Complex products: perpendicular vectors and rotation effects

[Ventrella]

My question is perhaps as much about the philosophy of math as it is about the specific tools of math: is perpendicularity and rotation integral and fundamental to the concept of multiplication - in all number spaces?

As I understand it, the product of complex numbers x = (a, ib) and y = (c, id) can be calculated as:

( (ac-bd), i(ad+bc) )

I noticed that this expression uses a particular mixing of the real and imaginary components of x and y, and it includes a negation in the real part. And I understand that this negation is related to the square root of -1. I have noticed a similarity between the multiplication of complex numbers and the operation for finding the perpendicular of a 2D vector (x, y) as (-y, x). The other perpendicular is (y, -x). It appears that negation of one component is critical for calculating perpendicularity.

Is there a way to express the concept of multiplication in a general way such that the behaviors of complex numbers is consistent with that of real (1-dimensional) numbers? Or do these properties of perpendicularity, negation, and rotation apply only to the higher number spaces (complex, quaternion, octonian, etc)? Might there be higher-order behaviors in the higher-dimensioned number spaces that are supersets of rotation and perpendicularity that our 3D brains cannot visualize or even comprehend?

Finally, what higher-order principle dictates that multiplication in higher-order number spaces should involve rotational effects? Some philosophers of mathematics would argue that these techniques are an invention of humans - an extrapolation of the concept of multiplication of real numbers to higher dimensions, which is arbitrary (although extremely useful and fully-consistent with all other mathematical concepts and operations).

 

[FactChecker]

My question is perhaps as much about the philosophy of math as it is about the specific tools of math: is perpendicularity and rotation integral and fundamental to the concept of multiplication - in all number spaces?

No. The definition of multiplication must allow it to represent rotations. The regular 2D vector space, R² does not have a multiplication that represents rotation. There is a dot product and a cross product but not a product that is a rotation. That is one of the great advantages of the complex plane over R². In three dimensions, the quaternions have operations for rotations, but the 3D vector space, R³, needs something like rotation matrices.

One thing to notice is that a multiplication which represents rotations in the complex plane has some tremendous advantages: Every rotation has an inverse, which is a rotation in the opposite direction. So there are always multiplicative inverses and division is defined. That allows a derivative based on (f(z)-f(z₀))/(z-z₀), which mimics the derivative of introductory calculus. This leads to very profound results.

PS. I like your fractal avatar.

 

[FactChecker]

finding the perpendicular of a 2D vector (x, y) as (-y, x). The other perpendicular is (y, -x). It appears that negation of one component is critical for calculating perpendicularity.

In R³, the vectors (1,1,0) and (0,0,1) are perpendicular. I don't see how your idea would apply to that. The general condition for two vectors to be perpendicular is that they have a zero dot product. The vectors A=(a₁, a₂, ..., a_n) and B = (b₁, b₂, ..., b_n) are perpendicular if and only if Σa_ib_i = 0. That works in any number of dimensions and can be generalized to orthogonal functions, and is a better way to think of "perpendicular" in equation form.

 

[Mark44]

the operation for finding the perpendicular of a 2D vector (x, y) as (-y, x).

The key point here is slope. The vector <x, y> has a slope of $\frac y x$. This is easy to see if you sketch the line segment from the origin to the point (x, y). In the plane, two line segments are perpendicular if their slopes are negative reciprocals of each other; that is, the slopes multiply to make -1. Clearly a slope of $\frac {x} {-y}$ will do, as will $\frac {-x} y$. The vector <-y, x> has a slope of $\frac x {-y}$.

The other perpendicular is (y, -x). It appears that negation of one component is critical for calculating perpendicularity.

The vector <y, -x> "points" in the opposite direction that <-y, x> does. Again, the key point for vectors in the plane is that their slopes are negative reciprocals.