$$\sum_{n=0}^{\infty}\frac{z^{kn}}{(kn)!}$$
Later edit: Is this correct?
$$e^{z\epsilon_p} = \sum_{n=0}^{\infty}\frac{z^n\epsilon_p^n}{n!}$$ $$\sum_{p=0}^{k-1}e^{z\epsilon_p} = \sum_{n=0}^{\infty}\frac{z^n}{n!}\sum_{p=0}^{k-1}\epsilon_p^n $$ $$\sum_{n=0}^{\infty}\frac{z^{kn}}{(kn)!} = \frac{\sum_{p=0}^{k-1}e^{\epsilon_p^nz}}{k} $$
$$\sum_{n=0} \frac{z^{kn}}{(kn)!}=$$ $$=\frac{1}{k}\sum_{l=0}^{k-1} e^{\omega_{k}^lz}$$
where $\omega_k=\exp(2\pi i/k)$
It can be considered a generalization of $\cosh(z)$ function to multicomplex numbers (For $k=2$ the series is the $\cosh(z)$, for $k=3$ the series is called cosexponential function : $cx(z)=1+z^3/3!+z^6/6!+...$)
