Csdrhrt

When it's $frac z{z-1}$ the answer in the discrete domain can be found in the general tables = $u[n]$, but for $frac z{z+1}$ I can't find the rule. Some examples say it's $(−1)^nu[n] $ or $a^kcos(kpi)$, which one is it and why?

To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.

You can write:

$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$

which is the $Z$ transform of $(−1)^nu[n]$

but also you can write

$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$

which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$

which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$

So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$

Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.

$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$