Consider the following equation:
\begin{equation*}
\mu^{3} = 1
\end{equation*}
The three solutions are
\begin{align*}
\mu_{0} &= 1, & \mu_{1} &= -\frac{1}{2}(1 - \sqrt{3} i) \equiv \mu, & \mu_{2} &= -\frac{1}{2}(1 + \sqrt{3} i) \equiv \nu
\end{align*}
Note that
\begin{align*}
\mu + \nu &= -1, & \mu \nu &= 1
\end{align*}
and
\begin{align*}
\mu^{2} &= \nu & \nu^{2} &= \mu
\end{align*}
Now consider the following three functions:
\begin{align*}
f_{0}(z) &= \exp{(\mu_{0}z)} = \exp{(z)}, \\
f_{1}(z) &= \exp{(\mu_{1}z)} = \exp{(\mu z)} = \exp{\left( -\frac{z}{2} \right)} \left[ \cos{\left( \frac{\sqrt{3}z}{2} \right)} + i \sin{\left( \frac{\sqrt{3}z}{2} \right)} \right], \\
f_{2}(z) &= \exp{(\mu_{2}z)} = \exp{(\nu z)} = \exp{\left( -\frac{z}{2} \right)} \left[ \cos{\left( \frac{\sqrt{3}z}{2} \right)} - i \sin{\left( \frac{\sqrt{3}z}{2} \right)} \right]
\end{align*}
First, you write \(f_{0}\) as an infinite series:
\begin{equation*}
f_{0}(z) = \sum_{n = 0}^{\infty} \frac{1}{\Gamma(n + 1)} z^{n} = p_{0}(z) + p_{1}(z) + p_{2}(z)
\end{equation*}
Here the \(p\)-functions are given by
\begin{align*}
p_{0}(z) &= \sum_{n = 0}^{\infty} \frac{1}{\Gamma(3n + 1)} z^{3n}, & p_{1}(z) &= \sum_{n = 0}^{\infty} \frac{1}{\Gamma(3n + 2)} z^{3n+1}, & p_{2}(z) &= \sum_{n = 0}^{\infty} \frac{1}{\Gamma(3n + 3)} z^{3n+2}
\end{align*}
You can also write \(f_{1}\) and \(f_{2}\) as an infinite series:
\begin{align*}
f_{1}(z) &= \sum_{n = 0}^{\infty} \frac{\mu^{n}}{\Gamma(n + 1)} z^{n} = p_{0}(z) + \mu p_{1}(z) + \nu p_{2}(z) \\
f_{2}(z) &= \sum_{n = 0}^{\infty} \frac{\nu^{n}}{\Gamma(n + 1)} z^{n} = p_{0}(z) + \nu p_{1}(z) + \mu p_{2}(z)
\end{align*}
The \(f\)-functions are complex exponential functions. You can write the \(p\)-functions in terms of the \(f\)-functions:
\begin{align*}
p_{0}(z) &= \frac{f_{0}(z) + f_{1}(z) + f_{2}(z)}{3} \\
p_{1}(z) &= \frac{2f_{0}(z) - f_{1}(z) - f_{2}(z) - i \sqrt{3} \left[ f_{1}(z) - f_{2}(z) \right]}{6} \\
p_{2}(z) &= \frac{2f_{0}(z) - f_{1}(z) - f_{2}(z) + i \sqrt{3} \left[ f_{1}(z) - f_{2}(z) \right]}{6}
\end{align*}
More explicitly:
\begin{align*}
p_{0}(z) &= \frac{1}{3} \exp{(z)} + \frac{2}{3} \exp{\left( -\frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} \\
p_{1}(z) &= \frac{1}{3} \exp{(z)} - \frac{1}{3} \exp{\left( -\frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} + \frac{1}{\sqrt{3}} \exp{\left( -\frac{z}{2} \right)} \sin{\left( \frac{\sqrt{3}z}{2} \right)} \\
p_{2}(z) &= \frac{1}{3} \exp{(z)} - \frac{1}{3} \exp{\left( -\frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} - \frac{1}{\sqrt{3}} \exp{\left( -\frac{z}{2} \right)} \sin{\left( \frac{\sqrt{3}z}{2} \right)}
\end{align*}
Note that all coefficients are real.
Similar to the exponential functions, the parity properties give rise to linearly-independent functions:
\begin{align*}
p_{0}(-z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(3n + 1)} z^{3n} \equiv q_{0}(z) \\
p_{1}(-z) &= -\sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(3n + 2)} z^{3n+1} \equiv -q_{1}(z) \\
p_{2}(-z) &= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{\Gamma(3n + 3)} z^{3n+2} \equiv q_{2}(z)
\end{align*}
More explicitly:
\begin{align*}
q_{0}(z) &= \frac{1}{3} \exp{(-z)} + \frac{2}{3} \exp{\left( \frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} \\
q_{1}(z) &= -\frac{1}{3} \exp{(-z)} + \frac{1}{3} \exp{\left( \frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} + \frac{1}{\sqrt{3}} \exp{\left( \frac{z}{2} \right)} \sin{\left( \frac{\sqrt{3}z}{2} \right)} \\
q_{2}(z) &= \frac{1}{3} \exp{(-z)} - \frac{1}{3} \exp{\left( \frac{z}{2} \right)} \cos{\left( \frac{\sqrt{3}z}{2} \right)} + \frac{1}{\sqrt{3}} \exp{\left( \frac{z}{2} \right)} \sin{\left( \frac{\sqrt{3}z}{2} \right)}
\end{align*}
When written out explicitly in terms of exponential and trigonometric functions, the \(p\)-functions and the \(q\)-functions do not look so impressive. But, I am still captivated by the fact that the three \(p\)-functions satisfy a cubic identity. Note that
\begin{equation*}
f_{0}(z) f_{1}(z) f_{2}(z) = 1
\end{equation*}
In terms of the \(p\)-functions, you get:
\begin{equation*}
[p_{0}(z)]^{3} + [p_{1}(z)]^{3} + [p_{2}(z)]^{3} - 3 p_{0}(z) p_{1}(z) p_{2}(z) = 1
\end{equation*}
Replacing \(z\) with \(-z\) gives
\begin{equation*}
[q_{0}(z)]^{3} - [q_{1}(z)]^{3} + [q_{2}(z)]^{3} + 3 q_{0}(z) q_{1}(z) q_{2}(z) = 1
\end{equation*}
Both of these cubic identities describe curves in 3-dimensional space. It turns out that these cubic identities follow from three bilinear relations involving both the \(p\)-functions and the \(q\)-functions:
\begin{align*}
p_{0}(z) q_{0}(z) + p_{1}(z) q_{2}(z) - p_{2}(z)q_{1}(z) &= 1 \\
p_{0}(z) q_{1}(z) - p_{1}(z) q_{0}(z) - p_{2}(z)q_{2}(z) &= 0 \\
p_{0}(z) q_{2}(z) - p_{1}(z) q_{1}(z) + p_{2}(z)q_{0}(z) &= 0
\end{align*}
These bilinear relations can be solved to write the \(q\)-functions in terms of the \(p\)-functions:
\begin{align*}
q_{0}(z) &= [p_{0}(z)]^{2} - p_{1}(z) p_{2}(z) & q_{1}(z) &= -[p_{2}(z)]^{2} + p_{0}(z) p_{1}(z) & q_{2}(z) &= [p_{1}(z)]^{2} - p_{0}(z) p_{2}(z)
\end{align*}
These are Wronskians. Alternatively, you can solve the for \(p\)-functions in terms of the \(q\)-functions:
\begin{align*}
p_{0}(z) &= [q_{0}(z)]^{2} + q_{1}(z) q_{2}(z) & p_{1}(z) &= [q_{2}(z)]^{2} + q_{0}(z) q_{1}(z) & p_{2}(z) &= [q_{1}(z)]^{2} - q_{0}(z) q_{2}(z)
\end{align*}
Again, these are Wronskians.
What about differentiation? Using the infinite series definition gives
\begin{align*}
\frac{d}{dz} p_{2}(z) &= p_{1}(z), & \frac{d}{dz} p_{1}(z) &= p_{0}(z), & \frac{d}{dz} p_{0}(z) &= p_{2}(z)
\end{align*}
and also
\begin{align*}
\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_{2}(z)
\end{align*}
Here are some argument-addition identities too:
\begin{align*}
p_{0}(w + z) &= p_{0}(w) p_{0}(z) + p_{1}(w) p_{2}(z) + p_{2}(w) p_{1}(z) \\
p_{1}(w + z) &= p_{0}(w) p_{1}(z) + p_{1}(w) p_{0}(z) + p_{2}(w) p_{2}(z) \\
p_{2}(w + z) &= p_{0}(w) p_{2}(z) + p_{1}(w) p_{1}(z) + p_{2}(w) p_{0}(z)
\end{align*}
These are analogous to the trigonometric/hyperbolic identities. The simplest way to derive these is to work with the matrix-valued function
\begin{align*}
\exp{(z A)} &= I p_{0}(z) + A p_{1}(z) + A^{2} p_{2}(z), & A^{3} &= I
\end{align*}
and use the property
\begin{equation*}
\exp{(w A)} \exp{(z A)} = \exp{(w A + z A)}
\end{equation*}
Unlike the trigonometric functions, the \(p\)-functions are not periodic.
