Complex numbers and vector multiplies continued

1. Sep 26, 2014

keldon

2. Sep 26, 2014

mathman

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

3. Sep 27, 2014

pwsnafu

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

4. Sep 27, 2014

keldon

Multiplication of two complex numbers, A and B.

5. Sep 28, 2014

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 $<a^2- b^2, ad+ bc>$.

6. Sep 28, 2014

mathman

|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.

7. Sep 29, 2014

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
$$1\sim \left(\begin{array}{cc} 1 & -0 \\ 0 & 1 \end{array}\right), \qquad i\sim \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right).$$

8. Oct 2, 2014

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.

9. Oct 2, 2014

WWGD

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

Last edited: Oct 2, 2014
10. Oct 2, 2014

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.

11. Oct 2, 2014

homeomorphic

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.

12. Oct 2, 2014

WWGD

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

13. Oct 2, 2014

homeomorphic

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.

14. Oct 3, 2014

FactChecker

Of course. You are right.

15. Apr 21, 2015

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)

16. Apr 21, 2015

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