Let \(f(z)\) be an analytic function on \({\mathbb C} \backslash \{ z_0 \}\), where \(z_0\) is a fixed point. Assume that \(f(z)\) is bijective from \({\mathbb C} \backslash \{ z_0 \}\) onto its image, and that \(f(z)\) is bounded outside \(D_r(z_0)\), where \(r\) is some fixed positive number. Show that there exist \(a, b, c, d \in \mathbb C\) with \(ad-bc \neq 0\), \(c \neq 0\) such that \(\displaystyle f(z) = \frac{az + b}{cz + d}\).
Solution
Since \(f\) is bounded for large \(|z|\), the singularity of \(f\) at \(\infty\) is removable. Hence \[ L:=\lim_{z\to\infty}f(z) \] exists and is finite.
The isolated singularity at \(z_0\) cannot be removable. If it were, \(f\) would extend to an entire function on \(\mathbb C\) with a finite limit at infinity, and therefore would be bounded and constant by Liouville’s theorem, contrary to injectivity. It also cannot be essential: by the Great Picard theorem, in every punctured neighborhood of \(z_0\) the function would assume all but at most one complex value infinitely often, again contradicting injectivity. Thus \(z_0\) is a pole.
Therefore \(f\) extends to a meromorphic function on the Riemann sphere with its only finite pole at \(z_0\), so \(f\) is rational. Because \(f\) is injective on \(\mathbb C\setminus\{z_0\}\), its degree is one: a rational map of degree \(d\ge2\) is generically \(d\)-to-one on the sphere. Hence \[ f(z)=\frac{az+b}{cz+d},\qquad ad-bc\ne0. \] Since \(f\) has the finite pole \(z_0\), the denominator is nonconstant, so \(c\ne0\). This is the required form.