Let \(p(z) = \frac{\prod_i^n z-a_i}{\prod_j^m z-b_j}\) with all \(a_i, b_j\) distinct. What is its topological degree?
Solution
1. Interpreting \(p\) as a rational map of the Riemann sphere,\[ p:\mathbb{CP}^1\longrightarrow\mathbb{CP}^1, \]its topological degree is\[ \boxed{\max\{n,m\}}. \]
Write \(P(z)=\prod_{i=1}^n(z-a_i)\) and \(Q(z)=\prod_{j=1}^m(z-b_j)\). Since all \(a_i,b_j\) are distinct, \(P\) and \(Q\) have no common factor. The rational map is represented in homogeneous coordinates by two homogeneous polynomials of the common degree \(d=\max(n,m)\): \[ [Z:W]\longmapsto \begin{cases} [W^{d-n}P_h(Z,W):W^{d-m}Q_h(Z,W)] \end{cases} \] after placing the necessary power of \(W\) in the lower-degree component. These homogeneous forms have no common zero on \(\mathbb{CP}^1\). A generic value therefore has exactly \(d\) preimages counted with multiplicity. Holomorphic maps preserve orientation at regular points, so the topological degree equals this algebraic count, namely \(d\).
2. In particular, if the intended map is instead the restriction of \(p\) to a large circle with values in \(\mathbb C^*\), its winding number is \(n-m\).
On a sufficiently large circle enclosing every zero and pole, the argument principle gives winding number equal to the number of zeros minus the number of poles, namely \(n-m\). This is a different notion from the degree of the rational self-map of \(S^2\) in 1.