How do I find $S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2 p}}{(2 p) !}+\frac{x^{3 p}}{(3 p) !}+\cdots$ by complex numbers, where $p \in N$? In my post on Quora, I had expressed the three series in terms of sine and cosine:
$$1+\frac{x^{3}}{3 !}+\frac{x^{6}}{6 !}+\frac{x^{9}}{9!}+\cdots \cdot=\frac{1}{3}\left[e^{x}+2 e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2} x\right)\right]\\
\frac{x^{2}}{2 !}+\frac{x^{5}}{5 !}+\frac{x^{8}}{8 !}+\cdots=\frac{1}{3}\left[e^{x}-e^{-\frac{x}{2}}\left(\sqrt{3} \sin \left(\frac{\sqrt{3}}{2} x\right)+\cos \left(\frac{\sqrt{3}}{2} x\right)\right)\right]\\
x+\frac{x^{4}}{4 !}+\frac{x^{7}}{7 !}+\cdots=
\frac{1}{3}\left[e^{x}+e^{-\frac{x}{2}}\left(\sqrt{3} \sin \left(\frac{\sqrt{3}}{2} x\right)-\cos \left(\frac{\sqrt{3}}{2} x\right)\right)\right].$$
They all have the same “length” $3$. I then think that the proof can be extended for a general series $S(p)$ with an arbitrary length $p$ and try to prove similarly that for any natunal number $p$,
$$
S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2 p}}{(2 p) !}+\frac{x^{3 p}}{(3 p) !}+\cdots=\frac{1}{p}\left(e^{x}+e^{\omega x}+e^{\omega^{2} x}+\cdots+e^{\omega^{n-1} x}\right) \tag*{(*)} 
$$
where $\omega$ is the complex $p$-th root of unity satisfying $$1+\omega+\omega^{2}+\cdots+\omega^{p-1}=0.$$
My question is how to prove (*) and express , in terms of sine and cosine, the series
$$e^{x}+e^{\omega x}+e^{\omega^{2} x}+\cdots+e^{\omega^{p-1} x}$$
 A: We are going to evaluate the series
$\displaystyle  S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2p}}{(2p) !}+\frac{x^{3p}}{(3p )!}+\cdots  $
Let’s start with the series
$ \displaystyle e^{z}=\sum_{x=0}^{\infty} \frac{z^{k}}{k !} \tag*{(*)} $
Now consider the complex $p^{th} $ root of unity $\omega=e^{\frac{ 2\pi}{p}i} $ satisfying
$$ 1+\omega+\omega^{2}+\cdots+\omega^{p-1} =0 \text{ and }\omega^{p}=1.$$
Putting $ z=x$ yields
$ \displaystyle e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} \tag*{(1)}$
Putting $ z=\omega x $ yields
$ \displaystyle e^{\omega x}=\sum_{k=0}^{\infty} \frac{\omega^{k} x^{k}}{k !}\tag*{(2)}$
Putting $\displaystyle z=\omega^{2} x $  yields
$ \displaystyle e^{\omega^{2} x}=\sum_{k=0}^{\infty} \frac{\omega^{2 k} x^{k}}{k !} \tag*{(3)}$
$$\vdots$$
Putting $\displaystyle z=\omega^{p-1} x $  yields
$ \displaystyle e^{\omega^{p-1} x}=\sum_{k=0}^{\infty} \frac{\omega^{(p-1)k} x^{k}}{k !} \tag*{(p-1)}$
$(1)+(2)+(3)+\cdots +(p-1)$ gives
$ \displaystyle \sum_{k=0}^{\infty} \frac{1+\omega^{k}+\omega^{2 k}+\cdots+\omega^{(p-1)k}}{k !}x^{k}=e^{x}+e^{\omega x}+e^{\omega^{2} x}+ \cdots +e^{\omega^{p-1} x}\tag*{}$
$ \begin{aligned} \displaystyle \because 1+\omega^{k}+\omega^{2 k}+\cdots+ \omega^{(p-1) k} &=\left\{\begin{array}{cl}\frac{1-\left(\omega^{p}\right)^{k}}{1-\omega^{p}} & \text { if } p\not | k \\ p & \text { if } p|k\end{array}\right. =\left\{\begin{array}{ll}0 & \text { if } p\not|k \\p & \text { if } p | k\end{array}\right.\\\displaystyle  \therefore \sum_{k=0}^{\infty} \frac{p}{(p k) !} x^{p k} &=e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \end{aligned} \tag*{}  $
$$ S(p)= \sum_{k=0}^{\infty} \frac{x^{p k}}{(p k) !} =\frac{1}{p} \left(e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \right)\tag*{(**)}  $$
When $p$ is odd,
$$
\begin{aligned}
S(p) &=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\
&=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{x e^{\frac{2 k \pi}{p}i} }+e^{x e^{\frac{2(p-k) \pi}{p} i}}\right)\right]\\& =\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(e^{i x \sin \frac{2k\pi}{p}}+e^{-i x \sin \frac{2k \pi)}{p}}\right)\right]\\ &=\frac{1}{p}\left[e^{x}+2 \sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right]
\end{aligned}
$$
When $p$ is even,
$$\begin{aligned}
S(p) &=\frac{1}{p}\left[e^{x}+e^{-x}+\sum_{k=1}^{\frac{p}{2} -1}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\ &=\frac{1}{p}\left[e^{x} +e^{-x} +2 \sum_{k=1}^{\frac{p}{2}-1} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right]
\end{aligned}$$
By the way,  by successive differentiation on $S(p)$, we can find, for any natural number $n<p$, the series,
$$
\frac{x^{n}}{n!}+\frac{x^{n+p}}{(n+p) !}+\frac{x^{n+2 p}}{(n+2 p) !}+\cdots
$$ whose length is $p$ and starting terms other than $1$.
