Euler's formula for scalar complex numbers is
\begin{equation*}
\exp{(iz)} = \cos{(z)} + i \sin{(z)}
\end{equation*}
Recall that
\begin{align*}
\cos{(z_{k})} &= 1 & \sin{(z_{k})} &= 0 & z_{k} &= 2k \pi & k &\in \mathbb{Z} \\
\cos{(z_{l})} &= -1 & \sin{(z_{l})} &= 0 & z_{l} &= (2l-1) \pi & l &\in \mathbb{Z} \\
\cos{(z_{m})} &= 0 & \sin{(z_{m})} &= 1 & z_{m} &= \frac{(4m+1) \pi}{2} & m &\in \mathbb{Z} \\
\cos{(z_{n})} &= 0 & \sin{(z_{n})} &= -1 & z_{n} &= \frac{(4n-1) \pi}{2} & n &\in \mathbb{Z}
\end{align*}
Thus, you have the exponential representations
\begin{align*}
1 &= \exp{[2 k \pi i]} & {-1} &= \exp{[(2l - 1) \pi i]} & i &= \exp{\left[ \frac{(4m+1) \pi i}{2} \right]} & {-i} &= \exp{\left[ \frac{(4n-1) \pi i}{2} \right]}
\end{align*}
Now consider taking the matrix exponential of a matrix proportional to the identity matrix \(I\):
\begin{equation*}
\exp{(izI)} = I \left[\cos{(z)} + i \sin{(z)}\right]
\end{equation*}
Thus, you have the following exponential representations:
\begin{align*}
I &= \exp{[2 k \pi i I]} & {-I} &= \exp{[(2l - 1) \pi i I]} & iI &= \exp{\left[ \frac{(4m+1) \pi i}{2}I \right]} & {-i}I &= \exp{\left[ \frac{(4n-1) \pi i}{2}I \right]}
\end{align*}
Let the matrix \(A\) be an involution:
\begin{equation*}
A^{2} = I
\end{equation*}
Then
\begin{equation*}
\exp{(iz A)} = I \cos{(z)} + i A \sin{(z)}
\end{equation*}
Now you have the following exponential representations:
\begin{align*}
I &= \exp{\left[ 2 k \pi i A \right]} & k &\in \mathbb{Z} \\
{-I} &= \exp{\left[ (2 l - 1) \pi i A \right]} & l &\in \mathbb{Z} \\
iA &= \exp{\left[ \frac{1}{2} (4m+1) \pi i A \right]} & m &\in \mathbb{Z} \\
{-iA} &= \exp{\left[ \frac{1}{2} (4n-1) \pi i A \right]} & n &\in \mathbb{Z}
\end{align*}
Using the above result, you can have another exponential representation of the identity:
\begin{align*}
I &= \exp{\left[ (2k - 1) \pi i I + (2l - 1) \pi i A \right]} & k, l &\in \mathbb{Z}
\end{align*}
and two exponential representations of the involution:
\begin{align*}
A &= \exp{\left[ \frac{1}{2}(4k-1) \pi i I + \frac{1}{2} (4l+1) \pi i A \right]} & k, l &\in \mathbb{Z} \\
A &= \exp{\left[ \frac{1}{2}(4m+1) \pi i I + \frac{1}{2} (4n-1) \pi i A \right]} & m, n &\in \mathbb{Z}
\end{align*}
You can define the principal values as the cases when \(k = l = m = n = 0\).