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. =)

I Refuse to Name This "Complex Analysis...Made Simple": 4

Hey guys! Having fun with your complex analysis assignment? =P

I've just got a few random comments that I wanted to make, and it's a bit random, but I think that you'll find it useful.

Comment 1: Taylor series/Laurent series expansions are AMAZEBALLS.

Most of the functions that you've encountered up until this point in life are super-super nice functions like \(\sin(x),\cos(x),\exp(x),\log(x),\frac{1}{1+x^n}\) which all have nice Taylor series expansions and what have you. And so when you see this absolutely incredible fact that "complex differentiability of a function in a neighbourhood" implies "(local) Taylor expansions", you mightn't have an appreciation for how "phwoar" this is. In particular, you mightn't have realised that a smooth (infinitely differentiable) real function mightn't have Taylor expansions! In fact, "most" (*) real functions don't have Taylor expansions. The annoying thing is actually writing down one of these smooth real functions. Soo, here's one that I prepared earlier: \begin{align} f:\mathbb{R}\rightarrow\mathbb{R},\, f(x)=\left\{\right. \begin{array}{cc} \exp(\frac{-1}{x^2}) & \text{ if }x>0, \\ 0 & \text{ if }x\leq 0. \end{array}\end{align} Now, I'm going to leave it as an exercise for you guys to check that this function is actually smooth everywhere along the real line (**). 

Now, assume for a second that this function had a Taylor expansion around \(x=0\). Well, the fact that \(f(x)\) is constant on the negative real line tells us that this Taylor expansion must just be \(0\), but then the fact that \(f(x)\) is non-zero on the positive real line contradicts this expansion. So, we see that this function has no Taylor expansion around \(x=0\).

So, at this point you might think that well, that was an okay example, but \(f(x)\) still has Taylor expansions at all the other points in its domain. Then, here's a function that's smooth but doesn't have a Taylor expansion ANYWHERE. (insert spooky music)

And so, you see, the fact that complex differentiability (and unlike the real case, you only need to check the first derivative) implies Taylor series expansions is TOTES AMAZEBALLS.


Comment 2: Question 20 of Practice Sheet 6 vs Question 24 of Practice sheet 8.

If you compare these two questions and the answers that you get for them (SPOILERS: YOU GET THE SAME ANSWER), you see that in fact: \begin{align} \int_0^{2\pi}\frac{\sin(t)}{(2+ \cos(t))^2}\mathrm{d}t=0.\end{align} This is annoyingly obvious in retrospect - you could have totally just solved this real integral explicitly to get \(0\) instead of spending 30~40 minutes labouring over Question 24. But in fact, the fact that this integral is \(0\) can be generalised somewhat to the following fact: \begin{align} \int_0^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t=0.\end{align} Let's prove this: \begin{align} &\int_0^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_\pi^{2\pi} \sin(t) f(\cos(t))\mathrm{d}t\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_{\pi}^{0} \sin(2\pi-\tau) f(\cos(2\pi-\tau))\mathrm{d}(2\pi-\tau),\\ &\text{ with the substitution }\tau=2\pi-t,\\ =&\int_0^{\pi} \sin(t) f(\cos(t))\mathrm{d}t+\int_\pi^{0} \sin(\tau) f(\cos(\tau))\mathrm{d}\tau=0.\end{align} In fact, this really shouldn't be a surprise to anyone who's played around with Fourier series, but it's still a cute fact. =)

Also, here's something-else nice about Question 20 that Andre Ratiu (the other tutor for this subject) pointed out to me: \begin{align} \int_0^{2\pi} \frac{1}{(2+\cos(t))^2}\mathrm{d}t=\int_0^{2\pi} \frac{1}{(2+\sin(t))^2}\mathrm{d}t \end{align} And here're two ways to see this (without doing the actual computation):

Method 1 (real analysis): the function \(\frac{1}{(2+\cos(t))^2}\) is periodic, and \(\frac{1}{(2+\sin(t))^2}\) is just a \(\frac{\pi}{2}\) shift of this. And you should be able to prove that for periodic function \(g(t)\) with period \(p\) that \begin{align}\int_0^p g(t)\mathrm{d}t=\int_b^{p+b} g(t) \mathrm{d}t\text{, for any }b\in\mathbb{R}.\end{align}

Method 2 (complex analysis): transform the two integrals into contour integrals around the unit disk on the complex plane and observe that the functions are rotations of each other by \(\frac{\pi}{2}\). It's the same idea, as the periodicity thingy above, but phrased complex analytically. =)

Orrite, that's...been a fruitful morning of procrastination for me...time to get back to doing...other stuff. =P

*: the space of functions with Taylor expansions around a certain point \(x_0\in\mathbb{R}\) is countably infinite dimensional, whereas the space of smooth real functions is uncountably infinite dimensional.

**: This is obvious everywhere except at \(x=0\). One hint that I could give you is to first show that all of the derivatives of \(\exp(-x^{-2})\) are in the form of \(P(x^{-1})\exp(-x^{-2})\), where \(P\) is a polynomial.

I Refuse to Name This "Complex Analysis...Made Simple": 3

Hey guys! Let us begin somewhat worryingly: with an apology.

...in the dark corridoors of my decrepit heart, a solitary ray of light lingers on. It shines upon my macabre soulscape, and whispers forth a faint, but grave exhortation: "...the truth! By God they deserve the truth!..."

That's right - you've been lied to! AND I AM SO SORRY. Specifically, you've been taught that a complex function \(f: U\rightarrow\mathbb{C}\) is a called analytic on \(U\) if it's complex-differentiable at every point on \(U\). BUT THIS IS A LIE! This is in fact the definition for holomorphicity. So, then, what the heck is an analytic function? Well, a function \(f:U\rightarrow\mathbb{C}\) is analytic if it can be locally written expressed as a convergent power series at every point in \(U\). So why the mixed up in nomenclature? To be honest, I have no idea - but I will say this: it turns out that the two criteria are equivalent - a function is holomorphic if and only if it's analytic!

*Phew* glad I got THAT off my chest. =P

There're a few cool consequences of this equivalence between holomorphic and analytic functions. Firstly, power series are infinitely complex-differentiable, so holomorphic functions are infinitely differentiable! This is kinda freaky, coz the definition of holomorphic only requires that \(f\) be complex-differentiable - it doesn't even say that the derivative should be continuous...and yet...we are, basking in its smug infinite differentiability.

Secondly, power series are basically just like beefed up polynomials, right? And we should know from highschool algebra that any two order \(n\) polynomials which agree on \(n+1\) points have to agree everywhere. Well, that observation also beefs up for analytic functions: any two analytic/holomorphic functions which agree on a countable collection of points with an accumulation (limit) point agree everywhere!

We can also use this characterisation to help us get a better feel for some of the results or phenomena that we see in complex analysis. You've seen from your tute questions that it is possible to have a vanishing contour integral of a function \(f\) around some \(\Gamma\) bounding a simply connected domain \(U\) so that \(f\) has a pole inside of \(U\). Take for example the following integral: \begin{align}\oint_{|z|=1}\frac{\mathrm{d}z}{z^2}=\int_0^{2\pi}\frac{i\exp(i\theta)}{\exp(2i\theta)}\mathrm{d}\theta=\int_0^{2\pi}i\exp(-i\theta)\mathrm{d}\theta=0.\end{align} So, is there a nice criterion for when such integrals will vanish? Well, let's assume (to make things easy) that the pole for \(f\) is positioned at \(0\) and that the pole is of order \(n\). This means that \(z^n f(z)\) is holomorphic at \(0\) and hence has a power series expansion at this point: \begin{align} z^nf(z)&=a_0+a_1z+a_2z^2+\ldots+a_{n-1}z^{n-1}+a_nz^n+\ldots\\ f(z)&=a_0z^{-n}+a_1z^{1-n}+a_2z^{2-n}+\ldots+a_{n-1}z^{-1}+a_n+\ldots\end{align} Now, doing a contour integral around the unit circle \(|z|=1\), we expect everything to vanish except possibly \(\oint_{|z|=1}a_{n-1}z^{-1}\mathrm{d}z\). This term contributes \(2\pi i a_{n-1}\) to the integral, and so a contour integral around a pole of a function vanishes precisely when the "power-series" has no \(z^{-1}\) term. In fact, the coefficient of the \(z^{-1}\) term is so important that we've given it a name: it's called the residue.

Soooooooooo...complex analysis...so simple, right? =)

I Refuse to Name This "Complex Analysis...Made Simple": 2

Complex differentiability, what's up with that, ammirite?

So, it turns out that if a complex function \(f: U\rightarrow \mathbb{C}\) is complex differentiable for an open plane region \(U\subset\mathbb{C}\), then \(f\) can be written as a power series around each point in \(U\). What is UP with that, ammirite??

Plus, once a function is complex differentiable ONCE, it's complex differentiable INFINITELY MANY TIMES. WHAT IS UP WITH THAT, AMMIRITE???

Wow...tough crowd, *awkward turtle time*. =P

Now, in order to show that a function is complex differentiable, you just need to write down its derivative (via first principles) and show that the limit exists. And here's where the complex case really differentiates itself from the real case. *insert smug smile for that pun*

Well, for one thing, we know that differentiability implies continuity. So, let's backtrack for a second and look at the continuity of a function. In the real case, to check continuity of a function \(g:\mathbb{R}\rightarrow\mathbb{R}\) at a point \(x_0\in\mathbb{R}\), we just need to check that the limit of \(g\) as we approach \(x_0\) from the left hand side of \(x_0\) agrees with what we get as we approach from the right hand side of \(x_0\). For complex functions from a planar domain, it's trickier. We've obviously got to check that the limit agrees in every which way, and there are uncountably infinitely many directions in which you can approach some point \(z_0=x_0+iy_0\in U=\mathrm{Dom}(f)\). And in fact, it's not even enough to just check the "linear directions"...as can be seen from the following example: \begin{align}f(x+yi)=\frac{x^2y}{x^4+y^2}.\end{align} Taking the limit as we approach \(0\) from any linear direction \(\alpha\in\mathbb{C}-\{0\}\), we get \(0\). Concretely written out: \begin{align}\lim_{t\to 0^+} f(t\alpha)=0.\end{align} However, taking the limit along a non-linear curve, we see that \begin{align}\lim_{t\to 0^+} f(t+t^2i)=\frac{1}{2}.\end{align} Thus, we see that this particular function \(f\) is not continuous at \(0\).

In order to check continuity in general, we really have to check every direction. This is done by going through an \(\epsilon-\delta\) convergence proof.

Um...ookay, maybe I'll talk more about complex differentiability next time. Or possibly when you guys start to look at them sweet sweet power series results!

I Refuse to Name This "Complex Analysis...Made Simple": 1

It's been AGES since I've blogged anything *flails limbs about, cracks knuckles*, let's start off with something nice and easy. Let's do some complex analysis! Caveat: this post is actually just recap of something that I said during the Tuesday tute (*) and something that I said to a few students who asked me about some big-picturification stuff after the Thursday tute.

So, if you're in either of my MAST30021 tutorials, you might recall question 1)c) from last week's tute:

"Find all solutions of \(e^z=-1\) with \(z\in\mathbb{C}\). How many solutions are there? Is the expected solution \(z=\log(-1)\) single-valued? How can we make the logarithm function single-valued?"

Now, you all saw that \(z=(2k+1)\pi i, k\in\mathbb{Z}\) is the solutions to \(z=\log(-1)\). And so, \(\log:\mathbb{C}-\{0\}\rightarrow\mathbb{C}\) is totes a multi-valued function. Now, we tend to like our functions single-valued, so how do we fix this? Well, we see that the problem is with the fact that \(e^{2\pi i}=1\), so when trying to solve for logarithms \(w=\log(z)\), given a solution \(w_0\), we can always just add or subtract \(2\pi i\) and get a new answer (and lather and repeat). The most straight-forward way to fix this is to force our logarithms to only spit out principal arguments. Let's motivate our eventual answer a little. We know that the usual real logarithm has the following algebraic property: \(\log(ab)=log(a)+log(b)\). Hence, by putting an arbitrary non-zero complex number \(z\) in polar form \(re^{i\theta}\), we expect that: \begin{align}\log(z)=\log(re^{i\theta})=\log(r)+\log(e^{i\theta})=\log(|z|)+arg(z).\end{align} And taking the principal argument, we define a function \begin{align}Log:\mathbb{C}-\{0\}&\rightarrow\{x+yi\in\mathbb{C}\mid y\in(-\pi,\pi]\}\subset\mathbb{C}\\ z&\mapsto \log|z|+Arg(z).\end{align}
Here's an attempt at illustrating what happens with this map:

I hope that the colours help you visualise what's happening in this map. Roughly, we're "cutting" \(\mathbb{C}-\{0\}\) along the negative real axis, and then stretching it out so that the region close to \(0\) stretches out and the region close to \(\infty\) pull in to form a horizontal strip.

Now, this is a perfectly reasonable fix to make a multi-valued function \(\log\) into a single-valued function. But it isn't...perfect, for the simple reason that this map isn't continuous along the negative real axis (this is an exercise, try it!). To restore continuity (in this course), we can remove the negative real axis from the domain - this is the branch cut stuff that you've seen in your slides.

Of course, there's something dissatisfying about this "fix" - it's not canonical. What I mean is that this fix isn't unique. See, instead of defining our principal arguments as \((-\pi,\pi]\), if we had taken \((0,2\pi]\) as our principal arguments, then our \(Arg\) and \(Log\) functions would be slightly different, and the branch cut that we needed to take (to get continuity of \(Log\)) would need to be the positive real line instead. Maybe you should try to take the principal arguments as \((\epsilon,2\pi+\epsilon]\) and see what branch cut you'd have to take?

So, there's totes a canonical fix for this. And the idea is to take that horizontal strip constituting the image of \(Log\) and to glue the top to the bottom. This makes for an infinitely long cylindrical pipe, which...doesn't quite fit on the complex plane, but at least makes \(Log\) a nice and continuous map without needing to take any branch cuts and stuff. This infinitely long cylinder is an example of what (some) mathematicians call Riemann surfaces, and they are at the heart of modern complex analysis.

Sooo...complex analysis, so simple right?  =P

*: Coz I sorta glossed over it, and I hope that this is expositionally clearer.