So I was asked to find the Laurent series of the expression around $z=2$: $$f(z)= \frac{z+3}{(z-2)^3}$$
I am aware that this can be decomposed as $$5\, \left( z-2 \right) ^{-3}+ \left( z-2 \right) ^{-2}$$ Is this the entire Laurent series of $f(z)$ around $z=2$. I am asking because in order to derive this I haven't even used the fact that expansion is being carried out about $2$. If so why?
".. about $z=2$" means you have to derive an expression of terms $(z-2)$. In this particular case, the fraction decomposition leads to
$$f(x) = \frac{z+3}{(z-2)^3} = \frac{1}{(z-2)^2} + \frac{5}{(z-2)^2}$$
which is already an expanded expression involving $(z-2)$ terms (of the form $(z-2)^n, n \in \mathbb Z$) , thus a Laurent expansion of the function $f(z)$. But, each Laurent expansion for a function is unique (check the adequate theorem in your notes). Thus this is the desired result.
Now, regarding to your comment for an expansion about $z=1$, this is a different case and you should work on your initial $f$ to create $(z-1)^n, n \in \mathbb Z$ terms.