How to find the sum of $\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}$? I'm really stuck with this problem and I hope some of you could give me a hint.
Consider the functions: $$f(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}, \quad f^{\prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}, \quad f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}, \quad f^{\prime \prime \prime}(z)=f(z)$$
I need to find the sum of each of these series.
My first thought was to show that:
$$f(z)+f^{\prime}(z)+f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !}=e^{z}$$
And then solve this differential equation.
However, officially I haven't had a course in differential equations. So is there another way I can solve this using basic Complex Analysis? I hope some of you can guide me in the right direction, what theorems could be useful?

In general, it is possible for us to extract components of power series by playing with roots of unity. Particularly, let's say
$$
f(z)=\sum_{n=0}^\infty c_nz^n
$$
converges absolutely for some $z\in\mathbb C$. If we set $\zeta=e^{2\pi i/q}$ then
$$
\frac1q\sum_{r=0}^{q-1}\zeta^{-ra}f(\zeta^rz)=\sum_{k=0}^\infty c_{qk+a}z^{qk+a}\tag1
$$
Consequently, we get
$$
\sum_{k=0}^\infty{z^{3k+1}\over(3k+1)!}=\frac13\sum_{r=0}^2 e^{-2\pi ir/3}\exp(e^{2\pi ir/3}z)
$$
Proof for (1):
Given $f$ converges absolutely at $z$, we can plug in the power series expansion for $f(z)$ to get
$$
\begin{aligned}
\sum_{r=0}^{q-1}\zeta^{-ra}\sum_{n=0}^\infty c_n\zeta^{rn}z^n
&=\sum_{n=0}^\infty c_nz^{n}\left(\sum_{r=0}^{q-1}\zeta^{r(n-a)}\right)
\end{aligned}
$$
If $n\equiv0\pmod q$, then
$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}=\sum_{r=0}^{q-1}1=q
$$
Otherwise we have $\zeta^{n-a}\ne1$:
$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}={\zeta^{q(n-a)}-1\over\zeta^{n-a}-1}=0
$$
Consequently, the purpose of sum of roots of unity in (1) is to extract coefficients from power series.
