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.

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