Why do we define the modulus of a complex number as we do?

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?

• Definitions are almost always made because something is used repeatedly and is worth defining. This is one such example. Apr 11 at 2:34
• 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: $\| z_1 z_2 \| = \| z_1 \| \| z_2 \|$. Apr 11 at 2:39
• 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.$That wouldn’t work with another metric. Apr 11 at 3:39 • Have you not studied vectors in$\Bbb R^2$or$\Bbb R^n$? Apr 11 at 3:52 • @TedShifrin I have. It is not immediately clear to me why we should treat$\Bbb C$as being isomorphic to$\Bbb R^2$. We certainly can do, but my question was why. Noah's answer gives excellent justification. Apr 11 at 13:02 2 Answers 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 $$x\in\mathbb{C}$$. • 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. Apr 11 at 3:42 • @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. Apr 11 at 20:44 • One wonders how you define continuity of a function$\Bbb{C} \to \Bbb{R}$without referring to the absolute value on$\Bbb{C}$itself. Apr 11 at 21:00 • @mechanodroid$\mathbb{C}$inherits the product topology from$\mathbb{R}$via the bijection$(a,b)\mapsto a+bi$. Apr 11 at 21:15 • @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$ℂ$? 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 $$\mathbf R$$. • +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. Apr 12 at 3:50 • Do you perhaps know how to describe all possible absolute values on$\Bbb{C}$, not necessarily extending the standard absolute value on$\Bbb{R}$? Apr 13 at 20:05 • @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. – KCd Apr 13 at 21:26 • 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) Apr 17 at 19:15 • 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.
– KCd
Apr 17 at 19:43