anonymous
  • anonymous
COMPLEX VARIABLE
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
what does that "*" represent?
anonymous
  • anonymous
* is equal to \(\times \)

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anonymous
  • anonymous
a cross-product?
anonymous
  • anonymous
no.., it just times
anonymous
  • anonymous
@gerryliyana solve for RHS first by using the values of dimensional vectors given in question
anonymous
  • anonymous
\[\vec{a}\times\vec{b}=\left|\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ u&v&0\\ x&y&0 \end{matrix}\right| =\hat{z}(uy-vx)\\ \vec{a}\cdot\vec{b}=ux+vy\\ ab = (u+iv)(x+iy)=ux+i(vx)+i(uy)+(i^2)(xy) \]
anonymous
  • anonymous
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)
anonymous
  • anonymous
\[ab=(uy-vx)+i(vx+uy)\]
anonymous
  • anonymous
its same with my answer @myko
anonymous
  • anonymous
yeah i think so
anonymous
  • anonymous
\[ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})\]
anonymous
  • anonymous
vx+uy =/= a.b @electrokid
anonymous
  • anonymous
so what the right answer??
anonymous
  • anonymous
oh yea.. my bad..
anonymous
  • anonymous
that "*" in the question must be a complex conjugate
anonymous
  • anonymous
then it seems you might have it right.
anonymous
  • anonymous
good idea!! oh my bad...,
anonymous
  • anonymous
yeah,"*" means conjugate
anonymous
  • anonymous
always make sure that people are talking the same math language
anonymous
  • anonymous
thank you @electrokid
anonymous
  • anonymous
@electrokid if "*" is complex conjugate then it will be rightly solved
anonymous
  • anonymous
i got it now.., thank you guys :)
anonymous
  • anonymous
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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