Why is the Euclidian norm used to measure complex numbers?

Question

Why is the Euclidian norm used to measure complex numbers?

The complex numbers are numbers (or more precisely, pairs of numbers), and I can't see why are they essentially connected to the Euclidian geometry. As far as a I can see, the only connection between Euclidian plane and pairs of numbers is that the last can be conjugated to the plane in very "neutral" way. But as much as i understand, the euclidian plane is part of a more general notion of manifold, and you can conjugate pairs of numbers to any 2 dimensional manifold, including non euclidian planes.

Are we using the euclidian norm ("modulus") only because we want some kind of way to measure complex numbers that "grows bigger as the complex numbers grows, no matter in what direction"? a way of measure which is "free of direction" (not taking into account the direction of which the quantity grows, but only its growth)?

This is my way to understand it so far. If I am right, we can basically use any similar norm to measure complex numbers "size" (independent of direction), not only the Euclidian norm, and they'll do the same work and fulfill their destiny.

Obviously there would be another but similar way to measure their "direction"..

So, am I right, and its just an convention to use that way to measure "size" and "direction" of complex numbers by the so called modulus and argument? Or am I wrong? thx for answering! :)

Answer 1

We want $|ab| = |a|\,|b|$. This fact comes up all of the time. (Especially in estimating geometric series.) Furthermore, we want $|\overline{a}| = |a|$. Just think about the geometry and this will be obvious.

Now, let's talk about the complex numbers as of the form $x+iy$ for real $x$ and $y$. We can already calculate that $$ |x+iy|^2=|(x+iy)^2|=|(x+iy)(x-iy)|=|(x^2-ixy+ixy+y^2)|=|x^2+y^2| $$

So $$ |x+iy| = \sqrt{x^2+y^2} $$

So we must use the euclidean norm if we want to satisfy even basic properties of a nice norm. It's more than a convention or a convenience. It's pretty much the only thing that'll work.

In the comments you ask whether or not we really need to use the euclidean norm or could we prove theorems in complex analysis without it. I'll address that here. Probably the only time to consider a strict increase in proof complexity is if the new argument generalizes better. For example, many simple proofs from Hilbert space theory can be made more complicated by dancing around the use of inner products. But this is good because you will get a proof that works in the (strictly more general) Banach space setting. (I recently came across a neat instance of this in defining trace class operators vs nuclear operators. It does happen and you're not crazy for asking.)

However, the complex numbers aren't like this. Anytime your working in the plane, you might as well assume you're working in $\mathbb{C}$. As above, $a\overline{a} = |a|^2$ so anytime you can do conjugation you might as well use the euclidean norm! So even it it were somehow possible to eliminate explicit references to the euclidean norm, you wouldn't be doing yourself any favors. It would not improve generality. Rather it would just obfuscate an otherwise plain detail.

Answer 2

Before I get into this, I will give you some background. Some people in some contexts use a sphere to represent complex numbers with one extra complex number included at infinity. That eliminates the hole at infinity. It's probably the projection of the Euclidean plane onto a sphere that preserves small shapes. But the regular complex plane is a group under addition unlike the one that includes an extra point at infinity. That was the background.

The actual reason for the Euclidean norm, I don't know. Here's one possible good reason to do it that way. The stereographic projection of the sphere is the projection that preserves small shapes. We could use the norm in the Stereographic representation of the sphere, that is the minimum distance along the surface of the sphere. But it's not as simple to describe as the Euclidean norm. The Poincaré disc model of the hyperbolic plane is the projection into Euclidean geometry that preserves small shapes. We also could use the norm in the Poincaré disk model of the hyperbolic plane, that is, the minimum distance between 2 points the hyperbolic plane. That also poses the same problem as the norm in the Stereographic projection of the sphere. Also, it excludes the points in the complex number plane at the edge and outside of the disk. I suppose the disc could be extended to be like another concave disc model of the hyperbolic plane that takes up all the space outside of the disc but the point that's equivalent to the point in the center of the original disc wouldn't exist. So the Euclidean norm is simpler than the Poincaré disc model of the hyperbolic plane.

Why do we use the sum of the products of each number and its complex conjugate instead of the sum of their squares? I don't know the actual reason. However, here's one possible good reason to do so. When you take the sum of the products of each number an its complex conjugate, the result is 0 if and only if all of the individual numbers are 0. That's not the case when you take the sum of the squares. For example, $1^2 + \mathbb{i}^2 = 0$