Tuesday, 25 November 2014

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. =/

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. =)