Finding the residue of a function at $z_0$?

I encountered a particular question that led me to question the definition that I was given for a residue, after reviewing the literature I simply want to confirm that my understanding is correct. Note that all the specifics in this question are only used as illustrations and I am not looking for answers on how to calculate a residue for a given function. The problem that led me to this conclusion is brought below. Also, I know there are many questions here concerning residues, however, they deal with examples and not with definitions, thus I believe this is not a duplicate.

Am I correct in understanding that the definition of a residue implies that the function has a similarity at the point $$z_0$$ and therefore in given and function (not in a series form) and asked to calculate the residue at $$z_0$$ the expansion must be done around $$z_0$$.

Given the function $$f(z) = \frac{3}{z-4} - \frac{2}{z-2}$$ , the question is to find the Laurent series expanded at $$z=1$$ in the domain $$2<|z-1|<4$$. We are then asked if we can use this to find the residue at $$z=1$$, and finally, we are asked for the residue at the point $$z=0$$.

It is clear to me that the point $$z=1$$ is not in the domain in which we expanded the series and therefore we can't find the residue there.

However, for the finding, the residue at zero is what led me to question my understanding.

The Laurent series expanded about $$z=1$$ doesn't have a singularity at $$z=0$$. The answer states that the function is rational in the domain and therefore the residue is zero.

• What does " A function has a similarity at the point $\;z_0\;$ " mean? Could it be you meant "singularity"? – DonAntonio Feb 8 at 9:52

If $$z_0$$ belongs to the domain of $$f$$, where $$f$$ is an analytical function, then the residue of $$f$$ at $$z_0$$ is automatically $$0$$, since the Laurent series of $$f$$ centered at $$z_0$$ is, in fact, a power series. In particular, the coefficient of $$(z-z_0)^{-1}$$ is $$0$$.
So, yes, the residue of your function $$f$$ both at $$0$$ and at $$1$$ is $$0$$.
The function $$f(z) = \frac{3}{z-4} - \frac{2}{z-2}$$ is holomorphic on $$D= \mathbb C \setminus \{2,4\}$$, hence $$f$$ has in $$1$$ and $$0$$ removable singularities. Therefore the residues in $$1$$ and $$0$$ are zero.