Below I've quoted Wikipedia's entry that relates the Z-Transform to the Laplace Transform. The part I don't understand is $z \ \stackrel{\mathrm{def}}{=}\ e^{s T}$; I thought $z$ was actually an element of $\mathbb{C}$ and thus would be $z \ \stackrel{\mathrm{def}}{=}\ Ae^{s T}$ (but then it would be different to the Laplace Transform...). I don't understand why the Z-Transform is not defined as: $$ X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] e^{-\omega n} $$ or something like that.

Z-transform

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of $$ z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \ $$ where $T = 1/f_s \ $is the sampling period (in units of time e.g., seconds) and $f_s \ $is the sampling rate (in samples per second or hertz)

Let $$ \Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T) $$ be a sampling impulse train (also called a Dirac comb) and $$ \begin{align} x_q(t) & \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T) \\ & = \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T) \end{align} $$ be the continuous-time representation of the sampled x(t) \ $$ x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \ $$ are the discrete samples of x(t) The Laplace transform of the sampled signal x_q(t) \ is $$ \begin{align} X_q(s) & = \int_{0^-}^\infty x_q(t) e^{-s t} \,dt \\ & = \int_{0^-}^\infty \sum_{n=0}^\infty x[n] \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] \int_{0^-}^\infty \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] e^{-n s T}. \end{align} $$ This is precisely the definition of the unilateral Z-transform of the discrete function $x[n] \ $. $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} $$ with the substitution of $z \leftarrow e^{s T} \ $.

Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal: $$ X_q(s) = X(z) \Big|_{z=e^{sT}} $$ The similarity between the Z and Laplace transforms is expanded upon in the theory of time scale calculus.

(Source: http://en.wikipedia.org/wiki/Laplace_transform#Laplace.E2.80.93Stieltjes_transform)

Here: http://en.wikipedia.org/wiki/Z-transform it says that $z \in \mathbb{C}$.