I Refuse to Name This "Complex Analysis...Made Simple": 5 (final)

Well, well well, wellity well well...I have no idea what to write about for this last blog entry. =/

Let's talk about...Riemann spheres!

So, at some point in this complex analysis course, you'll have learned that we really ought to think of the complex plane \(\mathbb{C}\) as lying on the Riemann sphere \(\hat{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}\). Doing so means that meromorphic functions \(f:\mathbb{C}\rightarrow\mathbb{C}\) (i.e.: they have poles) may be thought of as holomorphic functions \(f:\mathbb{C}\rightarrow\hat{\mathbb{C}}\). Moreover, if the behaviour of \(f(z)\) is nice enough at \(z=\infty\) (i.e.: no essential singularities), then \(f\) can be extended to a holomorphic function \(f:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}\).

One class of these holomorphic functions between Riemann spheres that you've encountered in this course are Mobius transformations: \begin{align} z\mapsto\frac{az+b}{cz+d}\text{, where }ad-bc\neq0.\end{align} You've already seen some nice facts about them. You know that they are conformal (definition: angle-preserving) transformations, that they take circles/lines to circles/lines and you know that they're bijections. In fact, they are precisely the collection of all bijective holomorphic maps between Riemann spheres.

Let's prove this!

Consider a bijective holomorphic function \(f:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}\). Since this is bijective, let \(a,b\in\mathbb{C}\) respectively denote the points which map to \(0=f(a)\) and \(\infty=f(b)\). We may assume by precomposing \(f\) with a Mobius transformation that \(b\neq\infty\). Now \(z=a\) must be an order 1 zero, or else there'd be a small neighbourhood around it where two points map to some \(\epsilon\) close to \(0\). Likewise we can argue that \(z=b\) must be an order 1 pole of \(f\) by considering the function \(f(\frac{1}{z})\). Thus, the function \begin{align}g(z)=\frac{(z-b)f(z)}{z-a}\end{align} is a holomorphic function on \(\hat{\mathbb{C}}\) with neither zeros nor poles. In particular, its absolute value function \(|f|\) is a continuous function on a compact domain (because spheres are totes closed and bounded) and hence must be bounded. Thus, by Liouville's theorem, the function \(g(z)\) is equal to some non-zero constant \(c\in\mathbb{C}\): \begin{align}f:z\mapsto\frac{cz-ac}{z-b}.\end{align} Finally, \(-bc+ac\neq0\) since \(c\neq0\) and \(a=b\) would contradict the bijectivity of \(f\). Thus, this is totes a Mobius transformation.

Note that using the same arguments, we can show that any holomorphic map between Riemann spheres takes the form of a rational function: \begin{align} \frac{a_0+a_1z+a_2z^2+\ldots+a_nz^n}{b_0+b_1z+b_2z^2+\ldots+b_mz^m}!\end{align}
In fact, I was going to play around with a few polynomials and count branch points and what not, leading us to the Riemann-Hurwitz formula; but it's getting late, and we really ought to give this semester a rest. =)