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. =)