I know that $\mathbb{C}$ is an unordered field and that (strictly non-real) complex numbers cannot be 'compared' in the sense that one is less than/greater than another.

However, we can compare real numbers; geometrically, this is because they lie on a straight line through the origin, $0+0i$, and one real is greater than another if it's 'further' along the real line than the other.

I'm wondering whether, given that two (or more) strictly complex (i.e. ones with nonzero real and imaginary parts) numbers can be compared given that they lie on a straight line through the origin. e.g. to me, it makes sense to say that $1+i<2+2i$, for instance, and that $i<10i,$ as one is a real multiple of the other.

Essentially, what I'm asking, is: can we compare complex numbers in any way other than comparing their moduli?


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    $\begingroup$ The partial ordering you described is exactly the same as the partial ordering given by the modulus. $\endgroup$ – oxeimon May 8 '14 at 20:02
  • $\begingroup$ Yes, this can be done. The definition you're looking for is:$z \leq w$ iff there exists $r \in [1,\infty)$ such that $rz=w$. This is a genuine partial order, and it is compatible with the multiplicative structure of $\mathbb{C}$ but not the additive structure. Contrary to oxeimon's comment, this is not the partial order induced on $\mathbb{C}$ by the modulus function, since that is a mere preorder; more precisely, its not antisymmetric. $\endgroup$ – goblin GONE May 31 '15 at 9:37
  • $\begingroup$ There is another interesting partial ordering of $\mathbb{C}$ given as follows: $z \leq w$ iff there exists $r \in [0,\infty)$ such that $z+r=w$. This is compatible with the additive structure of $\mathbb{C}$, and slightly compatible with the multiplicative. I leave it to you to work out the details. $\endgroup$ – goblin GONE May 31 '15 at 9:42

Yes, you can put a lexicographical ordering on the complex numbers, which makes it a totally ordered set. However, we cannot introduce a totally ordered relation on the complex numbers (as a field such that the field operations are compatible with the defined order) since, every ordered field is a formally real field.

  • $\begingroup$ What if we're comparing $a+bi$ and $c+di$, but $a<c$ and $b>d$? In this case, which is the bigger of the two? $\endgroup$ – beep-boop May 8 '14 at 19:30
  • $\begingroup$ in this case the second one is "bigger" $\endgroup$ – Math137 May 8 '14 at 19:31
  • $\begingroup$ Oh, so it's like comparing 'Alan Smith' and 'Joe Bloggs' in their position in the phone book? $\endgroup$ – beep-boop May 8 '14 at 19:33
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    $\begingroup$ Yes, it is exactly looks like ordering words in a dictionary. $\endgroup$ – Math137 May 8 '14 at 19:39
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    $\begingroup$ Yes it is. What you cannot define is a total order relation that is operation-compatible, which is not the case of the lexicographical order. Look at the wikipedia link you gave, they state explicitely that if the order used in the construction is total (corresponds to the usual order on $\mathbb{R}$ here, which is total), so is the lexicographical order. Actually, with axiom of choice, you can build a strict total order on a arbitrary set. But once again, in $\mathbb{C}$, it won't be operation compatible. If you're not convinced, please give me two non comparable elements ? $\endgroup$ – yago May 8 '14 at 19:47

Yes, it totally makes sense.

In order to use $\gt$, $\lt$ and $=$ the same way they are used on real numbers, a many-to-one relation that maps complex numbers into real numbers will suffice.

More generally, an order is nothing but a relation. A total order is a relation $R$ that is transitive, asymmetrical and connected.

Transitive: if $xRy$ and $yRz$ then $xRz$, e.g. if 3>2 and 2>1 then 3>1;

Asymmetrical: if $xRy$ then $yRx$ must be false, e.g. if 3 > 1 then 1 > 3 must be false;

Connected: Given any two terms $x$ and $y$, either $xRy$ or $yRx$.

Source: Russell, Bertrand. Introduction to Mathematical Philosophy, "Chapter IV The Definition of Order"

Example1: Who gets off the Ferris Wheel first

Given any two complex numbers $a=r_1e^{i\phi}$ and $b=r_2e^{i\psi}$,

we define $a>b$ as

$(\phi > \psi) $ or $(\phi=\psi$ and $r_1>r_2 )$

This means that the person who has travelled a greater degree from the starting point or, if the travelled degrees are the same, the person who is on the further end of the spoke gets off first.

Thus, $3e^{i\pi}> 2e^{i\pi}> 3e^{i\dfrac{\pi}{2}}$

Example2: Sailing in crosswinds enter image description here

Suppose you are sailing towards Northeast$(\dfrac{π}{4})$. You can produce various lifts by adjusting your sail, but some lifts are better than others because they produce more forward thrusts. Given a lift $fe^{i\theta}$, the forward thrust it produces is $fcos(θ−\dfrac{π}{4})$. This is the value you use to compare various lifts.

Edit: Depending on the field of terms, this comparison does not necessarily give rise to a total order. Different lifts can produce the same forward thrusts, thus the asymmetrical requirement is violated. Nevertheless, you can still compare lifts.

If you know Ruby programming language, you can customize the spaceship operator to sort complex numbers.

  • $\begingroup$ But, according to this approach, which term dominates out of the distance travelled, $r$, and the angle from the starting point, $\theta$? Say $r_1>r_2$ but $\theta_1<\theta_2$. In this case, which is bigger? Is there a way to get a weighted average of $\theta$ and $r$, in general? $\endgroup$ – beep-boop May 9 '14 at 0:07
  • $\begingroup$ The angle dominates. In your example, $r_2\angle\theta_2$ is greater. $\endgroup$ – George Chen May 9 '14 at 0:11
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    $\begingroup$ This is a very practical question, by the way. Imagine a ship detects a salvo of anti-ship missiles from all directions, the computerized phalanx would have to quickly decide which missile to shoot first. $\endgroup$ – George Chen May 9 '14 at 0:15
  • $\begingroup$ Do you have a formula to calculate the weighted average of $r$ and $\theta$? Or is it always the case that the angle dominates, even if $\theta_1$ is only slightly bigger than $\theta_2$, but $r_1<<r_2$? $\endgroup$ – beep-boop May 9 '14 at 0:28
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    $\begingroup$ The core idea is, in order to create a series, all you need is to construct a relation that is transitive, connected, asymetrical $\endgroup$ – George Chen May 9 '14 at 0:41

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