# What is cross product of two complex numbers?

1. Dec 1, 2008

### yungman

Let U=a+jb, V=c+jd.
What is U X V? Where "X" is the cross product. Can you explain?

2. Dec 1, 2008

### yungman

Let U=a+jb, V=c+jd.
What is U X V? Where "X" is the cross product. Can you explain?

3. Dec 1, 2008

### rock.freak667

UxV would give the area of the parallelogram formed by those two complex numbers.

4. Dec 1, 2008

### yungman

Can you please write out the steps to get the answer. I am very despirate!!!

Thanks a million!!

5. Dec 1, 2008

### rock.freak667

It basically works out as being

|a b|
|c d|

6. Dec 1, 2008

### flatmaster

I believe you treat the real axis and imaginary axis as two orthogonal directions. Imagine the real component has an implied unit vector sitting out front. Think of the imiganary number j as a unit vector pointing in the imaginary direction

7. Dec 1, 2008

### yungman

You mean U X V = ad-bcj so Re[U X V]=ad, Im[U X V]=bc ?

8. Dec 1, 2008

### yungman

You mean U X V = ad-bcj so Re[U X V]=ad, Im[U X V]=bc ?

9. Dec 1, 2008

### flatmaster

I believe that's correct. Immagine if you tried to cross two real numbers. You would expect to get exactly 0 which you do.

10. Dec 1, 2008

### flatmaster

What is stil ambiguious to me is that the result of a cross product should still be a vector that is perpendicular to both U and V. What way does this vector point?

11. Dec 1, 2008

### flatmaster

12. Dec 2, 2008

### Defennder

You mean the cross product of two vectors instead, I believe. On the Argand plane, you can treat the complex numbers as vectors. But then the area of parallelogram formed would be a real number (remember to take the absolute value as well), so you shouldn't have any imaginary part.

13. Dec 2, 2008

### yungman

14. Dec 2, 2008

### yungman

Thanks everybody.

15. Dec 2, 2008

### flatmaster

There's still something wierd, the source I cyted ended up giving the same intuative result I had as thinking of the real axes and imaginary axes as orthogonal basis, but the real component was still imiginary. IE

z = a+bi
w = c+di

16. Dec 2, 2008

### flatmaster

There are still some nice things that are comming out though. For instance, if two vectors point in the same direction in the real - imiginary plane, their cross is zero

z = a+bi
w = c+di

c = 2 a d = 2 b

z X w = adi - bci = a(2b)i - b(2a)i = 0

17. Dec 2, 2008

### yungman

I have been studying up the articles you gave me. The way I understand it is ZXW is pure imaginary.

That is the reason it is called complex product. There is no real part from the calculation.

The real product is Z.W=(Z'W+ZW')/2=ac+bd with no imaginary part.

Thanks for all your help, the link you gave was a good one. I already print out all the chapters and put it in my notes.

Good night

18. Dec 2, 2008

### yungman

Cross product of Complex numbers, is the book wrong or me?

This is the exact equations derivation from the “Field and Wave Electromagnetics” by David K. Cheng which have very very few errors. But this don’t make sense. Before I conclude that this is an error, let me run this by you guys/gals.

Re(A)=(A+A*)/2 Re(B)=(B+B*)/2
Re(A) X Re(B) = [(A+A*)/2] X [(B+B*)/2]
= (1/4)[(A X B* + A* X B) + (A X B + A* X B*)] (line 3).
= (1/2) Re(A X B* + A X B). (line 4).

I verified that A X B = (i/2)[(A* B) – (AB*)] which is pure imaginary.

If you look at (line 3) above,
A X B* = (i/2)[(A*B*) – (A*B)]
A* X B = (i/2)[(AB) – (A*B*)]
A X B = (i/2)[(A*B) – (AB*)]
A* X B* = (i/2)[(AB*) – (A*B)]
The sum of the four terms in (line 3) equal ZERO!!!

As seen on (line 4), (A X B* + A X B) is pure imaginary. So Re(A X B* + A X B) = ZERO!!

Am I missing something?
Thanks

19. Dec 2, 2008

### yungman

Cross product of Complex numbers, is the book wrong or me?

This is the exact equations derivation from the “Field and Wave Electromagnetics” by David K. Cheng which have very very few errors. But this don’t make sense. Before I conclude that this is an error, let me run this by you guys/gals.

Re(A)=(A+A*)/2 Re(B)=(B+B*)/2
Re(A) X Re(B) = [(A+A*)/2] X [(B+B*)/2]
= (1/4)[(A X B* + A* X B) + (A X B + A* X B*)] (line 3).
= (1/2) Re(A X B* + A X B). (line 4).

I verified that A X B = (i/2)[(A* B) – (AB*)] which is pure imaginary.

If you look at (line 3) above,
A X B* = (i/2)[(A*B*) – (A*B)]
A* X B = (i/2)[(AB) – (A*B*)]
A X B = (i/2)[(A*B) – (AB*)]
A* X B* = (i/2)[(AB*) – (A*B)]
The sum of the four terms in (line 3) equal ZERO!!!

As seen on (line 4), (A X B* + A X B) is pure imaginary. So Re(A X B* + A X B) = ZERO!!

Am I missing something?
Thanks

20. Dec 2, 2008

### D H

Staff Emeritus
Re: Cross product of Complex numbers, is the book wrong or me?

Kind of roundabout way to get to that result, but this is indeed a tautology for any two complex numbers A and B.

A*B-AB* is pure real because AB* = (A*B)*. However, your equality is not valid. Show your derivation.