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gerryliyana

  • 2 years ago

COMPLEX VARIABLE

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  1. gerryliyana
    • 2 years ago
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  2. electrokid
    • 2 years ago
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    what does that "*" represent?

  3. gerryliyana
    • 2 years ago
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    * is equal to \(\times \)

  4. electrokid
    • 2 years ago
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    a cross-product?

  5. gerryliyana
    • 2 years ago
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    no.., it just times

  6. niksva
    • 2 years ago
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    @gerryliyana solve for RHS first by using the values of dimensional vectors given in question

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

  8. myko
    • 2 years ago
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    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
    • 2 years ago
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    \[ab=(uy-vx)+i(vx+uy)\]

  10. gerryliyana
    • 2 years ago
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    its same with my answer @myko

  11. gerryliyana
    • 2 years ago
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    yeah i think so

  12. electrokid
    • 2 years ago
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    \[ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})\]

  13. myko
    • 2 years ago
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    vx+uy =/= a.b @electrokid

  14. gerryliyana
    • 2 years ago
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    so what the right answer??

  15. electrokid
    • 2 years ago
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    oh yea.. my bad..

  16. electrokid
    • 2 years ago
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    that "*" in the question must be a complex conjugate

  17. electrokid
    • 2 years ago
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    then it seems you might have it right.

  18. gerryliyana
    • 2 years ago
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    good idea!! oh my bad...,

  19. gerryliyana
    • 2 years ago
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    yeah,"*" means conjugate

  20. electrokid
    • 2 years ago
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    always make sure that people are talking the same math language

  21. gerryliyana
    • 2 years ago
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    thank you @electrokid

  22. niksva
    • 2 years ago
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    @electrokid if "*" is complex conjugate then it will be rightly solved

  23. gerryliyana
    • 2 years ago
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    i got it now.., thank you guys :)

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

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