Classical complex algebra uses the imaginary unit \(i\), satisfying the condition \(i^2=-1\). Hyperbolic algebra introduces another unit: \(\j\), for which \(\j^2=+1\). Considering these two units together leads to a four-dimensional commutative algebra, and then to a natural idempotent decomposition of space into two independent complex planes.

This decomposition is of interest not only from an algebraic point of view. It allows us to separate an object into two mutually orthogonal components, each of which preserves the structure of the ordinary complex plane. Thanks to this, many expressions containing the hyperbolic unit acquire a significantly simpler form, and the operations of multiplication, exponentiation, logarithmization, and the calculation of elementary functions are decomposed into two independent, complex problems.

In previous works, the idempotent basis was introduced as a convenient representation of hyperbolic algebra and was used in constructing geometric models of wave electricity. However, the question of its origin remained open: whether it is simply a fortunate choice of coordinate system or arises as an inevitable consequence of the very structure of the hyperbolic unit.

The goal of this work is to show that the idempotent basis \[ \tag{1} \left\{ \ep ,\; i\ep ,\; \em ,\; i\em \right\} \] is not an arbitrarily introduced coordinate system. It arises directly from the structure of the hyperbolic unit, and its real and complex directions can be expressed in terms of the logarithms \(\j\) and \(-\j\). Thus, the logarithmic properties of hyperbolic algebra lead to the natural construction of an idempotent basis.

This basis is actively used in the theory of wave electricity, where it allows for the independent description of two complex components of the objects under study. This work demonstrates that this representation has not only computational advantages but also an independent algebraic origin.

Consider two commuting units: \[ \tag{2} i^2=-1, \qquad \j^2=1, \qquad i\j=\j i. \] The common element of the algebra generated by them is of the form \[ \tag{3} Z=x_0+i x_1+\j x_2+i\j x_3, \qquad x_k\in\mathbb{R}. \] Therefore, the natural intermediate basis is \[ \tag{4} \left\{ 1,\;i,\;\j,\;i\j \right\}. \]

This basis demonstrates the four-dimensionality of the algebra, but does not yet reveal its internal structure. For this, we must use the condition \(\j^2=1\).

From the condition \[ \tag{5} \j^2=1 \] it follows \[ \tag{6} (1+\j)(1-\j)=0. \] This equality demonstrates the existence of two mutually orthogonal directions whose product is zero. We normalize them and introduce \[ \tag{7} \ep = \frac{1+\j}{2}, \qquad \em = \frac{1-\j}{2}. \]

Let's check their properties: \[ \tag{8} \ep^2 = \left( \frac{1+\j}{2} \right)^2 = \frac{1+2\j+\j^2}{4} = \frac{1+\j}{2} = \ep , \] \[ \tag{9} \em ^{\,2} = \left( \frac{1-\j}{2} \right)^2 = \em \] \[ \tag{10} \ep\em = \frac{(1+\j)(1-\j)}{4} = \frac{1-\j^2}{4} = 0. \]

Furthermore, \[ \tag{11} \ep +\em =1, \qquad \ep -\em =\j. \] Thus, the hyperbolic unit decomposes the unit of the algebra into two mutually orthogonal idempotents.

The relationship between the bases \(\{1,\j\}\) and \(\{\ep ,\em \}\) is defined by the transformation \[ \tag{12} \begin{pmatrix} \ep \\ \em \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1\\ \j \end{pmatrix}. \] The inverse transformation is \[ \tag{13} \begin{pmatrix} 1\\ \j \end{pmatrix} = \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} \ep \\ \em \end{pmatrix}. \]

Multiplying idempotents by the complex unit, we obtain \[ \tag{14} i\ep = \frac{i+i\j}{2}, \qquad i\em = \frac{i-i\j}{2}. \] Therefore, the intermediate basis \[ \left\{1,i,\j,i\j\right\} \] translates into an idempotent i-basis \[ \tag{15} \boxed{ \left\{ \ep ,\; i\ep ,\; \em ,\; i\em \right\}. } \]

The complete basis transformation can be written in matrix form: \[ \tag{16} \begin{pmatrix} \ep \\ i\ep \\ \em \\ i\em \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1&0&1&0\\ 0&1&0&1\\ 1&0&-1&0\\ 0&1&0&-1 \end{pmatrix} \begin{pmatrix} 1\\ i\\ \j\\ i\j \end{pmatrix}. \]

Let's write the general element of the algebra in the form \[ \tag{17} Z=z_1+\j z_2, \qquad z_1,z_2\in\mathbb{C}. \] Using the equalities \[ 1=\ep +\em , \qquad \j=\ep -\em , \] we obtain \[ \tag{18} Z = z_1(\ep +\em) + z_2(\ep -\em). \] After grouping \[ \tag{19} Z = \ep (z_1+z_2) + \em (z_1-z_2). \]

Denoting \[ \tag{20} A=z_1+z_2, \qquad B=z_1-z_2, \] we obtain the basic idempotent form \[ \tag{21} \boxed{ Z=\ep A+\em B, \qquad A,B\in\mathbb{C}. } \]

Since \[ \ep \em =0, \] the product of two elements \[ Z_1=\ep A_1+\em B_1, \qquad Z_2=\ep A_2+\em B_2 \] is equal to \[ \tag{22} Z_1Z_2 = \ep A_1A_2 + \em B_1B_2. \] Mixed products vanish. Therefore, the algebra decomposes into the direct sum of two independent complex planes: \[ \tag{23} \boxed{ \mathbb{C}\ep \oplus \mathbb{C}\em . } \]

For any idempotent \(P^2=P\), we have \[ \tag{24} e^{zP} = 1+P \sum_{n=1}^{\infty} \frac{z^n}{n!}. \] Since \[ \sum_{n=1}^{\infty} \frac{z^n}{n!} = e^z-1, \] we obtain \[ \tag{25} \boxed{ e^{zP}=1-P+Pe^z. } \] This equality allows us to directly relate the hyperbolic unit to the complex exponential.

Consider the exponential function in the \(\em \) plane: \[ \tag{26} e^{i\pi\em } = 1-\em + \em e^{i\pi}. \] Since \[ e^{i\pi}=-1, \] we have \[ \tag{27} e^{i\pi\em } = 1-2\em = \ep -\em = \j. \] Therefore, on the chosen principal branch \[ \tag{28} \boxed{ \ln\j = i\pi\em . } \]

Likewise \[ \tag{29} e^{i\pi\ep } = 1-\ep + \ep e^{i\pi} = 1-2\ep. \] But \[ 1-2\ep = -\ep +\em = -\j, \] That's why \[ \tag{30} \boxed{ \ln(-\j) = i\pi\ep . } \]

Add expressions (28) and (30): \[ \tag{31} \ln\j+\ln(-\j) = i\pi \left( \ep +\em \right). \] Since \[ \ep +\em =1, \] we obtain \[ \tag{32} \ln\j+\ln(-\j)=i\pi. \] From here, the complex unit is restored through two hyperbolic logarithms: \[ \tag{33} \boxed{ i = \frac{ \ln\j+\ln(-\j) }{\pi}. } \]

The difference of these same logarithms is \[ \tag{34} \ln(-\j)-\ln\j = i\pi \left( \ep -\em \right). \] Since \[ \ep -\em =\j, \] we obtain \[ \tag{35} \boxed{ i\j = \frac{ \ln(-\j)-\ln\j }{\pi}. } \]

Thus, the sum of the logarithms identifies the complex unit \(i\), and the difference identifies the mixed direction \(i\j\). Inverse relations have the form \[ \tag{36} \boxed{ \ln\j = \frac{\pi}{2} \left( i-i\j \right), } \] \[ \tag{37} \boxed{ \ln(-\j) = \frac{\pi}{2} \left( i+i\j \right). } \]

In matrix form, this transformation is written as \[ \tag{38} \begin{pmatrix} i\\ i\j \end{pmatrix} = \frac{1}{\pi} \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix} \begin{pmatrix} \ln\j\\ \ln(-\j) \end{pmatrix}. \] Inverse transformation: \[ \tag{39} \begin{pmatrix} \ln\j\\ \ln(-\j) \end{pmatrix} = \frac{\pi}{2} \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix} \begin{pmatrix} i\\ i\j \end{pmatrix}. \]

From formulas (28) and (30), the complex directions of idempotent planes follow directly: \[ \tag{40} \boxed{ i\ep = \frac{\ln(-\j)}{\pi}, \qquad i\em = \frac{\ln\j}{\pi}. } \] Dividing by \(i\) yields the idempotents themselves: \[ \tag{41} \boxed{ \ep = \frac{\ln(-\j)}{i\pi}, \qquad \em = \frac{\ln\j}{i\pi}. } \]

Therefore, the entire i-basis can be written in terms of the logarithms of hyperbolic units: \[ \tag{42} \boxed{ \left\{ \ep ,\; i\ep ,\; \em ,\; i\em \right\} = \left\{ \frac{\ln(-\j)}{i\pi},\; \frac{\ln(-\j)}{\pi},\; \frac{\ln\j}{i\pi},\; \frac{\ln\j}{\pi} \right\}. } \]

This equality shows that the real directions of the i-basis are given by the logarithms of \(\j\) and \(-\j\) divided by \(i\pi\), and the complex directions are given by the same logarithms divided by \(\pi\).

From equality \[ \ln\j=i\pi\em \] should \[ \tag{43} \j^b = e^{b\ln\j} = e^{i\pi b\em }. \] Using formula (25), we obtain \[ \tag{44} \boxed{ \j^b = \ep + \em e^{i\pi b}. } \]

The series expansion has the form \[ \tag{45} \j^b = 1+ \em \sum_{n=1}^{\infty} \frac{(i\pi b)^n}{n!}. \] Thus, the power function of the hyperbolic unit is defined as the usual complex exponential, localized in one of the two idempotent nplanes.

The complex logarithm is multivalued. Therefore, for \(\j\), branches are possible \[ \tag{46} \ln\j = i(\pi+2\pi n)\em , \qquad n\in\mathbb{Z}, \] and for \(-\j\) \[ \tag{47} \ln(-\j) = i(\pi+2\pi m)\ep , \qquad m\in\mathbb{Z}. \] In this paper, we use consistent principal branches \[ \tag{48} \ln\j=i\pi\em , \qquad \ln(-\j)=i\pi\ep . \] It is this choice that has minimal arguments and leads to formulas (33), (35), and (42).

The first complex plane has a basis \[ \tag{49} \left\{ \ep ,\; i\ep \right\}, \] and the second has a basis \[ \tag{50} \left\{ \em ,\; i\em \right\}. \] Due to the condition \[ \ep \em =0, \] these planes are algebraically orthogonal: their mixed products vanish.

In this case, the unit and hyperbolic unit are expressed as the sum and difference of two idempotent directions: \[ \tag{51} 1=\ep +\em , \qquad \j=\ep -\em . \] And the complex and mixed units are expressed as the sum and difference of logarithms: \[ \tag{52} i = \frac{\ln\j+\ln(-\j)}{\pi}, \qquad i\j = \frac{\ln(-\j)-\ln\j}{\pi}. \]

Thus, the two pairs of basis elements are related by the same symmetry: \[ \tag{53} \begin{aligned} 1&=\ep +\em , &\qquad \j&=\ep -\em ,\\ i&=i\ep +i\em , & i\j&=i\ep -i\em . \end{aligned} \] The I-basis is a decomposition of a four-dimensional algebra into symmetric and antisymmetric complex parts.

The obtained results can be represented as the following logical chain: \[ \tag{54} \boxed{ i^2=-1, \qquad \j^2=1 } \] \[ \Downarrow \] \[ \tag{55} \boxed{ \left\{1,i,\j,i\j\right\} } \] \[ \Downarrow \] \[ \tag{56} \boxed{ \ep = \frac{1+\j}{2}, \qquad \em = \frac{1-\j}{2} } \] \[ \Downarrow \] \[ \tag{57} \boxed{ \left\{ \ep , i\ep , \em i\em \right\} } \] \[ \Downarrow \] \[ \tag{58} \boxed{ \ln\j = i\pi\em , \qquad \ln(-\j) = i\pi\ep } \] \[ \Downarrow \] \[ \tag{59} \boxed{ i = \frac{\ln\j+\ln(-\j)}{\pi}, \qquad i\j = \frac{\ln(-\j)-\ln\j}{\pi} }. \]

Considering the complex unit \(i\) and the hyperbolic unit \(\j\) together initially leads to a four-dimensional basis \[ \left\{1,i,\j,i\j\right\}. \] However, factoring the condition \(\j^2=1\) shows that the idempotent decomposition is more natural \[ 1=\ep +\em , \qquad \j=\ep -\em . \]

From this, an i-basis arises \[ \left\{ \ep , i\ep , \em , i\em \right\}, \] which represents the original algebra as the direct sum of two independent complex planes.

The main result is that the complex directions of this basis can be obtained through the logarithms of hyperbolic units: \[ \ln\j=i\pi\em , \qquad \ln(-\j)=i\pi\ep . \] From this follow the symmetric formulas \[ i = \frac{\ln\j+\ln(-\j)}{\pi}, \qquad i\j = \frac{\ln(-\j)-\ln\j}{\pi}. \]

Thus, the complex unit \(i\) can be viewed as the sum of two logarithmic directions belonging to mutually orthogonal idempotent planes. The i-basis in this interpretation is not an artificial change of coordinates, but a natural consequence of the hyperbolic structure, the complex exponential, and the logarithm.