Maths - Dual Complex Arithmetic

Adding Dual Complex numbers

Just add the real and imaginary components independently as follows:

(a1 + i b1 + ε c1 +iε d1)+(a2 + i b2+ ε c2 +iε d2) = (a1+a2) + i (b1+b2) + ε (c1+c2) +iε (d1+d2)

Subtracting Dual Complex numbers

Just subtract the real and imaginary components independently as follows:

(a1 + i b1 + ε c1 +iε d1)-(a2 + i b2+ ε c2 +iε d2) = (a1-a2) + i (b1-b2) + ε (c1-c2) +iε (d1-d2)

Multiplying Dual Complex numbers

We put each quaternion in brackets and multiply out all the terms: (a1 + i b1 + ε c1 +iε d1)*(a2 + i b2+ ε c2 +iε d2). When we are multiplying the imaginary operators we use the multipication table:

Note that the order of multiplication is significant, in other words dc1 * dc2 is not necessarily equal to dc2 * dc1.

So back to the general case of multiplying any two quaternions, we can just expand out the terms and group as follows:

so the result is:

dc1 * dc2= a1*a2 - b1*b2 + i (a1*b2 + b1*a2) + ε (c1*a2 + d1*b2 + a1*c2 - b1*d2) +iε (d1*a2 - c1*b2 + b1*c2 + a1*d2)

Division

We don't tend to use the notation for division, since quaternion multiplication is not commutative we need to be able to distinguish between dc1*dc2^-1 and dc2^-1*dc1. So instead of a divide operation we multiply by the inverse.

Due to the zero terms in the multiplication table I don't think there is a general multiplicative inverse because information is lost by multiplication.

However, if we restrict ourselves to the linear rotation-displacement discussed above, then there must be an inverse because the operations they represent have an inverse. The complex number is always normalised so its inverse will be its conjugate and the inverse of the displacement is minus its value.

So, in this case, the inverse of:

a + b i + c ε + d iε

is:

a - b i - c ε - d iε

EuclideanSpace

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