Consider the following equation:
\begin{equation*}
\kappa^{4} = -1
\end{equation*}
The four solutions are
\begin{align*}
\kappa_{0} &= \frac{1 + i}{\sqrt{2}} \equiv \kappa, & \kappa_{1} &= -\frac{1 - i}{\sqrt{2}} = -\lambda, & \kappa_{2} &= -\frac{1 + i}{\sqrt{2}} = -\kappa, & \kappa_{3} &= \frac{1 - i}{\sqrt{2}} \equiv \lambda
\end{align*}
Note that
\begin{align*}
\kappa + \lambda &= \sqrt{2}, & \kappa - \lambda &= \sqrt{2}i, & \kappa \lambda &= 1, & \kappa^{2} &= i, & \lambda^{2} &= -i
\end{align*}
and
\begin{align*}
\kappa^{3} &= i \kappa = - \lambda, & \lambda^{3} &= -i \lambda = -\kappa
\end{align*}
Now consider the following four functions:
\begin{align*}
g_{0}(z) &\equiv \exp{(\kappa_{0} z)} = \exp{( \kappa z)} \\
g_{1}(z) &\equiv \exp{(\kappa_{1} z)} = \exp{( - \lambda z)} \\
g_{2}(z) &\equiv \exp{(\kappa_{2} z)} = \exp{( - \kappa z)} \\
g_{3}(z) &\equiv \exp{(\kappa_{3} z)} = \exp{( \lambda z)}
\end{align*}
More explicitly:
\begin{align*}
g_{0}(z) &= \exp{\left( \frac{z}{\sqrt{2}} \right)} \left[ \cos{\left( \frac{z}{\sqrt{2}} \right)} + i \sin{\left( \frac{z}{\sqrt{2}} \right)} \right] \\
g_{1}(z) &= \exp{\left( -\frac{z}{\sqrt{2}} \right)} \left[ \cos{\left( \frac{z}{\sqrt{2}} \right)} + i \sin{\left( \frac{z}{\sqrt{2}} \right)} \right] \\
g_{2}(z) &= \exp{\left( -\frac{z}{\sqrt{2}} \right)} \left[ \cos{\left( \frac{z}{\sqrt{2}} \right)} - i \sin{\left( \frac{z}{\sqrt{2}} \right)} \right] \\
g_{3}(z) &= \exp{\left( \frac{z}{\sqrt{2}} \right)} \left[ \cos{\left( \frac{z}{\sqrt{2}} \right)} - i \sin{\left( \frac{z}{\sqrt{2}} \right)} \right]
\end{align*}
First, you write \(g_{0}\) as an infinite series, split into four sums:
\begin{equation*}
g_{0}(z) = \sum_{n = 0}^{\infty} \frac{1}{\Gamma(n + 1)} \kappa^{n} z^{n} = q_{0}(z) + \kappa q_{1}(z) + i q_{2}(z) - \lambda q_{3}(z)
\end{equation*}
Here, the \(q\)-functions are given by
\begin{align*}
q_{0}(z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(4n + 1)} z^{4n} \\
q_{1}(z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(4n + 2)} z^{4n+1} \\
q_{2}(z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(4n + 3)} z^{4n+2} \\
q_{3}(z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(4n + 4)} z^{4n+3}
\end{align*}
You can also write the other \(g\)-functions in terms of the \(q\)-functions:
\begin{align*}
g_{1}(z) &= q_{0}(z) - \lambda q_{1}(z) - i q_{2}(z) + \kappa q_{3}(z) \\
g_{2}(z) &= q_{0}(z) - \kappa q_{1}(z) + i q_{2}(z) + \lambda q_{3}(z) \\
g_{3}(z) &= q_{0}(z) + \lambda q_{1}(z) - i q_{2}(z) - \kappa q_{3}(z)
\end{align*}
Note that this definition of the \(q\)-functions can be stated as a matrix equation:
\begin{equation*}
\begin{bmatrix}
g_{0}(z) \\
g_{1}(z) \\
g_{2}(z) \\
g_{3}(z)
\end{bmatrix} = \begin{bmatrix}
1 & \kappa & i & -\lambda \\
1 & -\lambda & -i & \kappa \\
1 & -\kappa & i & \lambda \\
1 & \lambda & -i & -\kappa
\end{bmatrix} \begin{bmatrix}
q_{0}(z) \\
q_{1}(z) \\
q_{2}(z) \\
q_{3}(z)
\end{bmatrix}
\end{equation*}
Inverting this matrix gives
\begin{equation*}
\begin{bmatrix}
q_{0}(z) \\
q_{1}(z) \\
q_{2}(z) \\
q_{3}(z)
\end{bmatrix} = \frac{1}{4} \begin{bmatrix}
1 & 1 & 1 & 1 \\
\lambda & -\kappa & -\lambda & \kappa \\
-i & i & -i & i \\
-\kappa & \lambda & \kappa & -\lambda
\end{bmatrix} \begin{bmatrix}
g_{0}(z) \\
g_{1}(z) \\
g_{2}(z) \\
g_{3}(z)
\end{bmatrix}
\end{equation*}
More explicitly:
\begin{align*}
q_{0}(z) &= \frac{g_{0}(z) + g_{1}(z) + g_{2}(z) + g_{3}(z)}{4} \\
q_{1}(z) &= \frac{\lambda g_{0}(z) - \kappa g_{1}(z) - \lambda g_{2}(z) + \kappa g_{3}(z)}{4} \\
q_{2}(z) &= \frac{-ig_{0}(z) +i g_{1}(z) -i g_{2}(z) +i g_{3}(z)}{4} \\
q_{3}(z) &= \frac{- \kappa g_{0}(z) + \lambda g_{1}(z) + \kappa g_{2}(z) - \lambda g_{3}(z)}{4}
\end{align*}
Even more explicitly:
\begin{align*}
q_{0}(z) &= \cos{\left( \frac{z}{\sqrt{2}} \right)} \cosh{\left( \frac{z}{\sqrt{2}} \right)} \\
q_{1}(z) &= \frac{1}{\sqrt{2}} \sin{\left( \frac{z}{\sqrt{2}} \right)} \cosh{\left( \frac{z}{\sqrt{2}} \right)} + \frac{1}{\sqrt{2}} \cos{\left( \frac{z}{\sqrt{2}} \right)} \sinh{\left( \frac{z}{\sqrt{2}} \right)} \\
q_{2}(z) &= \sin{\left( \frac{z}{\sqrt{2}} \right)} \sinh{\left( \frac{z}{\sqrt{2}} \right)} \\
q_{3}(z) &= \frac{1}{\sqrt{2}} \sin{\left( \frac{z}{\sqrt{2}} \right)} \cosh{\left( \frac{z}{\sqrt{2}} \right)} - \frac{1}{\sqrt{2}} \cos{\left( \frac{z}{\sqrt{2}} \right)} \sinh{\left( \frac{z}{\sqrt{2}} \right)}
\end{align*}
The parity properties are as follow:
\begin{align*}
q_{0}(-z) &= q_{0}(z), & q_{1}(-z) &= -q_{1}(z), & q_{2}(-z) &= q_{2}(z), & q_{3}(-z) &= -q_{3}(z) \\
q_{0}(iz) &= q_{0}(z), & q_{1}(iz) &= iq_{1}(z), & q_{2}(iz) &= -q_{2}(z), & q_{3}(iz) &= -iq_{3}(z) \\
q_{0}(-iz) &= q_{0}(z), & q_{1}(-iz) &= -iq_{1}(z), & q_{2}(-iz) &= -q_{2}(z), & q_{3}(-iz) &= iq_{3}(z)
\end{align*}
Note that
\begin{align*}
g_{0}(z) g_{2}(z) &= 1 & g_{1}(z) g_{3}(z) &= 1
\end{align*}
These equations are equivalent to the following two quadratic equations:
\begin{align*}
q_{0}^{2} - q_{2}^{2} + 2q_{1} q_{3} &= 1, & q_{1}^{2} - q_{3}^{2} - 2q_{0} q_{2} &= 0
\end{align*}
Thus, you can interpret the four \(q\)-functions as coordinates in a 4-dimensional space describing certain quadratic curves.
What about differentiation? Using the infinite series definition gives
\begin{align*}
\frac{d}{dz} q_{3}(z) &= q_{2}(z), & \frac{d}{dz} q_{2}(z) &= q_{1}(z), & \frac{d}{dz} q_{1}(z) &= q_{0}(z), & \frac{d}{dz} q_{0}(z) &= -q_{3}(z)
\end{align*}
This is similar to what we found for the cubic functions.