I am using the Z-transform for a part of my current research. I am new to the Z-transform, and somewhat scatter-brained at the moment, so, I would very much appreciate it if someone could tell me whether or not I messed up in the following computations. Note: this is a research question, not a homework question. I would like an answer of the form “yes, this is correct,” or, “no, this is incorrect; the correct formulas are [insert formulas here]”. I'm proof-reading (hah!) a proof I just wrote up, and would very much appreciate knowing what a different pair of eyes makes of this.
Anywho: let $\xi_{p}\overset{\textrm{def}}{=}e^{\frac{2\pi i}{p}}$ for every $p\in\mathbb{N}$, let $f\left(z\right)$ be an entire function ($f\left(z\right)=\sum_{n=0}^{\infty}a_{n}z^{n}$), let $\mathcal{Z}$ denote the (bilateral) Z-transform, and let $\phi\left[n\right]\overset{\textrm{def}}{=}\mathcal{Z}^{-1}\left\{ f\right\} \left[n\right].$ Let $\overset{6}{\equiv}$ denote congruence of integers modulo $6$. Also, let: $$n_{K}\overset{\textrm{def}}{=}\begin{cases} \frac{n}{K} & \textrm{if }n=Kr\textrm{ for some }r\in\mathbb{Z}\\ 0 & \textrm{if }n\notin K\mathbb{Z} \end{cases}$$ for all $K\in\mathbb{N}$. The three pertinent Z-transform rules I'm using from Wikipedia are:
$$\mathcal{Z}\left\{ \phi\left[n_{K}\right]\right\} \left(z\right) = f\left(z^{K}\right)$$ $$\mathcal{Z}\left\{ c^{n}\phi\left[n\right]\right\} \left(z\right)=f\left(\frac{z}{c}\right)$$ $$\mathcal{Z}\left\{ \phi\left[n+k\right]\right\} \left(z\right) = z^{k}f\left(z\right)$$
That being said, my questions are:
1) Does the inverse Z-transform $\phi$ of $f$ actually exist, and, moreover, is it unique?
2) Are the following three inverse Z-transforms computed correctly? $$\mathcal{Z}^{-1}\left\{ \frac{z^{2}}{2}\left(f\left(z^{6}\right)-f\left(-z^{6}\right)\right)\right\} \left[n\right] = \frac{1}{2}\phi\left[n_{6}+2\right]-\frac{1}{2}\left(-1\right)^{n_{6}}\phi\left[n_{6}+2\right]$$ $$\mathcal{Z}^{-1}\left\{ \frac{1}{6}\sum_{k=0}^{5}\xi_{6}^{-4k}f\left(\xi_{6}^{k}z^{2}\right)\right\} \left[n\right] = \frac{1}{6}\sum_{k=0}^{5}\xi_{6}^{-4k}\xi_{6}^{-kn_{2}}\phi\left[n_{2}\right]$$ $$\mathcal{Z}^{-1}\left\{ -\frac{1}{12}\sum_{j=0}^{11}\xi_{12}^{-8j}f\left(\xi_{12}^{j}z\right)\right\} \left[n\right] = -\frac{1}{12}\sum_{j=0}^{11}\xi_{12}^{-8j}\xi_{12}^{-jn}\phi\left[n\right]$$
3) Is the following argument valid (particularly with regard to the substitutions with $n$)?
Using (2), the inverse bilateral Z-transform of the functional equation:
$$\frac{z^{2}}{2}\left(f\left(z^{6}\right)-f\left(-z^{6}\right)\right)=\frac{1}{6}\sum_{k=0}^{5}\xi_{6}^{-4k}f\left(\xi_{6}^{k}z^{2}\right)-\frac{1}{12}\sum_{j=0}^{11}\xi_{12}^{-8j}f\left(\xi_{12}^{j}z\right)$$ is:
$$\frac{1}{2}\phi\left[n_{6}+2\right]-\frac{1}{2}\left(-1\right)^{n_{6}}\phi\left[n_{6}+2\right]=\frac{1}{6}\sum_{k=0}^{5}\xi_{6}^{-4k}\xi_{6}^{-kn_{2}}\phi\left[n_{2}\right]-\frac{1}{12}\sum_{j=0}^{11}\xi_{12}^{-8j}\xi_{12}^{-jn}\phi\left[n\right]$$
Then, replacing $n$ with $6n$:
$$6\phi\left[n+2\right]-6\left(-1\right)^{n}\phi\left[n+2\right] = 2\sum_{k=0}^{5}\xi_{6}^{-\left(3n+4\right)k}\phi\left[3n\right]-\sum_{j=0}^{11}\xi_{12}^{-2\left(3n+4\right)j}\phi\left[6n\right]$$
and hence:
$$\left(\frac{1-\left(-1\right)^{n}}{2}\right)\phi\left[n+2\right] = \begin{cases} \phi\left[3n\right]-\phi\left[6n\right] & \textrm{if }3n+4\overset{6}{\equiv}0\\ 0 & \textrm{else} \end{cases}$$
Then, since $3n+4\overset{6}{\equiv}1\textrm{ or }4$, I conclude that:
$$\left(\frac{1-\left(-1\right)^{n}}{2}\right)\phi\left[n+2\right] = 0,\textrm{ }\forall n\in\mathbb{Z}$$
Thus, $\phi\left[n\right]$ is zero for all odd $n$. Moreover, since $f\left(z\right)=\sum_{n=0}^{\infty}a_{n}z^{n}$ only has positive powers of $z$, it must be that $\phi$ is zero at every positive integer. $\checkmark$
Thanks a bunch for your help—and your time!