Transmission lines are often modelled using a two-port relationship
$$ \begin{bmatrix} V_2 \\ I_2 \end{bmatrix} = \begin{bmatrix} \cosh(x) & Z \sinh(x) \\ \frac{1}{Z} \sinh(x) & \cosh(x) \end{bmatrix} \begin{bmatrix} V_1 \\ I_1 \end{bmatrix}$$
where $I_n, V_n, Z, x \in \mathbb{C}$.
Without getting too deep in to the weeds of what these variables actually represent I'm interested in the structure of the matrix. It's almost a hyperbolic rotation matrix
$$ \begin{bmatrix} \cosh(x) & \sinh(x) \\ \sinh(x) & \cosh(x) \end{bmatrix}$$
although I've never seen a hyperbolic rotation matrix with a complex angle. But I don't think that matters.
The $Z$ and $\frac{1}{Z}$ terms change it to not be a hyperbolic rotation, though. I'm wondering if there's maybe some way to decompose the original matrix in to some combination of a hyperbolic rotation and some other elementary transformations? Like how for computer graphics transformations are often decomposed into rotation, scale, projection, etc. matrices.
Or is there perhaps some larger class of transformations which both a hyperbolic rotation and the original matrix belong beyond just 2D matrices with determinant 1?
It's easy to show that the matrices
$$A_{Z}=\begin{pmatrix}\cosh x&Z\sinh x\\\frac{1}{Z}\sinh x&\cosh x\end{pmatrix}$$
are isospectral, since they possess the same characteristic polynomial for all values $Z$. Since $\det A_Z=1$, the matrices are diagonalizable and one can show that there exists a matrix such that
$$A_{Z_2}=MA_{Z_{1}}M^{-1}$$
which, given the eigenvalue decomposition $A_Z=S_{Z}\Lambda S_{Z}^{-1}$ can be written as $M=S_{Z_1}S^{-1}_{Z_2}$. This yields the identity
$$A_Z=\begin{pmatrix}Z&0\\0&1\end{pmatrix}A_1\begin{pmatrix}1/Z&0\\0&1\end{pmatrix}$$
which shows that the two matrices are basically related by a simple change of basis:
$A_Z$ is basically $A_1$ written in the basis of vectors $\{(1/Z,0), (0,1)\}$, which means that all one needs to do is rescale all variables $V\to V/Z $.