There is one problem that I am solving at the moment and I am a bit confused. I will explain what I do, and what I am thinking, so if there is mistake, please point them out.
Problem : Find inverse z transform (bilateral) of function $F(z) = \frac{1}{(z-1)(z-2)(z-4)}$ with region of convergence $|z| > 4$
I create new function $G(z) = \frac{z^{n-1}}{(z-1)(z-2)(z-4)}$, and now if I am correct, there are three poles for $n>0$, all of them first order and using residue theorem, I can find $f(n)$ for $n>0$ and I can write $f(n) = result * \theta(n-1)$, where $\theta(n)$ is Heaviside theta function (http://reference.wolfram.com/language/ref/HeavisideTheta.html)
So, that should be first part of solution. For $n=0$ case, I have $f(n)=0$, so there is no second part of solution.
Now comes the problematic part. Since $z^{n-1}$ will go to the denominator for $n\leq 0$ and will introduce pole in $z=0$ with order $1,2,...$
How can I calculate this third part of $f(n)$ ? Since order of pole is changing, I don't know is it possible to use residue theorem ?