.27626pt_{2}F_{1}\left(1,\frac{1}{m};\frac{1}{m}+1;-z^{m}\right)-\frac{1}{2m}\left(\frac{\mathrm{atanh}\left(z^{m}\right)-z^{m}}{m}\right)\right\\}.\,\,(9)$

Thus, we obtain the following result:

######  Theorem 6.

Consider the conformal univalent map of the unit disk $\mathbb{D}$ onto a
domain convex in direction of real axis given by $(\ref{eqn: 6})$. Let the
dilatation function be given by $\omega(z)=z^{n}$ ,then the horizontal shear
of $F_{n}$ with dilatation $\omega$ is given by
$f_{n}(z)=h_{n}(z)+\overline{g_{n}(z)}$, where $h_{n}(z)$ and $g_{n}(z)$ are
given by (7) and (8) such that $f_{n}\in S_{H}^{0}$. In particular, when $n$
is an even positive integer, $f_{n}(\mathbb{D})$ lifts to a minimal graph
$(x_{1},x_{2},x_{3})$ where
$x_{1}=\operatorname{Re}\left\\{h(z)+g(z)\right\\}$,
$x_{2}=\operatorname{Im}\\{h(z)-g(z)\\}$ and $x_{3}$ is given by $(9)$.

The images of the unit disk under $F_{n}$ and $f_{n}$ for $n=2,3$ and $4$ are
shown in Fig: 5 as plots of the images of equally spaced radial segments and
concentric circles. The images of the resulting minimal surfaces obtained by
lifting $f_{n}$ in $\mathbb{R}^{3}$ are depicted in Fig: 6.

(a) $n=2$.

(b) $n=3$.

(c) $n=4$.

Figure 6: Minimal surface over harmonic shear $f_{n}$ with dilatation
$\omega=z^{2n}$ for $n=2,3,4.$

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