# Confused on the ROC of Z-transform

I'm given a specific function to find its inverse Z-transform. Specifically: $$X(z) = z^3 + \frac{1}{z - 2i} + \frac{1}{z+2i}, |z| > 2$$ Notice the $$|z| >2$$. Now what concerns me is the term $$z^3$$. It's a standard $$Z$$-transform that $$\delta[n-m] \stackrel{\mathcal{Z}}{\longleftrightarrow} z^{-m},$$ with its ROC being every $$z$$ except $$\infty$$ if $$m <0$$, as is the case here. So, doesn't this transform contradict the given fact that |z|>2? Because If I were to use that transform that'd mean that I would change the given ROC of $$X(z)$$ not to include $$\infty$$.

Since these problems are designed from other people couldn't it be that they just forgot that tiny detail and should've changed the starting ROC to $$2 < |z| < \infty$$? This is basically what I'm asking. If that's not the case, does there exist an inverse $$Z$$-transform of $$z^{3}$$ so that it will also include $$\infty$$ and not contradict the given known ROC?

I agree with you, it does not converge when $$z=\infty$$ or when $$z\le2$$ if we assume the inverse z-transform solution is anti-causal (as it would need to be from the $$z^3$$ term).
There is no inverse z-transform that also includes $$z=\infty$$ for $$X(z)=z^3$$, and given the inverse z-transform of this component alone is $$u[n+3]$$, which is non-causal, it can only exist if the ROC is less than $$\infty$$. The next two components with poles at $$\pm2j$$ would have a causal inverse z-transform if the $$ROC>2$$.