Preliminaries

Here, we introduce some of the basic results of the theory of bicomplex numbers. First, we define bicomplex numbers as follows:
$\mathbb {C}_{2}:=\{ a+bi_{1}+ci_{2}+dj:{i_1}^2={i_2}^2=-1, j^2=1\mbox{ and }i_2j=ji_2=-i_1,i_1j=ji_1=-i_2,i_2i_1=i_1i_2=j\}$ where $a,b,c,d\in\mathbb {R}$. The norm used on $\mathbb {C}_{2}$ is the Euclidean norm (also noted $\mid\mbox{ }\mid$) of $\mathbb {R}^{4}$. We remark that we can write a bicomplex number $a+bi_1+ci_2+dj$ as $(a+bi_1)+(c+di_1)i_2=z_1+z_2i_2$ where $z_1,z_2\in\mathbb {C}_{1}:=\{x+yi_1:{i_1}^{2}=-1\}$. Thus, $\mathbb {C}_{2}$ can be veiwed as the complexification of the usual complex numbers $\mathbb {C}_{1}$ and a bicomplex number can be seen as an element of $\mathbb {C}^{2}$. Moreover, the norm of the bicomplex number is the same as the norm of the associated element $(z_1,z_2)$ of $\mathbb {C}^{2}$. It is easy to see [8] that $\mathbb {C}_{2}$ is a commutative unitary ring with the following characterization for the noninvertible elements:

Proposition 1   Let $w=a+bi_1+ci_2+dj\in\mathbb {C}_{2}$. Then $w$ is noninvertible iff $(a=-d$ and $b=c)$ or $(a=d$ and $b=-c)$ iff ${z_1}^2+{z_2}^2=0$.

It is also important to know that every bicomplex number $z_1+z_2i_2$ has the following unique idempotent representation:

$$z_1+z_2i_2=(z_1-z_2i_1)e_1+(z_1+z_2i_1)e_2$$

where $e_1=\frac{1+j}{2}$ and $e_2=\frac{1-j}{2}$. This representation is very useful because: addition, multiplication and division can be done term-by-term. Also, an element will be noninvertible iff $z_1-z_2i_1=0$ or $z_1+z_2i_1=0$. The next definition will be useful to construct a natural ``disc" in $\mathbb {C}_{2}$.

Definition 1   We say that $X\subseteq\mathbb {C}_{2}$ is a $\mathbb {C}_{2}$-cartesian set determined by $X_1$ and $X_2$ if $X=X_{1}\times_e X_{2}:=\{z_1+z_2i_2\in\mathbb {C}_{2}:z_1+z_2i_2=w_1e_1+w_2e_2, (w_1,w_2)\in X_1\times X_2\}$.

In [8] it is shown that if $X_1$ and $X_2$ are domains of $\mathbb {C}_{1}$ then $X_1\times_e X_2$ is also a domain of $\mathbb {C}_{2}$. Then, a manner to construct a natural ``disc" in $\mathbb {C}_{2}$ is to take the $\mathbb {C}_{2}$-cartesian product of two discs in $\mathbb {C}_{1}$. Hence, we define the natural ``disc" of $\mathbb {C}_{2}$ as follows [8]: $D(0,r):=B^{1}(0,r)\times_{e} B^{1}(0,r)= \{z_1+z_2i_2:z_1+z_2i_2=w_1e_1+w_2e_2,\vert w_1\vert<r,\vert w_2\vert<r\}$ where $B^{n}(0,r)$ is the open ball of $\mathbb {C}^{n}_{1}\backsimeq\mathbb {C}^{n}$ with radius $r$.