## gerryliyana 2 years ago COMPLEX VARIABLE

1. gerryliyana

2. electrokid

what does that "*" represent?

3. gerryliyana

* is equal to $$\times$$

4. electrokid

a cross-product?

5. gerryliyana

no.., it just times

6. niksva

@gerryliyana solve for RHS first by using the values of dimensional vectors given in question

7. electrokid

$\vec{a}\times\vec{b}=\left|\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ u&v&0\\ x&y&0 \end{matrix}\right| =\hat{z}(uy-vx)\\ \vec{a}\cdot\vec{b}=ux+vy\\ ab = (u+iv)(x+iy)=ux+i(vx)+i(uy)+(i^2)(xy)$

8. myko

are you sure it's right? Because: a.b=ux+vy iz(axb)=i(uy-xv) and a*b=ux-vy+i(uy+vx) so it is not equal: ux-vy+i(uy+vx) =/= ux+vy+i(uy-xv)

9. electrokid

$ab=(uy-vx)+i(vx+uy)$

10. gerryliyana

its same with my answer @myko

11. gerryliyana

yeah i think so

12. electrokid

$ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})$

13. myko

vx+uy =/= a.b @electrokid

14. gerryliyana

so what the right answer??

15. electrokid

oh yea.. my bad..

16. electrokid

that "*" in the question must be a complex conjugate

17. electrokid

then it seems you might have it right.

18. gerryliyana

good idea!! oh my bad...,

19. gerryliyana

yeah,"*" means conjugate

20. electrokid

always make sure that people are talking the same math language

21. gerryliyana

thank you @electrokid

22. niksva

@electrokid if "*" is complex conjugate then it will be rightly solved

23. gerryliyana

i got it now.., thank you guys :)

24. gerryliyana

by the way.., if all of you have a facebook., add me on facebook http://www.facebook.com/gerry.resmiliyana thank you so much!