Complex numbers and vector multiplies continued

[keldon]

This is just a follow on from this thread, https://www.physicsforums.com/threads/complex-numbers-and-vector-multiplication.509944/

Basically I've noted that, in 2d at least that the complex multiplication of A and B is equal to (A dot conj(B)) + (A cross B)i

Would that then mean his initial suspicions were true?

 

[mathman]

Your assertion doesn't make sense. The "real" part is a number but the "imaginary" part is a vector.

 

[pwsnafu]

I have no idea what you mean by "complex multiplication of A and B".

 

[keldon]

I have no idea what you mean by "complex multiplication of A and B".

Multiplication of two complex numbers, A and B.

 

[HallsofIvy]

He is trying to treat the complex number, a+ bi, as a vector <a, b> and derive the product of two complex numbers using a combination of dot product and cross product. The product (a+ bi)(c+ di) is $a^2- b^2+ (ad+ bc)i$ so he wants a way to "multiply" <a, b> and <c, d> to get $<a^2- b^2, ad+ bc>$.

 

[mathman]

He is trying to treat the complex number, a+ bi, as a vector <a, b> and derive the product of two complex numbers using a combination of dot product and cross product. The product (a+ bi)(c+ di) is $a^2- b^2+ (ad+ bc)i$ so he wants a way to "multiply" <a, b> and <c, d> to get $<a^2- b^2, ad+ bc>$.

|ad+bc| = |AxB|, but there is no way to get the sign from the cross product.
The real part is ac-bd, while A.B = ac+bd.

 

[Hawkeye18]

The standard way is to represent a complex number $a+ib$ as a $2\times 2$ matrix
$$\left(\begin{array}{cc} a & - b \\ b & a \end{array}\right), $$
so you then just read the real and imaginary part off the first column.
The matrix multiplication then agrees with the multiplication of complex numbers.
Note that in this representation
$$1\sim \left(\begin{array}{cc} 1 & -0 \\ 0 & 1 \end{array}\right), \qquad
i\sim \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right).$$

 

[homeomorphic]

Yes, you're right. Complex multiplication DOES include both the dot product and cross product, under a suitable interpretation of the cross product. But there's a correction/clarification to be made:

conj(A) B = (a - ib)(c + id) = ac + bd + (ad-bc)i = A dot B +- |A cross B| i

Notice that it's the MAGNITUDE of A cross B, so it's a number, with the sign determined by whether it's up or down out of the plane. In the plane, you're not really losing any information from the cross product because there's a canonical direction it will be in, perpendicular to the plane (with up or down determined by the sign).

I'm not sure I would say he was "right" because his logic seems to be that since complex numbers can be interpreted as vectors and there are "only" two ways to multiply vectors, they must somehow be the same as dot products and cross products. That argument makes no sense. It is confusing the fact that there are only two ways to multiply vectors that are most commonly used with the fact that there only exist two ways of multiplying vectors. There are actually tons of different ways of multiplying different vectors, depending on what you mean by multiplication. If you put some restrictions on what you mean by multiplication, you could narrow it down a bit, although I'm not quite sure how to characterize it exactly (it depends on what dimension the vectors are in as well). The cross product is already a very weird and somewhat nasty form of "multiplication", being non-associative, so if you are going to include the cross product, you aren't going to be able to demand that "multiplication" means something that nice and well-behaved. Someone else might know a better way to frame it than I do, though. Anyway, his logic is a bit off, but it does happen that he's partly right.

Historically, the origins of the dot product and cross product are in the quaternions, which can be thought of as 4-dimensional complex numbers. There's a similar relationship between quaternion multiplication and the dot and cross products. So, what you've observed with complex numbers is a manifestation of this because complex numbers are a subset of the quaternions.

Visual Complex Analysis explains all this in detail.

 

[WWGD]

The standard way is to represent a complex number $a+ib$ as a $2\times 2$ matrix
$$\left(\begin{array}{cc} a & - b \\ b & a \end{array}\right), $$
so you then just read the real and imaginary part off the first column.
The matrix multiplication then agrees with the multiplication of complex numbers.
Note that in this representation
$$1\sim \left(\begin{array}{cc} 1 & -0 \\ 0 & 1 \end{array}\right), \qquad
i\sim \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right).$$

Nice point; notice too, that every analytic complex map can be described like this; this just uses the Jacobian and the Cauchy-Riemann equations.

 

[FactChecker]

I don't think that the two can be considered equivalent. A fundamental concept in multiplication of complex numbers is that their angles are added. The only information about angles in vector multiplication is the difference between the vector directions. There is no way to look at vector dot or cross products and determine the sum of the vector angles. The difference is fundamental and insurmountable.

 

[homeomorphic]

I don't think that the two can be considered equivalent. A fundamental concept in multiplication of complex numbers is that their angles are added. The only information about angles in vector multiplication is the difference between the vector directions. There is no way to look at vector dot or cross products and determine the sum of the vector angles. The difference is fundamental and insurmountable.

With a little tweak, it's not insurmountable, as I explained. Subtraction is a special case of addition. So, when you conjugate one of the complex numbers, you reverse one of the angles, so you are now dealing with the difference between the angles. So, with that tweak, as I showed, the real and imaginary parts do give you both the dot product and cross product. The catch is that it's complex multiplication plus conjugating one of the numbers (or, in other words, the complex inner product on C), not just complex multiplication alone.

 

[WWGD]

With a little tweak, it's not insurmountable, as I explained. Subtraction is a special case of addition. So, when you conjugate one of the complex numbers, you reverse one of the angles, so you are now dealing with the difference between the angles. So, with that tweak, as I showed, the real and imaginary parts do give you both the dot product and cross product. The catch is that it's complex multiplication plus conjugating one of the numbers (or, in other words, the complex inner product on C), not just complex multiplication alone.

Once you use conjugation you loose analyticity; not the end of the world, but it happens.

 

[homeomorphic]

Once you use conjugation you loose analyticity; not the end of the world, but it happens.

True. I don't think the OP would be bothered by that, though! lol

For that reason, the connection with vector products and quaternions isn't really a complex analysis issue--it's more of a complex arithmetic issue. Doesn't mean it's not enlightening. Sheds some light on quaternions, dot products, cross products, complex inner products and how they all related to each other. It's a good story to pursue if you want to understand quaternions better.

 

[FactChecker]

So, when you conjugate one of the complex numbers, you reverse one of the angles, so you are now dealing with the difference between the angles.

Of course. You are right.

 

[shreyakmath]

It can be approached using Linar Algebra

$z=a+ib=[a,b] \ \ \ w=c+id=[c,d] \\ z.w=(a+ib).(c+id)=ac+iad+ibc-bd=ac-bd+i(ad+bc) \\ =\begin{pmatrix} ac-bd\\ad+bc \end{pmatrix}=\begin{pmatrix} c &-d \\d &c \end{pmatrix}\begin{pmatrix} a\\b \end{pmatrix}\\ =[a,b]\bigodot [c,d]\\ Rot(z,\pi/2)=\begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a\\b \end{pmatrix}=i.z=[0,1]\bigodot [a,b]$
The $\bigodot$ is a vector product that is equivalent to multiplication of complex numbers (represented as 2-tuples)

 

[shreyakmath]

Moreover, the angles add. Consider the polar forms: $\\ z=[cos\theta_1 , sin\theta_1] \ \ \ w=[cos\theta_2,sin\theta_2] \\ z.w=[cos\theta_1 , sin\theta_1]\bigodot[cos\theta_2,sin\theta_2]\\ =\begin{pmatrix} cos\theta_2 &-sin\theta_2 \\ sin\theta_2 &cos\theta_2 \end{pmatrix}\begin{pmatrix} cos\theta_1\\ sin\theta_1 \end{pmatrix}\\ =\begin{pmatrix} cos(\theta_1+\theta_2)\\ sin(\theta_1+\theta_2) \end{pmatrix}$

Reference https://www.physicsforums.com/threa...tor-multiplies-continued.773049/#post-5083048