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! :)

  • $\begingroup$ @SimonS thx for the Answer Simon. I just want to make sure - There is no inherent connection between the complex numbers and this norm, we just like to use this norm becouse it establish a simple and comfortable connection between our norm and the exponential function or complex numbers, and we could use other norm but it would be just less comfortable to work with the exponent than? In the same sense that we can use other measurements than Radians for angles, but it would just make all the work longer? $\endgroup$ – tomers99 Jun 16 '15 at 2:34
  • $\begingroup$ Do you really mean "neutral" or did you mean "natural?" $\endgroup$ – Thomas Andrews Jun 16 '15 at 2:40
  • $\begingroup$ @ThomasAndrews natural, ofc :D $\endgroup$ – tomers99 Jun 16 '15 at 2:57

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.

  • $\begingroup$ Also, ideally, you want the norm to be the same on the reals. $\endgroup$ – Thomas Andrews Jun 16 '15 at 2:40
  • $\begingroup$ That's a consequence of $|a|=|\overline{a}|$, as $a=\overline{a}$ if $a$ is real. $\endgroup$ – Zach Stone Jun 16 '15 at 2:42
  • $\begingroup$ No, that just means that $|2|_{\mathbb C}=|2|_{\mathbb C}$, not that $|2|_{\mathbb C}=2$. $\endgroup$ – Thomas Andrews Jun 16 '15 at 2:44
  • $\begingroup$ That's right. I'm being silly. That is a useful addition. $\endgroup$ – Zach Stone Jun 16 '15 at 2:45
  • $\begingroup$ @Zach stone thx for you'r answer too. You two have convinced me that there are plenty of justified reasons to use that norm, which before seemd to me kinda artificial and not based on the internal nature of complex numbers but rather on the "coincidental" fact that our brain is bribed to the Euclidian geometry. One last question though. I was firstly really met with complex analysis today, and I saw a proof (of the fundamental theorom of the algebra) that intensively used the M. and Ar. of the complex numbers. Is there a way (possibly more complicated) to prove things in C.analysis without it? $\endgroup$ – tomers99 Jun 16 '15 at 2:47

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