Because the unilateral ZTransform $Z(f[n]) ) = \sum_{n=0}^{\infty}f[n]z^{-n}$ with $f[n]=-a^n u(-n-1)$ has $f[n]=0, n\geq0$ because the Heaviside function $u(-n-1) = 1$ only for $n<-1$. But this still does not address your original question, why the InverseZTransform disregards the area of definition given in the assumptions. A dull and unbelievable explanation could be: it knows it has to handle only input wich was generated by the unilateral `ZTransform[]` - but that's so absurd that I wish I hadn't typed it ... just for discussion: You know you can compute contour integrals with Mathematica specifying the path on your own.

@Udo Krause can you explain me why the unilateral z transform of the anticausal sequences does not exist? Thank you very much.

As Mathematica defines it InverseZTransform The inverse Z transform of a function `F(z)` is given by the contour integral `1/(2\[Pi] i) \[ContourIntegral]F(z)z^(n-1) \[DifferentialD]z`. it seems to me meaningless - or even inadequate - to define a region of convergence like `Abs[a] > Abs[z]` because the contour has to walk a closed path around the singularity $a = z$ in the complex plane. It seems that Mathematica disregards assumptions of such type: In[4]:= InverseZTransform[z/(z - a), z, n] Out[4]= a^n

[?] Inverse Z-Transform of z/(z - a) with different region of convergence?