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?
 A: 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 $x\in\mathbb{C}$.
A: 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 $\mathbf R$.
