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?
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I think you are not dealing with $i$ correctly in your multiplication. Note that $$(a + bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2.$$
Michael Albanese
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Maybe OP missed the little complex conjugate '*' and calculated $(a + i b)(a + i b)$ instead of $(a + i b)(a - i b)$. – Aaron Golden Oct 08 '13 at 04:26
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@AaronGolden That would get him a completely different answer though. – Dennis Meng Oct 08 '13 at 04:28
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@AaronGolden: In that case you get $(a+bi)(a+bi) = a^2 + 2abi - b^2$. – Michael Albanese Oct 08 '13 at 04:28
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Of course you're right. I wrote hastily. It seemed like an easy mistake to make that would flip the sign on $b^2$. But failing to handle $i^2$ right is the more likely mistake. – Aaron Golden Oct 08 '13 at 04:32
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The geometric meaning of the modulus is the length of the vector from the origin to the point in the complex plane. So it's just defined like the Euclidean distance to a point $(x, y)$, $\sqrt{x^2 + y^2}$. It just happens that "y component" is the imaginary part of the complex number $a + i b$, so you use $\sqrt{a^2 + b^2}$.
Aaron Golden
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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.
Jyrki Lahtonen
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Hi Jyrki Lahtonen, I was wondering how one goes from the mapping of the numbers to get to the stage of a determinant. In linear algebra courses one learns about a system of equations (which i believe is your mapping) and one learns how to compute the determinant of a matrix like structure. BUT how to we connect a linear mapping to a determinant? – Palu Nov 08 '18 at 05:31