What Are the Relations for the Polar Hypercomplex form $a + bi + cj + dk$? Olariu in "Complex Numbers in $N$ Dimensions" has polar hypercomplex numbers described by its generators as
\begin{gather}
\alpha^2 = \beta, \\
\beta^2 = 1, \\
\gamma^2 = \beta, \\
\alpha\beta =\beta\alpha = \gamma, \\
\alpha\gamma =\gamma\alpha = 1,\\
\beta\gamma = \gamma\beta = \alpha.
\end{gather} 
A commutative hypercomplex number can be expressed as a linear combination of real and non-real roots of $1$ and $-1$.  
My goal is to find an expression for these polar hypercomplex numbers as $a + bi + cj + dk$.  
Any help, reference would be greatly appreciated.
 A: I will attempt an answer at this point. First, I think you would like us to read section 3.4 which is found at pages 113-137. I make no claim to understand all those results, clearly the author has spent some time developing exponentials and trigonometric functions as well as studying the structure of zero-divisors and so forth. That said, the basic structure here is simply an algebra $\mathcal{A} = \mathbb{R} \oplus \alpha \mathbb{R} \oplus \beta \mathbb{R} \oplus \gamma \mathbb{R}$ where the multiplication is given by the relations in your post.
\begin{gather}
\alpha^2 = \beta, \\
\beta^2 = 1, \\
\gamma^2 = \beta, \\
\alpha\beta =\beta\alpha = \gamma, \\
\alpha\gamma =\gamma\alpha = 1,\\
\beta\gamma = \gamma\beta = \alpha.
\end{gather} 
You ask for an expression. I would start with $X=t+x\alpha+y\beta+ z\gamma$ this is a typical number in $\mathcal{A}$. We can multiply them as follows, suppose $A=a+b\alpha+c\beta+ d\gamma$ is another number in $\mathcal{A}$ then
\begin{align} AX &= (a+b\alpha+c\beta+ d\gamma)( t+x\alpha+y\beta+ z\gamma) \\
&= a( t+x\alpha+y\beta+ z\gamma)+b\alpha( t+x\alpha+y\beta+ z\gamma)+ \\
&\qquad +c\beta ( t+x\alpha+y\beta+ z\gamma)+ d\gamma ( t+x\alpha+y\beta+ z\gamma) \\
&= 1(at+dx+cy+bz)+\alpha(bt+ax+dy+cz) \\
&\qquad + \beta( ct+bx+ay+dz)+ \gamma( dt+cx+by+az)
\end{align}
Our notation here is that $e_1=1$ and $e_2 = \alpha$, $e_3=\beta$ and $e_4=\gamma$. It follows that I can read off a matrix representative of the number $A$ as follows:
$$ M_A = \left[ \begin{array}{cccc} a & d & c & b \\ b & a & d & c \\ c & b & a & d \\ d & c & b & a \end{array} \right] $$
The matrix above represents the linear map $L_A: \mathcal{A} \rightarrow \mathcal{A}$ defined by $L_A(X)=AX$ with respect to the basis of generators. Notice in particular,
$$ M_1 = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \
M_{\alpha} = \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \
M_{\beta}=\left[ \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] \
M_{\gamma}=\left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array} \right] $$
You can verify that $M_AM_B = M_{AB}$ hence $M: \mathcal{A} \rightarrow \mathbb{R}^{ 4 \times 4}$ provides an algebra homomorphism.
Interesting, this reminds me of $\mathcal{A}_2 = \mathbb{R} \oplus j\mathbb{R} \oplus j^2 \mathbb{R}\oplus j^3\mathbb{R}$ where $j^4=1$ which has matrix representatives of the following form:
$$ M_{a+bj+cj^2+dj^3} = \left[ \begin{array}{cccc} a  &  d  & c & b \\  b & a & d & c \\ c & b & a & d \\ d & c & b & a \end{array} \right]$$
Apparently, your algebra is isomorphic to the cyclotomic numbers of order $4$. Identify that $\alpha = j$, $\beta=j^2$ and $\gamma=j^3$. You can recover all your relations by imposing $j^4=1$. 

Woke up this morning and it occurred to me you might be after the isomorphism of $\mathbb{R} \oplus j\mathbb{R} \oplus j^2 \mathbb{R}\oplus j^3\mathbb{R}$ to $\mathbb{C} \oplus \mathcal{H}$ where $\mathbb{C}$ is the usual complex numbers and $\mathcal{H}$ are the hyperbolic numbers. The space $\mathcal{A}_3=\mathbb{C} \oplus \mathcal{H}$. I'll use $\mathbb{C} = \mathbb{R} \oplus i \mathbb{R}$ and $\mathcal{H} = \mathbb{R} \oplus \eta \mathbb{R}$ where $i^2=-1$ and $\eta^2=1$. Be carefull though, the identity for $\mathcal{A}_3$ in our current context is $(1,1)$
$$\mathcal{A}_3 = \{ (z_1,z_2) \ | \ z_1 \in \mathbb{C}, z_2 \in \mathcal{H} \}$$
In this notation $(0,\eta)^2 = (0, \eta^2)= (0,1)$ and $(i,0)^2 = (i^2, 0)= (-1,0)$. The isomorphism $\Psi$ from $\mathcal{A}_3$ to $\mathcal{A}_2$ will be fixed by its image on generator $j$. Clearly we need:
$$ \Psi(j) = (x_1+iy_1,x_2+\eta y_2)$$
such that $(x_1+iy_1,x_2+\eta y_2)^4=(1,1)$ but this is to ask:
$$ (x_1+iy_1)^4=1, \qquad (x_2+\eta y_2)^4=1 $$
One solution is given by:
$$ x_1 = 0, \ \ y_1 = 1, \ \ x_2=0, y_2 = 1$$
note $(i,\eta)^2 = (i^2,\eta^2) = (-1,1)$ and $(i,\eta)^4 = (i^4,\eta^4) = (1,1)$. Therefore, $\Psi(j) = (i,\eta)$ and you can cipher that: omitting the $\Psi$,
$$ j = (i,\eta) = \alpha $$
$$ j^2 = (-1,1) = \beta $$
$$ j^3 = (-i,\eta) = \gamma $$
so, you can use a polar representation in terms of sine and cosine for the copy of the complex numbers, however, as I suspected from the outset, there is a copy of the hyperbolic numbers implicit withing your algebra. Now you can use the relations above to make that explicit.
Incidentally, anytime you have a semi-simple real associative algebra which is commutative it will allow an isomorphism to a direct sum of copies of $\mathbb{C}$ and $\mathcal{H}$. If we allow noncommutativity then the algebra is isomorphic to a direct sum of matrix algebras over $\mathbb{R}$, $\mathbb{C}$ or the quaternions $\mathbb{H}$.
A: The answer is that it can't be done.  Because the ring is principal (as is described in answer 2) i.e. 
$a  + bi + cj + dk = A+ B \alpha +C \alpha^2 + D \alpha^3 $
there are no relations that would allow for 
$i^2 = \pm 1$, $j^2 = \pm 1$ and $k^2 = \pm 1$. 
