I am trying to get the z tranform from the fourier transform, so I am trying to get its equivalent in time to then, get the z transform, this is what I have: $$Y(w) = \left\{ \begin{array}{c l} 1 & 0\le |w|\le \frac{\pi}{2}\\ 0 & \frac{\pi}{2}\le |w|\le \pi\\ \end{array} \right. $$
$$V(w) = \left\{ \begin{array}{c l} 1 & \frac{\pi}{4}\le |w|\le \frac{\pi}{2}\\ 0 & 0\le |w|\le \frac{\pi}{4}; \frac{\pi}{2}\le |w|\le \pi\\ \end{array} \right. $$
So for the $Y(w)=u[f+ \frac{\pi}{2}]-u[f- \frac{\pi}{2}]$ I found that the in time it would be $Y[n]=\frac{sin(\frac{\pi n}{2})}{\pi n}$ and it Z transform of the $sin$ function would be $\frac{z^{-1}}{1+z^{-2}}$ but I don´t know what to do with $\frac{1}{\pi n}$ I am not sure if this is the right track since I would think there would be an easiest way to take the fourier expression to z expression, is there any equivalence for these rectangular or pulse functions?
Not every sequence that has a (discrete) Fourier transform has a $\mathcal{Z}$-transform, and not every sequence that has a $\mathcal{Z}$-transform has a Fourier transform. The sinc-sequence that you're referring to does indeed have a Fourier transform, but it does not have a $\mathcal{Z}$-transform.
If the $\mathcal{Z}$-transform of a sequence exists and if the region of convergence contains the unit circle, then the Fourier transform of the sequence simply is the $\mathcal{Z}$-transform evaluated at $z=e^{j\omega}$.