Sorry for not providing context here.
Suppose I have an output $Y(z)=\frac{z-1}{z}$ and input $X(z)=\frac{z^2+3z+2}{z^2}$ to yield a transfer function
$$T(z)=\frac{Y(z)}{X(z)}=\frac{\frac{z-1}{z}}{\frac{z^2+3z+2}{z^2}}=\frac{z-1}{z}\frac{z^2}{2z^2+3z+2}.$$
If you are finding the zeroes of this we have that $z=1$ is a zero.
My question is, is $z=0$ one also?
My problem is that I see a symbolic manipulation to
$$T(z)=\frac{z(z-1)}{2z^2+3z+2}$$
giving the impression that $T$ has a zero at $z=0$ but in my mind $T$ is not defined at $z=0$... although we do have
$$\lim_{z\rightarrow 0}T(z)=0.$$
In other words do we consider $z=a$ a zero of a transfer function if
$$\lim_{z\rightarrow a}T(z)=0\,\,??$$
Yes. In complex analysis it is customary (conventional) that all removable singularities are removed. And more generally, all analytic functions are extended to the largest domains that they can be by analytic continuation.
The transfer function, in this case, is more than just the formula used to define it.