Let $f: S^{2} \to S^{2}$ be a mapping given by the rational function $$\frac{a_{0} + a_{1} z + \cdots + a_{n} z^{n}}{b_{0} + b_{1} z + \cdots + b_{m} z^{m}}$$ where the numerator and denominator have no common roots and $a_{n} b_{m} \neq 0$. What is the degree of $f$? Take the definition of degree given in class.
The definition of degree, as given in class, is as follows:
Definition: Let $w$ be a regular value of the mapping $f$. Then define $$ \text{ind}_{w}(f) = \sum_{z \in f^{-1}(w)} n(z)$$ where $n(z) = 1$ if $\mathrm{d}f$ preserves orientation at $z$, and $n(z) = -1$ otherwise. $\text{ind}_{w}(f)$ is independent of $w$, and it is the degree of $f$.
I know that $\mathrm{d}f$ preserves orientation if the determinant of the matrix $\frac{\partial (u,v)}{\partial (x,y)}$ is positive, where $f = u + iv$. However, I am having trouble separating the rational mapping into its real and imaginary parts. Is there another way to check whether or not $\mathrm{d}f$ preserves orientation?
Any help/advice with this question is much appreciated.
There is no need to do so.
At all points $z_0$ where the denominator is not zero, $f$ is complex differentiable and therefore satisfies the Cauchy-Riemann equations: $$ \det \frac{\partial (u,v)}{\partial (x,y)}(z_0) = \begin{vmatrix} u_x(z_0) & u_y(z_0) \\ v_x(z_0) & v_y(z_0) \end{vmatrix} = \begin{vmatrix} u_x(z_0) & -v_x(z_0) \\ v_x(z_0) & u_x(z_0) \end{vmatrix} = u_x^2(z_0) + v_x^2(z_0) = |f'(z_0)|^2 $$ which means that $\mathrm{d}f$ is orientation preserving at all points where the derivative does not vanish.
Since $f'$ has only finitely many zeros, one can choose $w$ such that $f$ takes the value $w$ only with multiplicity one, i.e. $f'(z) \ne 0$ for all $z \in f^{-1}(w)$.
Then $\text{ind}_{w}(f)$ is just the number of solutions of $f(z) = w$, which can be determined as a function of the degrees of numerator and denominator.