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I'm taking an algebra class and we're examining the complex numbers. A lot of properties of $\mathbb{C}$ are new to me. I know that the complex numbers form a field, and I understand that they do not form an ordered field under the normal definition of order using $<$.

However, we have been given a definition of a norm on $\mathbb{C}$: the distance between a number $z \in \mathbb{C}$ and the origin (or the "length" of the number) is: $\sqrt{x^2+y^2}$, assuming $z=x+iy$. I don't really understand this. How can we have a "length" for a number, if we don't have an order? If there is no order, how can the plane even be constructed? We don't know if $1i<2i<3i$. How can we graph these as points of the "y" axis on the plane, if we don't know which comes after the other?

Perhaps I am confused about a more general question: how can a field that isn't ordered have a measure of "length", or have a "plane"?

user7240099
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    Think of complex numbers as vectors. You can't compare vectors how big one vector is w.r.t. another, are but you can compare their magnitudes – imranfat Oct 31 '17 at 21:11
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    When you say "We don't know if 1i < 2i < 3i", you're making a specific choice of collinear points. For any line in C we can define a natural ordering for the points on that line, but the problem is there is no consistent such ordering that will work for all lines simultaneously – Sort of Damocles Oct 31 '17 at 21:16
  • I think that is a very useful comparison and it makes a bit of sense, however with vectors we have both the x and y axis totally ordered (assuming we are working with $\mathbb{R}^2$). So while we can't compare how big one vector is w.r.t. another, we know that its y component is bigger than another vectors y component, and same goes for x component. With complex numbers, we can't have any meaning of order on the y axis, correct? – user7240099 Oct 31 '17 at 21:16
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    You can order the imaginary axis using the real coefficient, but you can't order the whole plane. Also distance is a real number so why do you need an ordering of the complexes anyway? – Randall Oct 31 '17 at 21:17
  • Well because we use distance when we define complex numbers in terms of polar co-ordinates. Since if $z=x+iy$, and we also have $\theta(z)$ be the "angle" of the complex number, then $z=|z|e^{i\cdot \theta (z)}$ where $|z|$ is the magnitude. How can we have magnitude being used here without an ordering of the plane? – user7240099 Oct 31 '17 at 21:21
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    One can define an order but is not consistent with $|\cdot|$. Why is not having a compatible order inconsistent with a metric? – copper.hat Oct 31 '17 at 21:21
  • Suppose for example we have a number in $\mathbb{C}$ with an x component of zero. So we get $z=iy$. Then we get that the length of $0+5y$ is greater than the length of $0+4y$. Yet we have no idea that the distance between $5y$ from the origin is "greater" then the distance between $4y$ from the origin because we don't even know if $5y>4y$, if that makes sense. Maybe $4y$ comes after $5y$ in the y axis. So how can we say something that is closer to the origin has a greater length? To your point, we can define an order and work off of that? Is that what they do when they define the norm? – user7240099 Oct 31 '17 at 21:37

2 Answers2

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Are the points in a plane naturally ordered?

Can you measure the distance between two points in the Euclidean plane?

Note that the Argand Diagram models complex numbers as points in a plane, where distances are natural, but there is no obvious linear order.

You know, surely, how to represent the points in a plane with co-ordinates. Each of the co-ordinate axes is ordered, but the points in the plane are not. You haven't thought of ordering points in a plane before.

The complex numbers give us a way of multiplying points in the plane which is compatible with the natural (vector-style) addition (distributive law, inverses except for zero). It turns out that this is a useful and natural definition of multiplication, which extends our normal understanding and solves a whole host of problems. We think of complex numbers as abstract entities on the whole, but the plane representation is behind a lot of useful applications.

Mark Bennet
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There is certainly a meaningful notion of order for the imaginary axis of the complex numbers and it is given by identifying it with the real line in the obvious manner. The same holds for any line in $\mathbb C$, just as in $\mathbb R ^2$. The problem - just as in the real plane - as user 'dbx' writes in the comments, is that these obvious orders on lines do not coalesce into a sensible order on the entire plane. This does not seem to bother you in the case of the real plane because you feel the axes are naturally ordered, but the picture is exactly the same for the complex numbers. Do not feel alienated by the imaginary unit $i$ - the pictures all stay the same.

The standard norm of a complex number comes from the Pythagorean theorem in the real plane, which lets you measure length by the difference between coordinates along orthogonal axes. It is these axes which have an obvious order underlying your intuition, not the entire plane. The order on the axes is what furnishes the notions of 'left, right, above, below'.

More fundamentally, the abstraction of "length of a vector" by the notion of a norm on a vector space does not depend in any way on an order on the set underlying the vector space or any subset thereof. The axioms for a norm comprise a list of synthetic properties you want "abstract length" to satisfy without concerning yourself with how to actually calculate it: You want

  1. the zero vector and only the zero vector to have zero length;
  2. the length of any vector is non-negative;
  3. the triangle inequality;
  4. nice behavior with "elongation" i.e multiplication by scalar.

So planning ahead, do not expect some order to underlie a general notion of norm.

This isn't so strange if you think about it: merely knowing one thing is larger than another says nothing about how large either thing is, and conversely, knowing a thing is longer than another says nothing about which is 'to the right' of which.

Arrow
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