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For a complex number $z = a+bi$, we say that its modulus is: $$|z|=\sqrt{a^2+b^2}$$

When we draw complex numbers in the Argand diagram, intuitively, this makes sense. But if we used a different projection for the diagram (i.e. a different metric for distance) then it wouldn't necessarily. Of course, complex numbers can also be written as:

$$z = re^{i\theta} = r(\cos\theta +i\sin\theta)$$

so an equivalent question could be, if this is what we define, why we define that:

$$|e^{i\theta}| = |\cos\theta + i\sin\theta| = 1$$

for all values of $\theta$, rather than just $\theta = n\pi$.

The answer may simply be that it is convenient to work with this definition. But is there a deeper reason? Are there any problems for which it is convenient to define things differently? And what would be the consequences if we did things differently?

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    $\begingroup$ Definitions are almost always made because something is used repeatedly and is worth defining. This is one such example. $\endgroup$ – CyclotomicField Apr 11 at 2:34
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    $\begingroup$ It seems quite natural to define the magnitude (or "norm") of a complex number $z$ to be the distance from $z$ to the origin, and Euclidean distance is probably the most intuitive way of measuring distance. One nice thing about this definition of the norm of $z$ is that norms are multiplicative: $\| z_1 z_2 \| = \| z_1 \| \| z_2 \|$. $\endgroup$ – littleO Apr 11 at 2:39
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    $\begingroup$ For power series, we have that the series has a radius of convergence $R$ such that the series converges for $|z|<R$ and diverges for |z|>R.$ That wouldn’t work with another metric. $\endgroup$ – Thomas Andrews Apr 11 at 3:39
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    $\begingroup$ Have you not studied vectors in $\Bbb R^2$ or $\Bbb R^n$? $\endgroup$ – Ted Shifrin Apr 11 at 3:52
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    $\begingroup$ @TedShifrin I have. It is not immediately clear to me why we should treat $\Bbb C$ as being isomorphic to $\Bbb R^2$. We certainly can do, but my question was why. Noah's answer gives excellent justification. $\endgroup$ – legionwhale Apr 11 at 13:02
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As CyclotomicField points out, it is a very convenient definition: regardless of whether we give it a name, the map $(a+bi)\mapsto \sqrt{a^2+b^2}$ comes up frequently.

However, we can indeed give an "intrinsic" motivation: there are a few basic assumptions which, when combined, identify the standard definition of modulus uniquely.

  • First, we have a "positivity" axiom: we want $\vert x\vert\ge 0$ for all $x$ and we want $\vert x\vert=0$ iff $x=0$.

  • Next, we have an "algebraic" axiom: thinking of a complex number as a unit vector scaled by a number (its modulus), we want the modulus function to be multiplicative: $\vert x\vert\vert y\vert$ should equal $\vert xy\vert$. Moreover, (real) scalar multiplication should play with the norm in the obvious way: $\vert \alpha x\vert=\vert\alpha\vert\vert x\vert$ (where the first "$\vert\cdot\vert$" refers to the usual absolute value function on $\mathbb{R}$); if you like, you can think of this as saying that the complex modulus should agree with the real modulus on real numbers.

  • Finally, we have a "topological" axiom: we want the map $\mathbb{C}\rightarrow\mathbb{R}:x\mapsto\vert x\vert$ to be continuous.

This turns out to be enough to identify the standard modulus function! The positivity and algebraic axioms alone tell us that $\vert 1\vert=1$ (since it must be nonzero yet equal to its square), and in turn that $\vert -1\vert=1$ (since it must be a nonnegative square root of $\vert 1\vert=1$), and in turn that $\vert i\vert=1$ (since it must be a nonnegative square root of $\vert-1\vert=1$), and so forth. In fact, this shows that $\vert e^{i\theta}\vert=1$ whenever $\theta$ is a rational multiple of $\pi$. And then the topological axiom finishes things off: by continuity we must have $\vert e^{i\theta}\vert=1$ for every $\theta$, and thinking about scalar multiplication pins down the value of $\vert x\vert$ for all $x\in\mathbb{C}$.

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    $\begingroup$ It is the hypotenuse of a right triangle. @legionwhale It is often referred ro as the Euclidean norm, because it is the basis of Euclidean geometry. $\endgroup$ – Thomas Andrews Apr 11 at 3:42
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    $\begingroup$ @user21820 I didn't add it since it's not necessary: I wanted to use as small a set of assumptions as possible. and we don't actually need the triangle inequality for that. $\endgroup$ – Noah Schweber Apr 11 at 20:44
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    $\begingroup$ One wonders how you define continuity of a function $\Bbb{C} \to \Bbb{R}$ without referring to the absolute value on $\Bbb{C}$ itself. $\endgroup$ – mechanodroid Apr 11 at 21:00
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    $\begingroup$ @mechanodroid $\mathbb{C}$ inherits the product topology from $\mathbb{R}$ via the bijection $(a,b)\mapsto a+bi$. $\endgroup$ – Noah Schweber Apr 11 at 21:15
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    $\begingroup$ @NoahSchweber: Okay I see your point! Is there an alternative minimal list of nice axioms that does include the triangle inequality and also uniquely identifies the euclidean norm as a natural norm for $ℂ$? $\endgroup$ – user21820 Apr 12 at 3:17
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Do you like the absolute value on $\mathbf R$? Well, it turns out that the formula $|a+bi| = \sqrt{a^2+b^2}$ is the unique absolute value on $\mathbf C$ extending the absolute value on $\mathbf R$. That is not why it was originally defined, but it provides an excellent reason that this function plays such a prominent role in work on the complex numbers. You don't need anything like polar decomposition ($re^{i\theta}$) to prove that uniqueness. It follows from $\mathbf R$ being complete and $\mathbf C$ being finite-dimensional over $\mathbf R$.

An "absolute value" on a field $F$ is an $\mathbf R$-valued function $|x|$ for $x \in F$ such that (i) $|x| \geq 0$ with equality if and only if $x = 0$, (ii), $|xy| = |x||y|$ for all $x$ and $y$ in $F$, and (iii) $|x+y| \leq |x| + |y|$ for all $x$ and $y$ in $F$. Using an absolute value on $F$ we get a metric on $F$ by $d(x,y) = |x-y|$ and that leads to associated topological and analytic ideas on $F$ if it has some nice properties for this metric (complete, locally compact, and so on).

You can apply this concept to a division ring in place of a field. In that spirit, the absolute value on quaternions, given by $|a+bi+cj+dk| = \sqrt{a^2+b^2+c^2+d^2}$, is the unique absolute value on the quaternions that extends the absolute value on $\mathbf R$.

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    $\begingroup$ +1. Note to the OP that, in my opinion at least, this is a strictly better axiomatization than the one I describe in the sense of mathematical utility; I picked my set of axioms for purely expository reasons. $\endgroup$ – Noah Schweber Apr 12 at 3:50
  • $\begingroup$ Do you perhaps know how to describe all possible absolute values on $\Bbb{C}$, not necessarily extending the standard absolute value on $\Bbb{R}$? $\endgroup$ – mechanodroid Apr 13 at 20:05
  • $\begingroup$ @mechanodroid there is no reasonable answer to that question: for every field extension $L/K$, an absolute value on $K$ always extends (using Zorn's lemma) to an absolute value on $L$. It is like asking for a description of all possible bases of $\mathbf R$ as a $\mathbf Q$-vector space. $\endgroup$ – KCd Apr 13 at 21:26
  • $\begingroup$ That said, one could hope for a classification of "tame" absolute values, e.g. ones which are Borel with respect to the usual topologies on $\mathbb{C}$ and $\mathbb{R}$ (and set theory provides even richer notions of "tameness"). This is often highly restrictive, though - for example, in a precise sense $\mathbb{R}$ has no "tame" (Hamel) bases as a $\mathbb{Q}$-vector space whatsoever. (@mechanodroid) $\endgroup$ – Noah Schweber Apr 17 at 19:15
  • $\begingroup$ It's highly restrictive, as you suggested. An absolute value on the nonzero numbers of $\mathbf R$ or $\mathbf C$ is a homomorphism $\mathbf R^\times \to (0,\infty)$ or $\mathbf C^\times \to (0,\infty)$ with extra properties, and a measurable homomorphism of locally compact groups is continuous (see mathoverflow.net/questions/120738/…). So the only absolute values on $\mathbf R$ and $\mathbf C$ are powers of the standard absolute values. $\endgroup$ – KCd Apr 17 at 19:43

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