# Regularity of Functions on the Reduced Quaternion Field in Clifford Analysis

Authors: Ji Eun Kim; Su Jin Lim; Kwang Ho Shon
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We define a new hypercomplex structure ofℝ
3 and a regular function with values in that structure. From the properties of regular functions, we research the exponential function on the reduced quaternion field and represent the corresponding Cauchy-Riemann equations in hypercomplex structures of ℝ
3.


---

## Body

## 1. Introduction

Meglihzon [1], Sudbery [2], and Fueter [3] demonstrated that there are three possible approaches (the Cauchy approach, Weierstrass approach, and Riemann approach) in the theories of functions that would generalize holomorphic functions with respect to several complex variables. Sudbery [2], Soucek [4], and Sommen [5] attempted to research the Cauchy approach using differential forms and differential operators in Clifford analysis. Fueter [3] and Naser [6] studied the properties of quaternionic differential equations as a generalization of the extended Cauchy-Riemann equations in the complex holomorphic function theory. Nôno [7–9] and Sudbery [2] gave a definition and the development of regular functions over the quaternion field. Ryan [10, 11] developed the theories of regular functions in a complex Clifford analysis using a generalization of the Cauchy-Riemann equation. Malonek [12] considered analogously the function theory of hypercomplex variables. He defined the hypercomplex differentiability for the existence of a function over the Clifford algebra and monogenicity based on a generalized Cauchy-Riemann system. Gotô and Nôno [13] and Koriyama et al. [14] dealt with differential operators with the derivative of regular functions in quaternion.We shall denote byℂ, ℝ, and ℤ, respectively, the field of complex numbers, the field of real numbers, and the set of all integers. We [15, 16] showed that any complex-valued harmonic function f
1 in a pseudoconvex domain D of ℂ
2
×
ℂ
2 has a hyperconjugate harmonic function f
2 in D such that the quaternion-valued function f
1
+
f
2
j is hyperholomorphic in D and gave a regeneration theorem in quaternion analysis in the view of complex and Clifford analysis. Further, we [17, 18] investigated the existence of the hyperconjugate harmonic functions of the octonion number system and some properties of dual quaternion functions.In this paper, we introduce the Fueter variables onℝ
3 and investigate a hypercomplex structure of ℝ
3. We define regular functions and obtain the representation of the corresponding Cauchy-Riemann equations for regular functions in the reduced quaternion field.


## 2. Preliminaries

A three-dimensional, noncommutative, and associative real field, called a ternary number system, is constructed by three base elementse
0, e
1, and e
2 which satisfy
(1)
e
0
2
=
1
,
e
1
2
=
e
2
2
=
-
1
,
e
1
e
2
=
-
e
2
e
1
,
e
0
¯
=
e
0
,
e
r
¯
=
-
e
r
(
r
=
1,2
)
.
In addition, let e
0 be the identity of a ternary number system and e
1 identifies the imaginary unit -
1 in the complex field, and
(2)
ℂ
(
𝕋
)
:
=
{
z
=
e
1
z
1
+
e
2
z
2
∣
z
1
,
z
2
∈
ℂ
}
,
where z
r
=
x
r
-
(
1
/
2
)
e
r
x
0
(
r
=
1,2
) and x
m
(
m
=
0,1
,
2
) are real variables. They satisfy the equations
(3)
z
r
w
k
¯
=
w
k
¯
z
r
¯
(
r
≠
k
)
,
where z
r
¯
=
x
r
+
(
1
/
2
)
e
r
x
0
(
r
=
1,2
), w
k
=
y
k
-
(
1
/
2
)
e
k
y
0, w
k
¯
=
y
k
+
(
1
/
2
)
e
k
y
0
(
k
=
1,2
), and y
m
(
m
=
0,1
,
2
) are real variables.For any two elementsz
=
e
1
z
1
+
e
2
z
2 and w
=
e
1
w
1
+
e
2
w
2 of ℂ
(
𝕋
), their product is given by
(4)
z
w
=
z
•
w
+
z
⊙
w
,
where the corresponding commutative inner product • satisfies
(5)
z
•
w
=
1
2
(
z
w
+
w
z
)
=
-
∑
r
=
1
2
‍
z
r
w
r
+
1
2
e
1
e
2
(
z
1
¯
w
2
-
w
2
¯
z
1
+
w
1
¯
z
2
-
z
2
¯
w
1
)
and the corresponding noncommutative outer product ⊙ satisfies
(6)
z
⊙
w
=
1
2
(
z
w
-
w
z
)
=
1
2
e
1
e
2
(
z
1
¯
w
2
+
w
2
¯
z
1
-
w
1
¯
z
2
-
z
2
¯
w
1
)
=
-
w
⊙
z
.
The conjugation z
*, the corresponding norm |
z
|, and the inverse z
-
1 of z in ℂ
(
𝕋
) are given by
(7)
z
*
=
e
1
¯
z
1
¯
+
e
2
¯
z
2
¯
,
|
z
|
2
=
z
z
*
=
z
•
z
*
=
∑
r
=
1
2
‍
z
r
z
r
¯
,
z
-
1
=
z
*
|
z
|
2
(
z
≠
0
)
.For any elementz in ℂ
(
𝕋
), we have the corresponding exponential function e
z denoted by
(8)
exp
⁡
(
z
)
=
exp
⁡
(
e
1
z
1
+
e
2
z
2
)
.Theorem 1.
Letz be an arbitrary number in ℂ
(
𝕋
). Then the corresponding exponential function exp
⁡
(
z
) of z in ℂ
(
𝕋
) is given as
(9)
exp
⁡
(
z
)
=
{
(
-
1
)
k
exp
⁡
(
x
0
)
exp
⁡
(
e
2
x
2
)
,
i
f
x
1
=
k
π
,
(
-
1
)
t
exp
⁡
(
x
0
)
exp
⁡
(
e
1
x
1
)
,
i
f
x
2
=
t
π
,
where k
,
t
∈
ℤ.
Furthermore, as hyperbolic functions, one has(10)
exp
⁡
⁡
(
z
)
=
{
(
-
1
)
k
exp
⁡
(
e
2
x
2
)
(
cosh
⁡
⁡
(
x
0
)
-
sinh
⁡
⁡
(
x
0
)
)
,
i
f
x
1
=
k
π
,
(
-
1
)
t
exp
⁡
(
e
1
x
1
)
(
cosh
⁡
⁡
(
x
0
)
-
sinh
⁡
⁡
(
x
0
)
)
,
i
f
x
2
=
t
π
,
where k
,
t
∈
ℤ.Proof.
For any elementz
=
e
1
z
1
+
e
2
z
2 of ℂ
(
𝕋
),
(11)
exp
⁡
(
z
)
=
exp
⁡
(
e
1
z
1
+
e
2
z
2
)
=
exp
⁡
(
e
1
z
1
)
exp
⁡
(
e
2
z
2
)
.
Since a scalar part of e
1
z
1 is (
1
/
2
)
x
0, a vector part of e
1
z
1 is e
1
x
1, and |
e
1
|
=
1, by [19],
(12)
exp
⁡
(
e
1
z
1
)
=
exp
⁡
(
x
0
2
)
{
cos
⁡
(
|
e
1
x
1
|
)
+
e
1
x
1
|
e
1
x
1
|
sin
(
|
e
1
x
1
|
)
}
=
exp
⁡
(
x
0
2
)
{
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
}
and, similarly, we have
(13)
exp
⁡
(
e
2
z
2
)
=
exp
⁡
(
x
0
2
)
{
cos
⁡
(
|
e
2
x
2
|
)
+
e
2
x
2
|
e
2
x
2
|
sin
(
|
e
2
x
2
|
)
}
=
exp
⁡
(
x
0
2
)
{
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
}
.
Then we have
(14)
exp
⁡
(
z
)
=
exp
⁡
(
x
0
2
)
{
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
}
×
exp
⁡
(
x
0
2
)
{
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
}
=
exp
⁡
(
x
0
)
{
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
}
×
{
cos
⁡
⁡
(
x
2
)
+
e
2
sin
⁡
(
x
2
)
}
=
exp
⁡
(
x
0
)
{
cos
⁡
(
x
1
)
cos
⁡
(
x
2
)
+
e
2
cos
⁡
(
x
1
)
×
sin
(
x
2
)
+
e
1
sin
(
x
1
)
cos
⁡
(
x
2
)
}
+
exp
⁡
(
x
0
)
e
1
e
2
sin
(
x
1
)
sin
(
x
2
)
.
Also, we obtain
(15)
exp
⁡
(
z
)
=
exp
⁡
(
e
2
z
2
+
e
1
z
1
)
=
exp
⁡
(
e
2
z
2
)
exp
⁡
(
e
1
z
1
)
=
exp
⁡
(
x
0
)
{
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
}
×
{
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
}
=
exp
⁡
(
x
0
)
{
cos
⁡
(
x
1
)
cos
⁡
(
x
2
)
+
e
2
cos
⁡
(
x
1
)
×
sin
⁡
(
x
2
)
+
e
1
sin
⁡
(
x
1
)
cos
⁡
⁡
(
x
2
)
}
+
exp
⁡
(
x
0
)
e
2
e
1
sin
(
x
1
)
sin
(
x
2
)
.
Since (15) has to be equal to (14), sin
(
x
1
)
sin
(
x
2
)
=
0, that is, sin
(
x
1
)
=
0 or sin
(
x
2
)
=
0. Therefore, x
1
=
k
π or x
2
=
t
π, and then cos
⁡
(
x
1
)
=
(
-
1
)
k or cos
⁡
(
x
2
)
=
(
-
1
)
t, where k
,
t
∈
ℤ. If x
1
=
k
π
(
k
∈
ℤ
), then
(16)
exp
⁡
(
z
)
=
exp
⁡
(
x
0
)
{
(
-
1
)
k
(
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
)
}
=
(
-
1
)
k
exp
⁡
(
x
0
)
exp
⁡
(
e
2
x
2
)
.
Similarly, if x
2
=
t
π
(
t
∈
ℤ
), then
(17)
exp
⁡
(
z
)
=
exp
⁡
(
x
0
)
{
(
-
1
)
t
(
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
)
}
=
(
-
1
)
t
exp
⁡
(
x
0
)
exp
⁡
(
e
1
x
1
)
.
Further, by the Euler formula and the addition rule of trigonometric functions,
(18)
exp
⁡
(
z
)
=
exp
⁡
(
e
1
z
1
+
e
2
z
2
)
=
exp
⁡
(
e
1
z
1
)
exp
⁡
(
e
2
z
2
)
=
(
cos
⁡
(
z
1
)
+
e
1
sin
(
z
1
)
)
(
cos
⁡
(
z
2
)
+
e
2
sin
(
z
2
)
)
=
{
cos
⁡
⁡
(
x
1
)
cos
⁡
(
e
1
x
0
2
)
+
sin
⁡
(
x
1
)
sin
(
e
1
x
0
2
)
+
e
1
(
sin
(
x
1
)
cos
⁡
(
e
1
x
0
2
)
-
cos
⁡
(
x
1
)
sin
(
e
1
x
0
2
)
)
}
·
{
cos
⁡
⁡
(
x
2
)
cos
⁡
(
e
2
x
0
2
)
+
sin
⁡
(
x
2
)
sin
(
e
2
x
0
2
)
+
e
2
(
sin
(
x
2
)
cos
⁡
(
e
2
x
0
2
)
-
cos
⁡
(
x
2
)
sin
(
e
2
x
0
2
)
)
}
.
Since cos
⁡
(
e
r
(
x
0
/
2
)
)
=
cosh
⁡
(
x
0
/
2
) and sin
(
e
r
(
x
0
/
2
)
)
=
e
r
sinh
⁡
(
x
0
/
2
)
(
r
=
1,2
), we have
(19)
exp
⁡
(
z
)
=
{
cos
⁡
⁡
(
x
1
)
cosh
⁡
(
x
0
2
)
+
e
1
sin
⁡
(
x
1
)
sinh
⁡
(
x
0
2
)
+
e
1
(
sin
(
x
1
)
cosh
⁡
(
x
0
2
)
-
e
1
cos
⁡
(
x
1
)
sinh
⁡
(
x
0
2
)
)
}
·
{
cos
⁡
(
x
2
)
cosh
⁡
(
x
0
2
)
+
e
2
sin
(
x
2
)
sinh
⁡
(
x
0
2
)
+
e
2
(
sin
(
x
2
)
cosh
⁡
(
x
0
2
)
-
e
2
cos
⁡
(
x
2
)
sinh
⁡
(
x
0
2
)
)
}
=
{
(
x
0
2
)
(
cos
⁡
⁡
(
x
1
)
+
e
1
sin
⁡
(
x
1
)
)
×
(
cosh
⁡
(
x
0
2
)
-
sinh
⁡
(
x
0
2
)
)
}
×
{
(
x
0
2
)
(
cos
⁡
⁡
(
x
2
)
+
e
2
sin
⁡
(
x
2
)
)
×
(
cosh
⁡
(
x
0
2
)
-
sinh
⁡
(
x
0
2
)
)
}
=
(
cos
⁡
⁡
(
x
1
)
+
e
1
sin
⁡
(
x
1
)
)
(
cos
⁡
⁡
(
x
2
)
+
e
2
sin
⁡
(
x
2
)
)
×
(
cosh
⁡
(
x
0
2
)
-
sinh
⁡
(
x
0
2
)
)
2
.
Since
(20)
(
cosh
⁡
(
x
0
2
)
-
sinh
⁡
(
x
0
2
)
)
2
=
cosh
⁡
2
(
x
0
2
)
+
sinh
⁡
2
(
x
0
2
)
-
2
cosh
⁡
2
(
x
0
2
)
sinh
⁡
2
(
x
0
2
)
=
1
+
2
cosh
⁡
⁡
(
x
0
)
-
1
2
-
sinh
⁡
⁡
(
x
0
)
=
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
,
we obtain
(21)
exp
⁡
(
z
)
=
exp
⁡
(
e
1
z
1
)
exp
⁡
(
e
2
z
2
)
=
(
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
)
×
(
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
and, similarly,
(22)
exp
⁡
(
z
)
=
exp
⁡
(
e
2
z
2
)
exp
⁡
(
e
1
z
1
)
=
(
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
)
×
(
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
.
Since (22) has to be equal to (21), sin
(
x
1
)
sin
(
x
2
)
=
0, that is, sin
(
x
1
)
=
0 or sin
(
x
2
)
=
0. Therefore, x
1
=
k
π or x
2
=
t
π, and then cos
⁡
(
x
1
)
=
(
-
1
)
k or cos
⁡
(
x
2
)
=
(
-
1
)
t, where k
,
t
∈
ℤ. If x
1
=
k
π  (
k
∈
ℤ
), then
(23)
exp
⁡
(
z
)
=
(
cos
⁡
(
x
2
)
+
e
2
sin
(
x
2
)
)
×
(
-
1
)
k
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
=
(
-
1
)
k
exp
⁡
(
e
2
x
2
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
.
Similarly, if x
2
=
t
π  (
t
∈
ℤ
), then
(24)
exp
⁡
(
z
)
=
(
cos
⁡
(
x
1
)
+
e
1
sin
(
x
1
)
)
×
(
-
1
)
t
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
=
(
-
1
)
t
exp
⁡
(
e
1
x
1
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
.Remark 2.
By Theorem1 and the properties of the Euler formula, if x
1
=
k
π, then we can write
(25)
exp
⁡
(
z
)
=
(
-
1
)
k
exp
⁡
(
e
2
x
2
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
=
(
-
1
)
k
exp
⁡
(
e
2
x
2
-
x
0
)
=
(
-
1
)
k
exp
⁡
(
e
2
F
2
¯
)
,
also, if x
2
=
t
π, then
(26)
exp
⁡
(
z
)
=
(
-
1
)
t
exp
⁡
(
e
1
x
1
)
(
cosh
⁡
(
x
0
)
-
sinh
⁡
(
x
0
)
)
=
(
-
1
)
t
exp
⁡
(
e
1
x
1
-
x
0
)
=
(
-
1
)
t
exp
⁡
(
e
1
F
1
¯
)
,
where k
,
t
∈
ℤ and F
r
¯
=
x
r
+
e
r
x
0
(
r
=
1,2
) are the conjugate Fueter variables of F
r
=
x
r
-
e
r
x
0
(
r
=
1,2
) (see [20]).LetΩ be an open subset of ℝ
3 and let a function f
(
a
) be defined by the following form on Ω with values in ℂ
(
𝕋
):
(27)
f
:
Ω
⟶
ℂ
(
𝕋
)
,
satisfying
(28)
a
=
(
x
0
,
x
1
,
x
2
)
∈
Ω
⟼
f
(
a
)
=
e
1
f
1
(
x
0
,
x
1
,
x
2
)
+
e
2
f
2
(
x
0
,
x
1
,
x
2
)
∈
ℂ
(
𝕋
)
,
where f
r
=
u
r
-
(
1
/
2
)
e
r
u
0, f
r
¯
=
u
r
+
(
1
/
2
)
e
r
u
0
(
r
=
1,2
) and u
m
(
m
=
0,1
,
2
) are real-valued functions.From the chain rule, we use the following differential operators:(29)
∂
∂
A
:
=
2
∂
∂
x
0
-
1
2
e
1
∂
∂
x
1
-
1
2
e
2
∂
∂
x
2
=
-
e
1
∂
∂
z
1
-
e
2
∂
∂
z
2
,
∂
∂
A
*
=
2
∂
∂
x
0
+
1
2
e
1
∂
∂
x
1
+
1
2
e
2
∂
∂
x
2
=
e
1
∂
∂
z
1
¯
+
e
2
∂
∂
z
2
¯
,
where
(30)
∂
∂
z
1
=
1
2
∂
∂
x
1
+
e
1
∂
∂
x
0
,
∂
∂
z
2
=
1
2
∂
∂
x
2
+
e
2
∂
∂
x
0
,
∂
∂
z
1
-
=
1
2
∂
∂
x
1
-
e
1
∂
∂
x
0
,
∂
∂
z
2
-
=
1
2
∂
∂
x
2
-
e
2
∂
∂
x
0
in ℂ
(
𝕋
). We have the following equations:
(31)
f
r
∂
∂
z
r
=
∂
f
r
∂
z
r
,
f
r
∂
∂
z
r
¯
=
∂
f
r
∂
z
r
¯
(
r
=
1,2
)
,
f
1
¯
∂
∂
z
2
=
(
∂
f
1
∂
z
2
¯
)
¯
,
f
2
¯
∂
∂
z
1
=
(
∂
f
2
∂
z
1
¯
)
¯
,
f
1
¯
∂
∂
z
2
¯
=
(
∂
f
1
∂
z
2
)
¯
,
f
2
¯
∂
∂
z
1
¯
=
(
∂
f
2
∂
z
1
)
¯
,
and then, the operator ∂
/
∂
A operates to f as follows:
(32)
∂
f
∂
A
=
(
-
e
1
∂
∂
z
1
-
e
2
∂
∂
z
2
)
(
e
1
f
1
+
e
2
f
2
)
=
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
+
e
1
e
2
(
∂
f
1
∂
z
2
¯
-
∂
f
2
∂
z
1
¯
)
,
∂
f
∂
A
*
=
(
e
1
∂
∂
z
1
¯
+
e
2
∂
∂
z
2
¯
)
(
e
1
f
1
+
e
2
f
2
)
=
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
+
e
1
e
2
(
∂
f
2
∂
z
1
-
∂
f
1
∂
z
2
)
,
f
∂
∂
A
=
(
e
1
f
1
+
e
2
f
2
)
(
-
e
1
∂
∂
z
1
-
e
2
∂
∂
z
2
)
=
f
1
∂
∂
z
1
+
f
2
∂
∂
z
2
+
e
1
e
2
(
f
2
¯
∂
∂
z
1
-
f
1
¯
∂
∂
z
2
)
=
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
+
e
1
e
2
{
(
∂
f
2
∂
z
1
¯
)
¯
-
(
∂
f
1
∂
z
2
¯
)
¯
}
,
f
∂
∂
A
*
=
(
e
1
f
1
+
e
2
f
2
)
(
e
1
∂
∂
z
1
¯
+
e
2
∂
∂
z
2
¯
)
=
-
f
1
∂
∂
z
1
¯
-
f
2
∂
∂
z
2
¯
+
e
1
e
2
(
f
1
¯
∂
∂
z
2
¯
-
f
2
¯
∂
∂
z
1
¯
)
=
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
+
e
1
e
2
{
(
∂
f
1
∂
z
2
)
¯
-
(
∂
f
2
∂
z
1
)
¯
}
.
Thus, we have a corresponding Laplacian in the reduced quaternion ℂ
(
𝕋
):
(33)
Δ
a
=
∂
2
∂
A
∂
A
*
=
∂
2
∂
A
*
∂
A
=
4
∂
2
∂
x
0
2
+
1
4
∂
2
∂
x
1
2
+
1
4
∂
2
∂
x
2
2
.Remark 3.
Let Ω be an open set ofℝ
3. From the definition of the differential operators in ℂ
(
𝕋
), we have
(34)
∂
∂
A
•
f
=
1
2
(
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
+
f
1
∂
∂
z
1
+
f
2
∂
∂
z
2
)
+
1
2
e
1
e
2
(
f
2
¯
∂
∂
z
1
-
∂
f
2
∂
z
1
¯
+
∂
f
1
∂
z
2
¯
-
f
1
¯
∂
∂
z
2
)
=
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
+
1
2
e
1
e
2
×
{
(
∂
f
2
∂
z
1
¯
)
¯
-
(
∂
f
1
∂
z
2
¯
)
¯
-
∂
f
2
∂
z
1
¯
+
∂
f
1
∂
z
2
¯
}
,
∂
∂
A
⊙
f
=
1
2
(
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
-
f
1
∂
∂
z
1
-
f
2
∂
∂
z
2
)
+
1
2
e
1
e
2
(
f
1
¯
∂
∂
z
2
+
∂
f
1
∂
z
2
¯
-
∂
f
2
∂
z
1
¯
-
f
2
¯
∂
∂
z
1
)
=
1
2
e
1
e
2
{
(
∂
f
1
∂
z
2
¯
)
¯
+
∂
f
1
∂
z
2
¯
-
∂
f
2
∂
z
1
¯
-
(
∂
f
2
∂
z
1
¯
)
¯
}
,
∂
∂
A
*
•
f
=
-
1
2
(
∂
f
1
∂
z
1
¯
+
∂
f
2
∂
z
2
¯
+
f
1
∂
∂
z
1
¯
+
f
2
∂
∂
z
2
¯
)
+
1
2
e
1
e
2
(
∂
f
2
∂
z
1
-
f
2
¯
∂
∂
z
1
¯
+
f
1
¯
∂
∂
z
2
¯
-
∂
f
1
∂
z
2
)
=
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
+
1
2
e
1
e
2
×
{
∂
f
2
∂
z
1
-
(
∂
f
2
∂
z
1
)
¯
+
(
∂
f
1
∂
z
2
)
¯
-
∂
f
1
∂
z
2
}
,
∂
∂
A
*
⊙
f
=
1
2
(
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
+
f
1
∂
∂
z
1
¯
+
f
2
∂
∂
z
2
¯
)
+
1
2
e
1
e
2
(
∂
f
2
∂
z
1
+
f
2
¯
∂
∂
z
1
¯
-
f
1
¯
∂
∂
z
2
¯
-
∂
f
1
∂
z
2
)
=
1
2
e
1
e
2
{
∂
f
2
∂
z
1
+
(
∂
f
2
∂
z
1
)
¯
-
(
∂
f
1
∂
z
2
)
¯
-
∂
f
1
∂
z
2
}
and, therefore,
(35)
∂
f
∂
A
=
∂
∂
A
•
f
+
∂
∂
A
⊙
f
,
∂
f
∂
A
*
=
∂
∂
A
*
•
f
+
∂
∂
A
*
⊙
f
.
Similarly, we have
(36)
f
•
∂
∂
A
=
1
2
(
f
1
∂
∂
z
1
+
f
2
∂
∂
z
2
+
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
)
+
e
1
e
2
(
f
1
¯
∂
∂
z
2
-
f
2
¯
∂
∂
z
1
+
∂
f
2
∂
z
1
¯
-
∂
f
1
∂
z
2
¯
)
=
∂
f
1
∂
z
1
+
∂
f
2
∂
z
2
+
1
2
e
1
e
2
×
{
(
∂
f
1
∂
z
2
¯
)
¯
-
(
∂
f
2
∂
z
1
¯
)
¯
+
∂
f
2
∂
z
1
¯
-
∂
f
1
∂
z
2
¯
}
,
f
⊙
∂
∂
A
=
1
2
(
f
1
∂
∂
z
1
+
f
2
∂
∂
z
2
-
∂
f
1
∂
z
1
-
∂
f
2
∂
z
2
)
+
1
2
e
1
e
2
(
f
1
¯
∂
∂
z
2
-
f
2
¯
∂
∂
z
1
-
∂
f
2
∂
z
1
¯
+
∂
f
1
∂
z
2
¯
)
=
1
2
e
1
e
2
{
(
∂
f
1
∂
z
2
¯
)
¯
-
(
∂
f
2
∂
z
1
¯
)
¯
-
∂
f
2
∂
z
1
¯
+
∂
f
1
∂
z
2
¯
}
,
f
•
∂
∂
A
*
=
1
2
(
-
f
1
∂
∂
z
1
¯
-
f
2
∂
∂
z
2
¯
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
)
+
1
2
e
1
e
2
(
f
1
¯
∂
∂
z
2
¯
-
f
2
¯
∂
∂
z
1
¯
+
∂
f
2
∂
z
1
-
∂
f
1
∂
z
2
)
=
-
∂
f
1
∂
z
1
¯
-
∂
f
2
∂
z
2
¯
+
1
2
e
1
e
2
×
{
(
∂
f
1
∂
z
2
)
¯
-
(
∂
f
2
∂
z
1
)
¯
+
∂
f
2
∂
z
1
-
∂
f
1
∂
z
2
}
,
f
⊙
∂
∂
A
*
=
1
2
(
-
f
1
∂
∂
z
1
¯
-
f
2
∂
∂
z
2
¯
+
∂
f
1
∂
z
1
¯
+
∂
f
2
∂
z
2
¯
)
+
1
2
e
1
e
2
(
f
1
¯
∂
∂
z
2
¯
-
f
2
¯
∂
∂
z
1
¯
-
∂
f
2
∂
z
1
+
∂
f
1
∂
z
2
)
=
1
2
e
1
e
2
{
(
∂
f
1
∂
z
2
)
¯
-
(
∂
f
2
∂
z
1
)
¯
-
∂
f
2
∂
z
1
+
∂
f
1
∂
z
2
}
and, therefore,
(37)
f
∂
∂
A
=
f
•
∂
∂
A
+
f
⊙
∂
∂
A
,
f
∂
∂
A
*
=
f
•
∂
∂
A
*
+
f
⊙
∂
∂
A
*
.Definition 4.
LetΩ be an open set in ℝ
3 and for any element a in Ω. A function f
(
a
) is said to beL(R)-regular on Ω if the following conditions are satisfied:(i)
f
r
(
r
=
1,2
) are continuously differential functions on Ω, and
(ii)
∂
f
(
a
)
/
∂
A
*
=
0
(
f
(
a
)
(
∂
/
∂
A
*
)
=
0
) on Ω.In particular, the equation∂
f
/
∂
A
*
=
0 of Definition 4 is equivalent to
(38)
∂
∂
A
*
•
f
(
a
)
=
-
∂
∂
A
*
⊙
f
(
a
)
.
Moreover, (38) is equivalent to the following system:
(39)
∂
f
1
∂
z
1
¯
=
-
∂
f
2
∂
z
2
¯
,
∂
f
1
∂
z
2
=
∂
f
2
∂
z
1
.
The above system is a corresponding Cauchy-Riemann system in ℂ
(
𝕋
).Remark 5.
From the multiplications ofℂ
(
𝕋
), the equation f
(
∂
/
∂
A
*
)
=
0 of Definition 4 is equivalent to
(40)
∂
∂
A
*
•
f
(
a
)
=
∂
∂
A
*
⊙
f
(
a
)
.
Also, the above equation (40) is equivalent to the following system:
(41)
∂
f
1
∂
z
1
¯
=
-
∂
f
2
∂
z
2
¯
,
f
1
¯
∂
∂
z
2
¯
=
f
2
¯
∂
∂
z
1
¯
⟺
(
∂
f
1
∂
z
2
)
¯
=
(
∂
f
2
∂
z
1
)
¯
.
Further, the above system (41) is also a corresponding Cauchy-Riemann system in ℂ
(
𝕋
). Since the system (39) is equivalent to the system (41), we say that f
(
a
) of Definition 4 is a regular function on Ω
⊂
ℝ
3. When the function f
(
a
) is either anL-regular function or anR-regular function on Ω
⊂
ℝ
3, we simply say that f
(
a
) is a regular function on Ω
⊂
ℝ
3.


## 3.  Properties of Regular Functions with Values inℂ
(
𝕋
)

We define the derivativef
′
(
a
) of f
(
a
) by the following:
(42)
f
′
(
a
)
:
=
∂
f
(
a
)
∂
A
.Proposition 6.
LetΩ be an open set in ℝ
3 and let a function f
(
a
) be a regular function defined on Ω. Then
(43)
f
′
(
a
)
=
-
2
e
r
(
∂
f
∂
z
r
-
∂
f
∂
z
r
¯
)
=
4
∂
f
∂
x
0
=
-
e
1
∂
f
∂
x
1
-
e
2
∂
f
∂
x
2
(
r
=
1,2
)
.Proof.
From the definition of a regular function(
:
(
∂
f
/
∂
A
*
)
=
0
), we have
(44)
∂
f
1
∂
z
1
¯
=
-
∂
f
2
∂
z
2
¯
,
∂
f
1
∂
z
2
=
∂
f
2
∂
z
1
.
Therefore,
(45)
∂
∂
z
•
f
=
∂
f
1
∂
z
1
¯
+
2
e
1
∂
u
1
∂
x
0
+
∂
u
0
∂
x
0
+
∂
f
2
∂
z
2
¯
+
2
e
2
∂
u
2
∂
x
0
+
∂
u
0
∂
x
0
+
1
2
e
1
e
2
(
f
2
¯
∂
∂
z
1
¯
+
2
e
1
∂
u
2
∂
x
0
+
e
2
e
1
∂
u
0
∂
x
0
-
∂
f
2
∂
z
1
)
+
1
2
e
1
e
2
(
2
e
1
∂
u
2
∂
x
0
-
e
1
e
2
∂
u
0
∂
x
0
+
∂
f
1
∂
z
2
-
2
e
2
∂
u
1
∂
x
0
+
e
2
e
1
∂
u
0
∂
x
0
-
f
1
¯
∂
∂
z
2
¯
-
2
e
2
∂
u
1
∂
x
0
-
e
1
e
2
∂
u
0
∂
x
0
)
=
4
∂
f
∂
x
0
+
1
2
e
1
e
2
(
∂
f
2
∂
z
1
)
¯
-
1
2
e
1
e
2
(
∂
f
1
∂
z
2
)
¯
,
∂
∂
z
⊙
f
=
1
2
e
1
e
2
×
(
f
1
¯
∂
∂
z
2
¯
+
2
e
2
∂
f
1
∂
x
0
+
∂
f
1
∂
z
2
-
2
e
2
∂
f
1
∂
x
0
-
∂
f
2
∂
z
1
+
2
e
1
∂
f
2
∂
x
0
-
f
2
¯
∂
∂
z
1
¯
-
2
e
1
∂
f
2
∂
x
0
)
=
2
e
1
e
2
(
e
1
∂
f
1
∂
x
0
-
e
1
∂
f
1
∂
x
0
-
e
2
∂
f
2
∂
x
0
+
e
2
∂
f
2
∂
x
0
)
-
1
2
e
1
e
2
(
∂
f
2
∂
z
1
)
¯
+
1
2
e
1
e
2
(
∂
f
1
∂
z
2
)
¯
=
0
.
Hence, we obtain the equation
(46)
∂
f
∂
z
=
∂
∂
z
•
f
+
∂
∂
z
⊙
f
=
4
∂
f
∂
x
0
.
Similarly, by calculating the derivative f
′
(
z
) of f
(
z
),
(47)
∂
∂
z
•
f
=
∂
f
1
∂
x
1
+
∂
f
2
∂
x
2
+
1
2
e
1
e
2
{
(
∂
f
2
∂
z
1
)
¯
-
(
∂
f
1
∂
z
2
)
¯
+
∂
f
1
∂
x
2
-
∂
f
2
∂
x
1
}
,
∂
∂
z
⊙
f
=
(
e
1
∂
f
2
∂
x
1
-
e
1
∂
f
2
∂
x
1
-
e
2
∂
f
1
∂
x
2
+
e
2
∂
f
1
∂
x
2
)
-
1
2
e
1
e
2
(
∂
f
2
∂
z
1
)
¯
+
1
2
e
1
e
2
(
∂
f
1
∂
z
2
)
¯
.
Therefore, we have the equation
(48)
∂
f
∂
z
=
∂
∂
z
•
f
+
∂
∂
z
⊙
f
=
-
e
1
(
e
1
∂
f
1
∂
x
1
+
e
2
∂
f
2
∂
x
1
)
-
e
2
(
e
2
∂
f
2
∂
x
2
+
e
1
∂
f
1
∂
x
2
)
=
-
e
1
∂
f
∂
x
1
-
e
2
∂
f
∂
x
2
.
Further, using the same procedure, we obtain the equations
(49)
∂
f
∂
z
=
-
2
e
r
(
∂
f
∂
z
r
-
∂
f
∂
z
r
¯
)
(
r
=
1,2
)
.Proposition 7.
LetΩ be an open set in ℝ
3. If f
(
a
) is a regular function on Ω, then we have
(50)
∂
n
f
∂
A
n
=
4
n
∂
n
f
∂
x
0
n
,
where n is a positive integer.Proof.
Sincef is a regular function on Ω with values in ℂ
(
𝕋
), by Definition 4,
(51)
∂
∂
A
*
(
4
∂
f
∂
x
0
)
=
4
∂
∂
x
0
(
∂
f
∂
A
*
)
=
0
.
Hence, ∂
f
/
∂
x
0 is a regular function with values in ℂ
(
𝕋
). From Proposition 6, we have
(52)
∂
2
f
∂
A
2
=
∂
∂
A
(
∂
f
∂
A
)
=
4
2
∂
2
f
∂
x
0
2
.
By repeating the above process, we can obtain the equation
(53)
∂
n
f
∂
A
n
=
4
n
∂
n
f
∂
x
0
n
.We let(54)
□
a
=
∑
r
=
1
2
‍
∂
2
∂
z
r
∂
z
r
¯
=
2
∂
2
∂
x
0
2
+
1
4
e
1
∂
2
∂
x
1
2
+
1
4
e
2
∂
2
∂
x
2
2
on an open set Ω in ℝ
3.Theorem 8.
LetΩ be an open set in ℝ
3. If f is a regular function on Ω, then the following equation holds true:
(55)
□
a
f
(
a
)
=
-
1
8
∂
2
f
(
a
)
∂
A
2
.Proof.
Sincef is a regular function on Ω, we have the following system:
(56)
4
∂
u
0
∂
x
0
=
∂
u
1
∂
x
1
+
∂
u
2
∂
x
2
,
∂
u
2
∂
x
1
=
∂
u
1
∂
x
2
,
4
∂
u
r
∂
x
0
=
-
∂
u
0
∂
x
r
(
r
=
1,2
)
.
By the definition of □
a, we have
(57)
□
a
f
=
(
2
∂
2
∂
x
0
2
+
1
4
e
1
∂
2
∂
x
1
2
+
1
4
e
2
∂
2
∂
x
2
2
)
(
u
0
+
e
1
u
1
+
e
2
u
2
)
=
2
∂
2
u
0
∂
x
0
2
+
2
e
1
∂
2
u
1
∂
x
0
2
+
2
e
2
∂
2
u
2
∂
x
0
2
-
∂
2
u
1
∂
x
0
∂
x
1
+
e
1
∂
2
u
0
∂
x
0
∂
x
1
-
∂
2
u
2
∂
x
0
∂
x
2
+
e
2
∂
2
u
0
∂
x
0
∂
x
2
=
-
2
∂
2
u
0
∂
x
0
2
-
2
e
1
∂
2
u
1
∂
x
0
2
-
2
e
2
∂
2
u
2
∂
x
0
2
=
-
2
∂
2
f
∂
x
0
2
.
From Proposition 7, we have ∂
2
f
/
∂
A
2
=
4
2
(
∂
2
f
/
∂
x
0
2
). Hence, by calculating and comparing the above polynomials, we obtain that □
a
f is equal to -
(
1
/
8
)
(
∂
2
/
∂
A
2
)
f.Next, we consider a differential form(58)
ω
=
4
d
x
1
∧
d
x
2
-
e
1
d
x
0
∧
d
x
2
+
e
2
d
x
0
∧
d
x
1
.Theorem 9.
LetΩ be an open set in ℝ
3 and let U be any domain on Ω with a smooth distinguished boundary b
U such that U
⊂
Ω. If f is a regular function on Ω, then one has
(59)
∫
b
U
‍
ω
f
=
0
,
where ω
f is the reduced quaternionic product of the form ω on the function f
(
a
).Proof.
Sinceω
f
=
4
f
d
x
1
∧
d
x
2
-
e
1
f
d
x
0
∧
d
x
2
+
e
2
f
d
x
0
∧
d
x
1, we have
(60)
d
(
ω
f
)
=
4
∂
f
∂
x
0
d
x
0
∧
d
x
1
∧
d
x
2
+
e
1
∂
f
∂
x
1
d
x
0
∧
d
x
1
∧
d
x
2
+
e
2
∂
f
∂
x
2
d
x
0
∧
d
x
1
∧
d
x
2
=
4
∂
(
e
1
f
1
+
e
2
f
2
)
∂
x
0
d
I
+
e
1
∂
(
e
1
f
1
+
e
2
f
2
)
∂
x
1
d
I
+
e
2
∂
(
e
1
f
1
+
e
2
f
2
)
∂
x
2
d
I
=
{
(
4
∂
u
0
∂
x
0
-
∂
u
1
∂
x
1
-
∂
u
2
∂
x
2
)
+
e
1
(
4
∂
u
1
∂
x
0
+
∂
u
0
∂
x
1
)
+
e
2
(
4
∂
u
2
∂
x
0
+
∂
u
0
∂
x
2
)
+
e
1
e
2
(
∂
u
2
∂
x
1
-
∂
u
1
∂
x
2
)
}
d
I
,
where d
I
=
d
x
0
∧
d
x
1
∧
d
x
2 in U. From the corresponding Cauchy-Riemann system (39) for f
(
a
) in ℂ
(
𝕋
), we have the system (56). Hence, d
(
ω
f
)
=
0 and, therefore, by Stokes theorem, we obtain the following result:
(61)
∫
b
U
‍
ω
f
=
∫
U
‍
d
(
ω
f
)
=
0
.



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*Source: 654798-2014-03-20.xml*