Complex 'Extensions'

June 07, 2021 | 5 minutes, 43 seconds


This post relies on my previous post on principal roots - or in this post, particularly the \(\mathfrak{P}_k(z)\) - aka the Principal Equivalent - function. If you haven't given it a read, I'd strongly advise doing so.

In the previous post I sort of extended our idea of principal roots and piecewise functions to the complex plane. That is, I took a complex number and raised it to some power, then took the first principal p root of that. This is also a demonstration that our typical power laws don't traditionally hold over the complex numbers. With that being said though, it does show you a sort of mapping of complex numbers to a single particular region. That is, we rotate every complex number until it's within the region we want - a sort of principal region, if you will.

Suppose we now opt to focus on the very thing we start with - the absolute value of a number. I briefly discussed last time that for all real values, \(\left|x\right|=\mathfrak{P}_{2k}(x)\) for integer k greater than 0. Suppose that our goal is to preserve the idea that we split our numbers into a few regions - the negatives, the zero(es) and the positives. Suppose we also want to preserve the ideas that build off the absolute value, such as the maximum function and the minimum function. What we therefore do is consider the simplest but also most appropriate function and idea over the complex numbers - that \(\left|x\right|=\mathfrak{P}_2(x)\).

(Disclaimer - the 'complex absolute value' is generally accepted as the norm of the complex number, while the absolute value on \(\mathbb{R}\) is a one dimensional norm. So when I say 'complex absolute value' in this post - I'm talking about the absolute value function extended to the complex numbers as previously defined... as not-entirely-correct as it is.)

Our goal now: To create new functions - sort of extensions - that exist on \(\mathbb{C}\).



\[ \mathfrak{P}_2(z) = \begin{cases} z & \operatorname{Arg}(z)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right] \\ -z & \operatorname{Arg}(z)\in\left(-\pi,-\frac{\pi}{2}\right]\cup\left(\frac{\pi}{2},\pi\right] \end{cases}\]


We have previously derived the formula for the maximum of two real numbers. That is, \(\max(a,b)=\frac{1}{2}\left(a+b+\left|a-b\right|\right)\).

If we have a maximum function on \(\mathbb{C}\) - let's call it \(\operatorname{maxW}\) - then using the definition of \(\left|z\right|\) (as above, not the norm) we have \(\operatorname{maxW}\left(z,w\right)=\frac{1}{2}\left(z+w+\mathfrak{P}_2\left(z-w\right)\right)\).

In piecewise form, we have:

\[ \operatorname{maxW}\left(z,w\right) = \begin{cases} z & \operatorname{Arg}(z-w)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right] \\ w & \operatorname{Arg}(z-w)\in\left(-\pi,-\frac{\pi}{2}\right]\cup\left(\frac{\pi}{2},\pi\right] \end{cases}\]

It is here we realise that had we used \(\mathfrak{P}_4(z)\) as the definition of the complex absolute value, for example, we would have gotten values that are not equal to either \(z\) nor \(w\) - and therefore not preserving the idea we wanted. With that being said, with this definition, it becomes apparent after experimentation that this function we've created almost seems lexicographic on the complex numbers. We'll come back to that later.


We can similarly define minimum:

\[ \operatorname{minW}\left(z,w\right) = \begin{cases} w & \operatorname{Arg}(z-w)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right] \\ z & \operatorname{Arg}(z-w)\in\left(-\pi,-\frac{\pi}{2}\right]\cup\left(\frac{\pi}{2},\pi\right] \end{cases}\]

Alternative form

Consider \(\operatorname{maxW}(z,w) = z\). This implies that \(\operatorname{Arg}(z-w)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right]\), which occurs when either \(\operatorname{Re}(z-w)>0 \implies \operatorname{Re}(z)>\operatorname{Re}(w)\) or \(\operatorname{Re}(z-w)=0\land \operatorname{Im}(z-w)\geq 0 \implies \operatorname{Re}(z)=\operatorname{Re}(w) \land \operatorname{Im}(z)\geq\operatorname{Im}(w)\).

Similarly consider \(\operatorname{maxW}(z,w) = w\). We have that \(\operatorname{Re}(z)<\operatorname{Re}(w)\) or \(\operatorname{Re}(z)=\operatorname{Re}(w)\land \operatorname{Im}(z)\leq\operatorname{Im}(w)\).

Putting this together we simply have:

\[ \operatorname{maxW}(z,w) = \begin{cases} z & \operatorname{Re}(z)>\operatorname{Re}(w) \\ w & \operatorname{Re}(z)<\operatorname{Re}(w) \\ z & \operatorname{Re}(z)=\operatorname{Re}(w) & \operatorname{Im}(z)\geq\operatorname{Im}(w) \\ w & \operatorname{Re}(z)=\operatorname{Re}(w) & \operatorname{Im}(z)\leq\operatorname{Im}(w) \end{cases}\]

By our observation before, this is pretty much lexicographic (if such a silly notion on the complex numbers exists). A similar construction can be made for \(\operatorname{minW}\) - such an exercise is left to you if you're reading this.

It is however important to note that we aren't defining a specific ordering of complex numbers - e.g. \(z\geq w\) - in the traditional sense. These are again extensions of functions normally found on \(\mathbb{R}\), which is pretty cool (in my opinion). What is cooler is that we stumbled into one of the simplest ideas of ordering (that is, if I were to ask you how you would "order" the complex numbers, lexicographic ordering would likely be your answer or one of your answers) - through a fairly simple extension of our pre-existing piecewise functions. Also notice that the properties of our original maximum and minimum functions are fully preserved by these extensions.


Attempt to solve for \(x\in\mathbb{R}\) such that \(\operatorname{max}\left(x,x^2+1\right)=0\).

By a typical argument, \(x^2+1\geq x,\ \forall x\in\mathbb{R}\) (proof is given by letting \(f(t)=t^2-t+1\) and going from there), and therefore \(x^2+1 = 0\). There are clearly no solutions in the reals to this equation, but over the complex numbers, possible results could be \(x=i\) and \(x=-i\).

So then instead solving for \(z\in\mathbb{C}\) such that \(\operatorname{maxW}(z,z^2+1)=0\). We can argue the domain of \(\operatorname{Arg}\left(z-z^2-1\right)\) - which constitutes a proof of the solution - but it is easier to simply try each case. That is, suppose \(z=0\) firstly - this doesn't yield a valid solution. Then suppose \(z^2+1=0\implies z\in\left\{-i,i\right\}\). If we set \(z\) to each value appropriately, we find that only \(-i\) satisfies this equation. So \(\operatorname{maxW}(z,z^2+1)=0\implies z=-i\) because \(\operatorname{maxW}(-i,0)=0\).

A similar question could be something like \(\operatorname{maxW}\left(z^2,z^2+1\right)=0\) or another \(\operatorname{maxW}\left(z+1,z^2+1\right)=0\). Have a go at trying to work these out as an exercise.