"We can introduce the concept of order of zero at infinity of a function $f(z)$ by considering the order of the zero at $z=0$ of the function $f(1/z)$. calculate the order of all the zero in $\overline{\mathbb{C}}$ of the functions:"
where $\overline{\mathbb{C}}$ is the extended complex plane which includes a point at $\infty$.
So, $$\frac{(z+1)^2}{z^3}$$ now i believe this to have a finite zero of order 2 and a pole of order 3. we know that for a meromorphic function that number of zeros in $\overline{\mathbb{C}}$ = the number of poles and so that suggests we have a single zero at infinity right?
$$f(1/z) = (1/z+1)^2z^3 = \frac{(1+z)^2}{z^2}z^3 = (1+z)^2z$$ and consider z = 0,
If this is correct then moving on, if we have a rational function say $f(z) = \frac{q_m (z)}{p_m (z)}$ where $$q_m (z) = \prod_{j \in [1,m]}(z-a_j)$$ and $$p_n(z)= \prod_{j \in [1,n]}(z-b_j)$$
where both $a_j$ and $b_j$ are all distinct. Considering a contour $\gamma$ in $\mathbb{C}$ such that all the points $(a_j)_{j \in [1,m]}$ and $(b_j)_{j \in [1,n]}$
then we apply the residue theorem to the integral, $$\int_{\gamma}\frac{f'(z)}{f(z)} dz = 2 \pi i (N-P)$$ where N is the number of zeros in f, counting multiplicity and P is the number of poles in f, counting multiplicity
Which since they're all distinct would be $(m-n)$ yes? or...we could consider the points at infinity in which case, we'd be looking at the poles and zeros of $$f(1/z) = \frac{q_m (1/z)}{p_m (1/z)} = \frac{\prod_{j \in [1,m]}(1/z-a_j)}{\prod_{j \in [1,n]}(1/z-b_j)} = \frac{\prod_{j \in [1,m]}(1-za_j)/z^{m}}{\prod_{j \in [1,n]}(1-zb_j)/z^n} = \frac{\prod_{j \in [1,m]}(1-za_j)z^{n}}{\prod_{j \in [1,n]}(1-zb_j)z^m}$$ which as we're only considering $z \rightarrow 0$ in this case we have a single pole (order m) and a single zero (order n) at $z=0$ for $f(1/z): z \rightarrow 0$
in which case $$\int_{\gamma}\frac{f'(z)}{f(z)} dz = 2 \pi i (P-N)$$
i understand this isnt a proof but this is mostly for my understanding as sometimes the more conceptual questions dont allow for as much physical manipulation...so am i correct in my thinking?
Thank you for taking the time to help.