Why is the modulus of a complex number $a^2+b^2$? Why is the modulus not $\sqrt{a^2-b^2}$? Carrying out standard multiplication this would be the result-why is this not the case?  I know viewing the complex plane you can easily define the sum as being the distance to the points, but what meaning does $\sqrt{a^2-b^2}$ have? 

Let's consider the mapping $z\mapsto (a+bi)z$ from the complex plane to itself. This is an $\Bbb{R}$-linear mapping. It maps the basis vectors thusly $1\mapsto a+bi$ and $i\mapsto -b+ai$. So the determinant is (dun, dunn, dunnn,...)
$$
\left|\begin{array}{rr}a&-b\\b&a\end{array}\right|=a^2+b^2.
$$
From linear algebra, or vector calculus even (think Jacobians), you hopefully remember that this determinant is the scaling factor of areas. So if you have some (measurable) subset $S$ of the complex plane, then the area of $zS$ is $(a^2+b^2)$ times the area of $S$. Taking the square root normalizes this to a 1-dimensional quantity.
It is true that in other number fields this determinant of a multiplication by a constant takes a different form. For example in the field $\Bbb{Q}(\sqrt d)$, $d$ a square-free integer, we get $a^2-db^2$ for the determinant of the mapping
$x\mapsto (a+b\sqrt d)x$, where $x=u+v\sqrt d, u,v\in\Bbb{Q}$. May be you have seen that somewhere, and it confused you a bit? Anyway, to get $a^2-b^2$, you would need $d=1$, but that is a square and won't give you an extension field.
A glaring problem with $a^2-b^2$ as any kind of a "norm" is that it can go to zero even, when $(a,b)\neq(0,0)$. 
My main point was that this concept of a (squared) norm as the determinant of the linear mapping given by multiplication is one that generalizes to all number systems.
