Suppose we have a rational function $f(z)=\frac{p(z)}{q(z)}$ on the Riemann sphere, where $p,q$ are polynomials with no repeated roots, would the degree simply be the sum of the polynomial degrees of $p$ and $q$? (Sorry, I'm being silly here, I forgot the definition of degrees using local degrees.)
I'm thinking that if we consider local degrees, for each root $z_i$ of $p(z)$ of multiplicity $k$, the local degree at $z_i$ should be $k$ as $f(z)$ basically just wraps around $k$ times around $z_i$.
If $z_i$ is instead a root of $q(z)$, then the local degree of $(z-z_i)^{-k}$ should still be $k$, since we are mapping $0$ to $\infty$ which is a reflection, causing $(-1)\times(-k)=k$.
Is my reasoning correct?