# Why Z-transform is considered as separate transform?

The mathematical formula of the Laplace and Z transforms are same with just one difference. I.e. in the first we use $t$ for continuous-time signal and in the latter uses $n$ for discrete-time signal. I don't think that there are any other differences.

While discussing Fourier transform for continuous-time (normally analog) signals, we use the continuous Fourier transform and for discrete-time (most often digital) signals we use the discrete-time Fourier transform (DTFT).

So my question is why the Z transform is considered a separate transform? Why it is not named the "Discrete Laplace Transform" ?

There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)
$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$
with some $T>0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$ with $f_n=f(nT)$.
In practice it is not convenient to have the factor $e^{sT}$ appear everywhere. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:
$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{\infty}f_nz^{-n}$$
The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.