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?

  • 12
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Have you not studied vectors in $\Bbb R^2$ or $\Bbb R^n$? $\endgroup$ – Ted Shifrin Apr 11 at 3:52
  • 2
    $\begingroup$ @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

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 $x\in\mathbb{C}$.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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 $\mathbf R$.

  • 5
    $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.