Taylor series expansion having Z^-1

I'm getting error for 1/z as it takes 0 values for taylor series expansion while taking z transform

Answer 1

[Walter Roberson]

You cannot take a taylor series at one of its singularities -- so do not expand the taylor series around z = 0

You can take taylor series near a singularity, but if you do so then you will probably need a fairly high order to get any accuracy.

Could you confirm that z^-1 is intended to represent 1/z in the context? Is the context a laplace transform, in which case 1/z might be representing a derivative ?