Let $f(z)=\frac{p(z)}{q(z)}:U\to \hat{\mathbb{C}}$ be a non-constant rational function. Then for any $a\in \hat{\mathbb{C}}$, the size of the fiber at $a$ doesn't 1;;depend on our choice of $a$ (when counted with multiplicity).

In particular, we need only show that the number of zeros of $f$ is equal to 1;; the number of poles. To do this, the key part of our function being rational is that we may apply a transformation of the form $f(c_{0}z+\frac{b_0}{z})$ and still get a rational function, so that we can assume $f(\mathrm{\infty })\ne 0,\mathrm{\infty }$.