Order of a meromorphic function at $\infty$ Let $f$ be meromorphic on the whole extended complex plane $\overline{\Bbb C}$.
Now my teacher wrote
$$
\text{ord}_{\infty}f=-\frac1{2\pi i}\int_{|z|=R}\frac{f'(\zeta)}{f(\zeta)}\,d\zeta
$$
and I can't understand if it is a definition or we can derive this from the fact that, reasonably we can define
$$
\text{ord}_{\infty}f:=\text{ord}_0g\;\;\;,
$$
where $g(z):=f(1/z)$.
In the latter case, we should be able to prove that
$$
\text{ord}_0g=-\frac1{2\pi i}\int_{|z|=R}\frac{f'(\zeta)}{f(\zeta)}\,d\zeta
$$
but I'm not able to prove this last equality.
Can someone shade a light please?
 A: Recall, by the argument principle, that
$$
\frac 1{2\pi i}\int_{|z|=R}\frac{g'(\zeta)}{g(\zeta)}d\zeta=N-P
$$
where $N$ is the number of zeros of $g$ inside the contour (counted with multiplicity), and $P$ is the number of poles of $g$ inside the contour (where a pole of order $n$ is counted $n$ times). Since $f$ is meromorphic on $\overline{\mathbb C}$, the zeros and poles of $g$ are isolated, so there exists some small $R>0$ such that $\{z:|z|<R\}$ includes only the pole of $g$ at $0$. Hence, the integral for this $R$ will yield $-\text{ord}_{0}g$. This means
$$
\text{ord}_{\infty}f=-\frac 1{2\pi i}\lim_{R\to 0}\int_{|z|=R}\frac{g'(\zeta)}{g(\zeta)}d\zeta.
$$
Note that
$$
g'=\frac d{dz}(f^{-1})=-\frac{f'}{f^2}\implies \frac{g'}{g}=-f\cdot \frac{f'}{f^2}=-\frac{f'}{f}
$$
and so
$$
\text{ord}_{\infty}f=\frac 1{2\pi i}\lim_{R\to 0}\int_{|z|=R}\frac{f'(\zeta)}{f(\zeta)}d\zeta.
$$
It seems you have a negative sign added, which makes sense if your convention is that $\frac 1{z^2}$ has a pole of order $-2$ at the origin, rather than $2$.
