Q:

Matrix representation of complex numbers in exponential form

Do there exist matrices M and P for this equation? Or perhaps M and P dont need to be matrices? I saw this and this question after googling which made me wonder about whether the exponential form of complex numbers would still work.
$
z = a\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}+b\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} = \begin{pmatrix} a&b\\ -b&a \end{pmatrix} = Me^P
$
$
$
$
$
Addition:
I just found this article which I dont really understand, but where they solve a case similar to this. The thing I dont understand in this article is how you exponentiate with a matrix and what that alpha is doing there. Also I dont understand that in order to make this work they have to make the matrix $\Phi$ a complex matrix, looks like it defeats the purpose of having a matrix representation of complex numbers. Not sure whether to append this to my question or add as answer, because I still dont understand it.
They say,
$
z = \begin{pmatrix} a&jb\\ j\alpha^2b&a \end{pmatrix} = aI+b\Phi
$
Where,
$
\Phi = \begin{pmatrix} 0&j\\ j\alpha^2&0 \end{pmatrix}
$
For which they say, 
$
e^{\phi\Phi}=cos(\alpha\phi)I+\frac{1}{\alpha}sin(\alpha\phi)\Phi
$

A:

Show that $z=\rho e^{i\theta}$ with $i\theta=\begin{pmatrix}0&\theta\\-\theta&0\end{pmatrix}$. Calculate $e^{i\theta}$ and deduce $\theta,\rho$.

