If $X(z)$ is the Z-transform of a discrete timeserie $x(t)$, what is the Z-transform of $x(t)+k$ where $k$ is a constant?
From the linearity property of the Z-transform I would expect it to be $X(z) + \mathcal{Z}(k)$, where $\mathcal{Z}(k)$ is the Z-transform of $k$. That, in turn, would be $k$ times the Z-transform of $1$.
In different tables I find the following expression for the Z-transform of $1$: $\frac{z}{z-1}$, but if I do the calculation by hand, (i.e., compute $\sum_{-\infty}^{\infty}1^{-n}$) I obtain $0$. I suspect the former expression to be in fact the Z-transform of the unit step.
Which is correct? Is $\mathcal{Z}(x(t)+k)$ equal to $\mathcal{Z}(x(t)$?