# On Hyperbolic Complex Numbers
## 2 The Basic Hyperbolic Complex Algebraic Structure
### Geometry  
Let us call  
\[C_{h}(r)=\{(x,y)^{T}\in\mathbb{R}^{2}:x^{2}-y^{2}=r^{2}\}\]a hyperbolic circle of hyperbolic radius \(r>0\) and  
\[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]  
its \(\mathfrak{X}\)-sector in short, see Figure 1.  
If \(z_{1}\in C_{h}^{+}(r)\) and \(z_{2}\in C_{h}^{+}(1)\), then \(z_{1}\odot_{h}z_{2}\in C_{h}^{+}(r)\). The Lie group on \(C_{h}^{+}(r)\) is thus generated by the multiplications with the elements from the unit half-circle \(C_{h}^{+}(1)\). In other words, if the function \(Q_{h}:\mathbb{R}^{2}\to\mathbb{R}\) is defined by \(Q_{h}(x,y)=x^{2}-y^{2}\), then multiplication of \(z_{1}=(x,y)^{T}\) by any \(z_{2}\in C_{h}^{+}(1)\) leaves \(Q_{h}\) invariant. Definition 5: The hyperbolic coordinate transformation \(Hyp_{\mathfrak{X}(\mathfrak{Y})}:\mathbb{R}^{+}\times\mathbb{R}\to\mathfrak{X }(\mathfrak{Y})\), \((x,y)^{T}=Hyp_{\mathfrak{X}(\mathfrak{Y})}(r,\psi)\), is accordingly defined as  
\[x=r\cosh\psi,y=r\sinh\psi\text{ or }x=r\sinh\psi,y=r\cosh\psi. \tag{7}\]  
Here, the notation \(\mathfrak{X}(\mathfrak{Y})\) indicates whether the signed area content \(\psi\) is measured counterclockwise between \((1,0)^{T}\) and \((x,y)^{T}\in\mathfrak{X}\) or clockwise between \((0,1)^{T}\) and \((x,y)^{T}\in\mathfrak{Y}\), see Figure 2.  
The union of all \(\mathfrak{X}\)-sectors \(C_{h}^{+}(r),r>0\) represents a disjoint decomposition of the cone \(\mathfrak{X}\), see Figure 2.  
The transformation \(Hyp_{\mathfrak{X}}\) therefore has an a.e. uniquely determined inverse \(Hyp_{\mathfrak{X}}^{-1}:\mathfrak{X}\to\mathbb{R}^{+}\times\mathbb{R}\), which is given by \(Hyp^{-1}(x,y)=(r,\psi)\) with  
\[r=\sqrt{x^{2}-y^{2}},\psi=\ln\sqrt{\frac{x+y}{x-y}}. \tag{8}\]  
Figure 1: Hyperbolic half–circles in the event space \(\mathfrak{X}\). **Remark 1**. : _With \(sign(\psi)=sign(y)\), the quantity \(\psi\cdot r^{2}\) equals two times the signed area content of the hyperbolic sector \(S(\mathsf{o},P,A)\)\(=r\bullet S(\mathsf{o},u,\epsilon)\) the boundary of which is marked with the help of the points \(\mathsf{o}=(0,0)^{T},P=(x,y)^{T}\in C_{\mathsf{H}}^{+}(r)\) and \(A=(r,0)^{T}\), see Figure 3._  
For an elementary geometric proof of this remark, we refer to Appendix A. **Definition 6**. : _The geometrical hyperbolic vector product of \(z_{1}=Hypx(\mathfrak{P})(r_{1},\psi_{1})\) and \(z_{2}=Hypx(\mathfrak{P})(r_{2},\psi_{2})\) being both either from \(\mathfrak{X}\) or from \(\mathfrak{Y}\) is defined as_  
\[Hypx(\mathfrak{P})(r_{1},\psi_{1})\ast_{h}Hypx(\mathfrak{P})(r_{2},\psi_{2} )=Hypx(\mathfrak{P})(r_{1}r_{2},\psi_{1}+\psi_{2}).\]  
\[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]0 \(\psi\) equals two times the signed area content of the hyperbolic sector \(S(\mathsf{o},u,\epsilon)\) with \(u=\frac{z}{r}\) and where \(z=(x,y)^{T}=Hypx(r,\psi)\). \[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]1 Signed area contents (**left**) and disjoint cone decomposition (**right**). The expression on the right side of the last equation is always uniquely defined because due to Remark 1 the sum \(\psi_{1}+\psi_{2}\) represents the finite content of a bounded hyperbolic sector (triangle), which itself is a subset of an unbounded set having infinite area content, see \[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]2  
Let \(0<t_{1}<t_{2},\varrho>0,P_{1}=Hypx(\varrho,t_{1}),P_{2}=Hypx(\varrho,t_{2})\) and let  
\[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]3  
denote a hyperbolic sector (or hyperbolic triangle) spanned by the points \(\omega,P_{1},P_{2}\), see \[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]4 Additionally, let \(T_{\psi},\psi>0\) be a transformation defined by  
\[C_{h}^{+}(r)=C_{h}(r)\cap\{(x,y)^{T}\in\mathbb{R}^{2}:x>0\}\]5  
Then,  
\[T_{\psi}S(\omega,Hypx(\varrho,t_{1}),Hypx(\varrho,t_{2})) =\{r\bigg{(} \begin{array}{c}\cosh\psi\\ \sinh\psi\end{array} \bigg{)}*_{h}\bigg{(} \begin{array}{c}\cosh t\\ \sinh t\end{array} \bigg{)},0<r<\varrho,t_{1}<t<t_{2}\}\] \[=\{r\bigg{(}